Before any new identity, the book quietly leans on one old friend: the distance formula. For two points P(x1, y1) and Q(x2, y2), the distance is PQ = √((x2 − x1)² + (y2 − y1)²). It is really Pythagoras in disguise — the horizontal gap and the vertical gap are the two shorter sides of a right triangle, and PQ is the hypotenuse. Keep this picture in mind, because the whole fundamental law of this chapter is proved by measuring one distance in two different ways.
Example 1 (first part). Find AB for A(3, 8) and B(5, 6). The horizontal gap is 5 − 3 = 2 and the vertical gap is 6 − 8 = −2. Squaring kills the minus sign: AB = √(2² + (−2)²) = √(4 + 4) = √8 = 2√2.
Example 1 (second part). For P(cos x, cos y) and Q(sin x, sin y), the gaps are sin x − cos x and sin y − cos y. Squaring each: (sin x − cos x)² = sin²x − 2 sin x cos x + cos²x = 1 − 2 sin x cos x, because sin² + cos² = 1. The y-part gives 1 − 2 sin y cos y the same way. Adding: PQ² = 2 − 2(sin x cos x + sin y cos y), so PQ = √(2 − 2(cos x sin x + cos y sin y)). Notice how the identity sin² + cos² = 1 did all the heavy lifting.
اس باب کی بنیاد ایک پرانا فارمولا ہے: دو نقطوں P(x₁, y₁) اور Q(x₂, y₂) کے درمیان فاصلہ PQ = √((x₂ − x₁)² + (y₂ − y₁)²) ہوتا ہے۔ یہ دراصل فیثاغورث کا قضیہ ہے — افقی اور عمودی فاصلے قائمہ زاویہ مثلث کے دو ضلعے ہیں اور PQ وتر ہے۔
مثال ۱: نقاط A(3, 8) اور B(5, 6) کے لیے افقی فرق 2 اور عمودی فرق −2 ہے، لہٰذا AB = √(4 + 4) = √8 = 2√2۔ دوسرے حصے میں P(cos x, cos y) اور Q(sin x, sin y) کے لیے مربع کھولنے اور sin² + cos² = 1 استعمال کرنے سے PQ = √(2 − 2(cos x sin x + cos y sin y)) حاصل ہوتا ہے۔
Here is the single most important result of the chapter. For any angles α and β, cos(α − β) = cos α cos β + sin α sin β. The book proves it with a lovely piece of geometry. Take a unit circle (radius 1) with centre O. Mark A at angle α and B at angle β on the circle, so A = (cos α, sin α) and B = (cos β, sin β). Also mark C = (1, 0) and D at angle α − β, so D = (cos(α − β), sin(α − β)).
Now look at triangles OAB and OCD. Both have two sides of length 1 (radii) and the angle between them is α − β in both cases. By SAS, the triangles are congruent, so the third sides are equal: |AB| = |CD|. Measure both with the distance formula. |AB|² = (cos α − cos β)² + (sin α − sin β)², and |CD|² = (cos(α − β) − 1)² + sin²(α − β). Expand both, use sin² + cos² = 1 wherever it appears, and the two expressions collapse to the same thing only if cos(α − β) = cos α cos β + sin α sin β. That is the whole trick: one length, measured twice.
📌 Book note: the picture is drawn for α > β > 0, but the formula is true for all values of α and β, positive, negative or zero.
Example 2. Find sin(5π/12) without tables. Write sin(5π/12) = cos(π/2 − 5π/12) = cos(π/12), and π/12 = π/3 − π/4. By the fundamental law: cos(π/3 − π/4) = cos(π/3)cos(π/4) + sin(π/3)sin(π/4) = (1/2)(√2/2) + (√3/2)(√2/2) = (√2 + √6)/4, which is the same as (√3 + 1)/(2√2). So sin(5π/12) = (√3 + 1)/(2√2).
اس باب کا سب سے اہم نتیجہ یہ ہے: cos(α − β) = cos α cos β + sin α sin β۔ ثبوت کی ترکیب خوبصورت ہے: واحد دائرے پر زاویوں α اور β والے نقطے A اور B لیں، اور C = (1, 0) اور زاویہ α − β والا نقطہ D لیں۔ دونوں مثلثوں OAB اور OCD میں دو ضلعے رداس (یعنی 1) ہیں اور درمیانی زاویہ دونوں میں α − β ہے، اس لیے SAS سے دونوں مثلثیں مطابق ہیں اور |AB| = |CD|۔ دونوں فاصلوں پر فاصلے کا فارمولا لگا کر پھیلائیں اور sin² + cos² = 1 استعمال کریں تو یہی بنیادی قانون نکل آتا ہے۔
📌 کتاب کا نوٹ: شکل α > β > 0 کے لیے بنائی گئی ہے، لیکن فارمولا α اور β کی تمام قیمتوں کے لیے درست ہے۔
مثال ۲: sin(5π/12) کے لیے پہلے اسے cos(π/12) لکھیں، پھر π/12 = π/3 − π/4 رکھ کر بنیادی قانون لگائیں تو جواب (√3 + 1)/(2√2) آتا ہے۔
Once the fundamental law is in hand, ten useful formulae fall out of it just by choosing clever values or replacing β by −β. Work through them slowly — each one is a one-line substitution, and together they are the toolkit for the rest of the chapter:
(i) Put α = π/2 in the fundamental law: cos(π/2 − β) = cos(π/2)cos β + sin(π/2)sin β = 0 + sin β. (ii) cos(π/2 + α) = −sin α. (iii) sin(π/2 + α) = cos α. These three say that shifting by a right angle swaps sine and cosine, with a sign to watch.
(iv) Replace β by −β in the fundamental law, using cos(−β) = cos β and sin(−β) = −sin β: cos(α + β) = cos α cos β − sin α sin β. (v) sin(α + β) = sin α cos β + cos α sin β. (vi) sin(α − β) = sin α cos β − cos α sin β. The last two come from sin(α + β) = cos(π/2 − (α + β)) and the fundamental law — or just remember the pattern: sine keeps its sign, cosine flips it.
(vii) cos(2π − θ) = cos θ and (viii) sin(2π − θ) = −sin θ: a full turn changes nothing for cosine, and flips sine. (ix) Dividing the sine formula by the cosine formula: tan(α + β) = (tan α + tan β)/(1 − tan α tan β). (x) Likewise tan(α − β) = (tan α − tan β)/(1 + tan α tan β). For the tan formulae, divide top and bottom by cos α cos β and watch tan appear everywhere.
بنیادی قانون ہاتھ میں آ جائے تو صرف β کی جگہ −β رکھنے یا α = π/2 لینے سے دس مفید فارمولے نکل آتے ہیں۔ انھیں ایک ایک کر کے سمجھیں:
(i) cos(π/2 − β) = sin β، (ii) cos(π/2 + α) = −sin α، (iii) sin(π/2 + α) = cos α — یعنی قائمہ زاویہ کا اضافہ sine اور cosine کو آپس میں بدل دیتا ہے۔ (iv) cos(α + β) = cos α cos β − sin α sin β، (v) sin(α + β) = sin α cos β + cos α sin β، (vi) sin(α − β) = sin α cos β − cos α sin β۔
(vii) cos(2π − θ) = cos θ اور (viii) sin(2π − θ) = −sin θ — مکمل چکر cosine کو نہیں بدلتا۔ (ix) tan(α + β) = (tan α + tan β)/(1 − tan α tan β) اور (x) tan(α − β) = (tan α − tan β)/(1 + tan α tan β) — یہ sine والے فارمولے کو cosine والے پر تقسیم کر کے اور اوپر نیچے cos α cos β سے کاٹ کر بنتے ہیں۔
An allied angle is any angle of the form n × 90° ± θ, where n is an integer. The whole section is really one device, and the book states it plainly. First, look at the multiple of 90°: if it is an odd multiple (90°, 270°, …), the ratio changes to its co-ratio — sin becomes cos, tan becomes cot, sec becomes cosec, and so on. If it is an even multiple (180°, 360°, …), the ratio stays the same. Second, decide the sign: work out which quadrant the allied angle lands in (taking θ acute), and give the result the sign that the original ratio has in that quadrant.
📌 Book device (remember it): odd multiple of 90° → co-ratio; even multiple → same ratio; sign from the quadrant. The book works this out in two cases: (a) sine of odd-multiple angles, and (b) cosine of even-multiple angles — for example cos(π − θ) = −cos θ (π is an even multiple, so cos stays cos; π − θ lies in quadrant II, where cos is negative), while cos(2π − θ) = cos θ (quadrant IV, cos positive).
Example 3. Without tables: (i) sin 225° = sin(180° + 45°) = −sin 45° = −1/√2 (quadrant III, sin negative). (ii) tan 600° = tan(540° + 60°) = tan(6 × 90° + 60°) = tan 60° = √3 (even multiple, tan stays tan; 600° lands where tan is positive). (iii) cot(−225°) = −cot 225° = −cot(180° + 45°) = −cot 45° = −1. ⚠️ Book misprint flagged: the book's worked line prints “−(cot 45°) = 1” here — that last step is wrong, since cot 45° = 1 and the minus sign survives, giving −1. (iv) cosec(−420°) = −cosec 420° = −cosec(360° + 60°) = −cosec 60° = −2/√3.
Example 4. Simplify [sin(180° − θ) cos(360° − θ) tan(90° + θ)] / [sin(90° − θ) cos(180° + θ) tan(270° − θ)]. Replace each piece: sin(180° − θ) = sin θ, cos(360° − θ) = cos θ, tan(90° + θ) = −cot θ, sin(90° − θ) = cos θ, cos(180° + θ) = −cos θ, tan(270° − θ) = cot θ. So the fraction is (sin θ · cos θ · (−cot θ)) / (cos θ · (−cos θ) · cot θ) = (−sin θ)/(−cos θ) = tan θ. Everything cancels except one sin over one cos — that is the whole point of the exercise.
ملحقہ زاویہ وہ زاویہ ہے جو n × 90° ± θ کی صورت میں ہو، جہاں n کوئی صحیح عدد ہو۔ پورا سبق ایک ہی ترکیب پر چلتا ہے: اگر 90° کا اضعاف طاق ہو (90°، 270° وغیرہ) تو نسبت اپنی ہم نسبت (co-ratio) میں بدل جاتی ہے — sin کی جگہ cos، tan کی جگہ cot وغیرہ۔ اگر اضعاف جفت ہو (180°، 360° وغیرہ) تو نسبت وہی رہتی ہے۔ پھر علامت اس ربع سے طے کریں جس میں ملحقہ زاویہ پڑتا ہے، اور وہ علامت لگائیں جو اصل نسبت کی اس ربع میں ہوتی ہے۔
📌 کتاب کی ترکیب: طاق اضعاف → ہم نسبت، جفت اضعاف → وہی نسبت، علامت ربع سے۔
مثال ۳: (i) sin 225° = −1/√2، (ii) tan 600° = √3، (iii) cot(−225°) = −1۔ ⚠️ کتاب کی غلطی: کتاب کی حل شدہ سطر میں آخر میں “= 1” چھپا ہے جو غلط ہے، کیونکہ cot 45° = 1 اور منفی کی علامت باقی رہتی ہے، اس لیے جواب −1 ہے۔ (iv) cosec(−420°) = −2/√3۔
مثال ۴: دی گئی لمبی کسر میں ہر جز کو ملحقہ زاویوں کے اصول سے بدلیں تو اوپر sin θ · cos θ · (−cot θ) اور نیچے cos θ · (−cos θ) · cot θ رہ جاتا ہے، سب کچھ کٹ کر tan θ بچتا ہے۔
This section is where the formulae earn their keep, through Examples 5 to 10. Example 5 proves sin(α + β) sin(α − β) = sin²α − sin²β = cos²β − cos²α. Expand both factors with the sine addition/subtraction formulae and the cross terms cancel, leaving sin²α cos²β − cos²α sin²β; replace cos²β by 1 − sin²β (and then sin²α by 1 − cos²α) and both forms appear. It is a good one to memorise, because it turns a product of sines into a difference of squares.
Example 6 finds all six ratios of 105° by writing 105° = 60° + 45° and using the addition formulae. Example 7 shows (cos 11° + sin 11°)/(cos 11° − sin 11°) = tan 56°: divide top and bottom by cos 11°, giving (1 + tan 11°)/(1 − tan 11°), which is exactly tan(45° + 11°) by the tan addition formula. Example 8 is a quadrant workout: cos α = −7/25 with α in quadrant II gives sin α = 24/25; tan β = 12/5 with β in quadrant III gives sin β = −12/13 and cos β = −5/13. Then sin(α + β) = (24/25)(−5/13) + (−7/25)(−12/13) = −36/325, and cos(α + β) = (−7/25)(−5/13) − (24/25)(−12/13) = 323/325. Sine negative, cosine positive — so α + β lies in quadrant IV.
Example 9 uses a triangle, where α + β + γ = 180°: then tan(α + β) = tan(180° − γ) = −tan γ, and the same idea gives tan α + tan β + tan γ = tan α tan β tan γ. Example 10 writes 3 sin θ + 4 cos θ in the form r sin(θ + φ). Set 3 = r cos φ and 4 = r sin φ; squaring and adding gives r² = 25, so r = 5, and dividing gives tan φ = 4/3. Hence 3 sin θ + 4 cos θ = 5 sin(θ + φ) with tan φ = 4/3. ⚠️ Book slip flagged: the book prints “tan⁻¹φ = 4/3” at the end of this example; it means tan φ = 4/3.
اس حصے میں فارمولوں کا عملی استعمال ہے۔ مثال ۵: sin(α + β) sin(α − β) = sin²α − sin²β = cos²β − cos²α — دونوں کو کھولیں تو درمیانی رقمیں کٹ جاتی ہیں اور مربعوں کا فرق بچتا ہے۔ مثال ۶: 105° = 60° + 45° لکھ کر اس کی چھوں نسبتیں نکالی جاتی ہیں۔ مثال ۷: اوپر نیچے cos 11° سے تقسیم کریں تو (1 + tan 11°)/(1 − tan 11°) بنتا ہے جو tan(45° + 11°) = tan 56° ہے۔
مثال ۸: cos α = −7/25 (ربع دوم) سے sin α = 24/25، اور tan β = 12/5 (ربع سوم) سے sin β = −12/13، cos β = −5/13۔ پھر sin(α + β) = −36/325 اور cos(α + β) = 323/325 — sine منفی اور cosine مثبت، اس لیے α + β ربع چہارم میں ہے۔ مثال ۹: مثلث میں α + β + γ = 180° سے tan(α + β) = −tan γ اور tan α + tan β + tan γ = tan α tan β tan γ۔ مثال ۱۰: 3 sin θ + 4 cos θ = 5 sin(θ + φ) جہاں tan φ = 4/3۔ ⚠️ کتاب کی لغزش: کتاب میں آخر میں “tan⁻¹φ = 4/3” چھپا ہے، درست مطلب tan φ = 4/3 ہے۔
There is nothing new to memorise here — just put β = α in the addition formulae you already have. (i) sin(α + α) = sin α cos α + cos α sin α, so sin 2α = 2 sin α cos α. (ii) cos(α + α) = cos α cos α − sin α sin α, so cos 2α = cos²α − sin²α. Then, using sin²α + cos²α = 1, the same result wears two more disguises: replace sin²α by 1 − cos²α to get cos 2α = 2cos²α − 1, or replace cos²α by 1 − sin²α to get cos 2α = 1 − 2sin²α. Three forms, one identity — pick whichever form kills the terms in front of you. (iii) tan(α + α) = (tan α + tan α)/(1 − tan α · tan α), so tan 2α = 2tan α/(1 − tan²α).
Two rearrangements of the cosine forms are used so often that they deserve a line of their own: 2cos²α = 1 + cos 2α and 2sin²α = 1 − cos 2α. These are the tools that lower a square to a first power, which is exactly what Example 13 and Exercise 10.3 Q4 do with cos⁴θ and sin⁴θ.
یہاں کچھ نیا یاد کرنے کی ضرورت نہیں — جمع والے فارمولوں میں β کی جگہ α رکھ دیں۔ (i) sin 2α = 2 sin α cos α، (ii) cos 2α = cos²α − sin²α، اور sin²α + cos²α = 1 سے اس کی دو اور صورتیں: cos 2α = 2cos²α − 1 اور cos 2α = 1 − 2sin²α۔ تین صورتیں، ایک ہی شناخت — جو صورت سامنے والی رقمیں ختم کرے وہی چنیں۔ (iii) tan 2α = 2tan α/(1 − tan²α)۔
دو تبدیلیاں بار بار کام آتی ہیں: 2cos²α = 1 + cos 2α اور 2sin²α = 1 − cos 2α — یہ مربع کو پہلی قوت میں اتار دیتی ہیں، اور یہی مثال ۱۳ اور مشق ۱۰.۳ کے سوال ۴ میں cos⁴θ اور sin⁴θ کے ساتھ ہوتا ہے۔
Half-angle formulae come from the double-angle cosine forms, read backwards. Put α/2 in place of α in cos 2α = 2cos²α − 1 and cos 2α = 1 − 2sin²α, and solve: cos(α/2) = ±√((1 + cos α)/2) and sin(α/2) = ±√((1 − cos α)/2). The ± is not decoration — you must pick the sign from the quadrant where α/2 actually lies. For tangent, dividing gives tan(α/2) = sin α/(1 + cos α) = (1 − cos α)/sin α, and this form has no ± problem because the sign is already carried by sin α.
Triple-angle formulae come from writing 3α as 2α + α and using the addition formulae plus the double-angle results: sin 3α = 3 sin α − 4 sin³α, cos 3α = 4 cos³α − 3 cos α, and tan 3α = (3 tan α − tan³α)/(1 − 3 tan²α). The book's proof of tan 3α is worth following once: tan(2α + α) with tan 2α substituted, then a single compound fraction tidied up.
Example 11 proves (sin θ + sin 2θ)/(1 + cos θ + cos 2θ) = tan θ: write sin 2θ = 2 sin θ cos θ and cos 2θ = 2cos²θ − 1, factor sin θ(1 + 2cos θ) on top and cos θ(1 + 2cos θ) below, and cancel. Example 12 shows sin 2θ = 2tan θ/(1 + tan²θ) and cos 2θ = (1 − tan²θ)/(1 + tan²θ) by writing 2 sin θ cos θ over 1 = sin²θ + cos²θ and dividing top and bottom by cos²θ. Example 13 reduces cos⁴θ: cos²θ = (1 + cos 2θ)/2, so cos⁴θ = (1 + cos 2θ)²/4 = (1 + 2cos 2θ + cos²2θ)/4, and cos²2θ = (1 + cos 4θ)/2 finishes it: cos⁴θ = (3 + 4cos 2θ + cos 4θ)/8.
نصف زاویے کے فارمولے دگنے زاویے والے cosine فارمولوں کو الٹا پڑھنے سے بنتے ہیں: α کی جگہ α/2 رکھ کر حل کریں تو cos(α/2) = ±√((1 + cos α)/2) اور sin(α/2) = ±√((1 − cos α)/2)۔ یہ ± محض سجاوٹ نہیں — علامت اس ربع سے طے کریں جس میں α/2 واقع ہو۔ tangent کے لیے tan(α/2) = sin α/(1 + cos α) = (1 − cos α)/sin α، جس میں ± کا مسئلہ نہیں کیونکہ علامت sin α میں ہی موجود ہے۔
تگنے زاویے کے لیے 3α کو 2α + α لکھیں: sin 3α = 3 sin α − 4 sin³α، cos 3α = 4 cos³α − 3 cos α، tan 3α = (3 tan α − tan³α)/(1 − 3 tan²α)۔ مثال ۱۱: (sin θ + sin 2θ)/(1 + cos θ + cos 2θ) = tan θ — اوپر sin θ(1 + 2cos θ) اور نیچے cos θ(1 + 2cos θ) بنتا ہے، کٹ جاتا ہے۔ مثال ۱۲: sin 2θ اور cos 2θ کی tan θ والی صورتیں۔ مثال ۱۳: cos⁴θ = (3 + 4cos 2θ + cos 4θ)/8۔
Add and subtract the four addition/subtraction formulae in pairs and four product identities drop out: 2 sin α cos β = sin(α + β) + sin(α − β), 2 cos α sin β = sin(α + β) − sin(α − β), 2 cos α cos β = cos(α + β) + cos(α − β), and −2 sin α sin β = cos(α + β) − cos(α − β). Now comes the substitution that turns these around: put α + β = P and α − β = Q, so α = (P + Q)/2 and β = (P − Q)/2. The same four lines become the sum-to-product identities: sin P + sin Q = 2 sin((P+Q)/2) cos((P−Q)/2), sin P − sin Q = 2 cos((P+Q)/2) sin((P−Q)/2), cos P + cos Q = 2 cos((P+Q)/2) cos((P−Q)/2), and cos P − cos Q = −2 sin((P+Q)/2) sin((P−Q)/2). One substitution, eight formulae — and the minus sign in the last one is the one students forget most, so circle it.
Example 14: 2 sin 7θ cos 3θ = sin(7θ + 3θ) + sin(7θ − 3θ) = sin 10θ + sin 4θ. Example 15: sin 19° cos 11° + sin 71° sin 11° = 1/2 — take half of everything, use the product formulae, and cos 82° = cos(90° − 8°) = sin 8° makes the middle terms cancel. Example 16: sin 5x + sin 7x = 2 sin 6x cos(−x) = 2 sin 6x cos x. Example 17: cos θ + cos 3θ + cos 5θ + cos 7θ = 4 cos θ cos 2θ cos 4θ — pair them up, factor 2cos θ, and apply the sum formula once more. Example 18: cos 20° cos 40° cos 80° = 1/8, by writing the product as (1/4)(4 cos 20° cos 40° cos 80°) and collapsing pairs with 2cos A cos B until only 1/2 is left.
جمع و تفریق والے چار فارمولوں کو جوڑوں میں جمع اور تفریق کریں تو حاصل ضرب والی چار شناختیں نکلتی ہیں: 2 sin α cos β = sin(α + β) + sin(α − β)، 2 cos α sin β = sin(α + β) − sin(α − β)، 2 cos α cos β = cos(α + β) + cos(α − β)، −2 sin α sin β = cos(α + β) − cos(α − β)۔ اب α + β = P اور α − β = Q رکھیں تو α = (P + Q)/2 اور β = (P − Q)/2، اور وہی چار سطریں مجموعے سے حاصل ضرب والی شناختیں بن جاتی ہیں — آخری والی میں منفی کی علامت سب سے زیادہ بھولی جاتی ہے، اسے گول کر لیں۔
مثال ۱۴: 2 sin 7θ cos 3θ = sin 10θ + sin 4θ۔ مثال ۱۵: sin 19° cos 11° + sin 71° sin 11° = 1/2 — cos 82° = sin 8° سے درمیانی رقمیں کٹ جاتی ہیں۔ مثال ۱۶: sin 5x + sin 7x = 2 sin 6x cos x۔ مثال ۱۷: cos θ + cos 3θ + cos 5θ + cos 7θ = 4 cos θ cos 2θ cos 4θ۔ مثال ۱۸: cos 20° cos 40° cos 80° = 1/8۔
Below are all questions of Exercises 10.1, 10.2, 10.3 and 10.4 from Unit 10 of the official book (printed pages 185, 189–191, 194–195, 198–199), each solved step by step. The book has no multiple-choice section, no review exercise and no inquisitive/numerical section in this unit, so none is invented here. Two book slips are flagged inside the answers: Example 3(iii) on printed page 184 (a wrong final sign) and Exercise 10.4 Q4(i) (false as printed — shown honestly, not “fixed” by inventing a new angle).
Counting rule used here: one numbered question = one answer block; its parts (i), (ii), … are solved inside that block. That gives 6 + 15 + 5 + 10 = 36 blocks, covering 123 parts in total.
(i) cos is even, so cos(−1230°) = cos 1230°. Now 1230° = 3 × 360° + 150°, so cos 1230° = cos 150° = cos(180° − 30°) = −cos 30° = −√3/2. (ii) tan is odd, so tan(−1035°) = −tan 1035°. And 1035° = 2 × 360° + 315°, with tan 315° = tan(360° − 45°) = −tan 45° = −1. So the answer is −(−1) = 1. (iii) 1140° = 3 × 360° + 60°, so sec 1140° = sec 60° = 2. (iv) cosec is odd: cosec(−690°) = −cosec 690°. 690° = 360° + 330°, and cosec 330° = cosec(360° − 30°) = −cosec 30° = −2. So the answer is −(−2) = 2. (v) 1320° = 3 × 360° + 240°, and cot 240° = cot(180° + 60°) = cot 60° = 1/√3. (vi) cos(−240°) = cos 240° = cos(180° + 60°) = −cos 60° = −1/2.
(i) cos 168° = cos(180° − 12°) = −cos 12°. (ii) sin 192° = sin(180° + 12°) = −sin 12°. (iii) cos 333° = cos(360° − 27°) = cos 27°. (iv) tan 213° = tan(180° + 33°) = tan 33°. (v) cos(−435°) = cos 435° = cos(360° + 75°) = cos 75° = cos(90° − 15°) = sin 15°. (vi) sin 219° = sin(180° + 39°) = −sin 39°. (vii) tan(−597°) = −tan 597°; 597° = 360° + 237°, and tan 237° = tan(180° + 57°) = tan 57° = cot 33°. So the answer is −cot 33°. (viii) cos(−111°) = cos 111° = cos(90° + 21°) = −sin 21°. (ix) sin(−390°) = −sin 390° = −sin(360° + 30°) = −sin 30°.
(i) sin(180° + α) = −sin α and sin(90° − α) = cos α, so the left side is (−sin α)(cos α) = −sin α cos α. ✓ (ii) sin 810° = sin(720° + 90°) = 1; sin 630° = sin(360° + 270°) = sin 270° = −1; cos 135° = −√2/2; sin 225° = −√2/2. So LHS = (1)(−1) + (−√2/2)(−√2/2) = −1 + 1/2 = −1/2. ✓ (iii) tan 150° = −1/√3 and cot 330° = cot(360° − 30°) = −√3, so their product is 1. sec 135° = −√2 and cosec 225° = −√2, so 2 sec 135° cosec 225° = 2 × 2 = 4. Hence LHS = 1 − 4 = −3. ✓ (iv) sin 210° = −1/2, cos 240° = −1/2, tan 225° = 1, cot 225° = 1. Sum = −1/2 − 1/2 + 1 + 1 = 1. ✓
(i) Top: tan(180° + α) = tan α and cot(90° − α) = tan α, so the numerator is tan²α. Bottom: sin(360° − α) = −sin α and cos(270° + α) = sin α, so the denominator is −sin²α. The fraction is tan²α/(−sin²α) = −(sin²α/cos²α)/sin²α = −1/cos²α = −sec²α. ✓ (ii) sin(π + θ) = −sin θ, so sin²(π + θ) = sin²θ. tan(3π/2 + θ) = −cot θ. cot(3π/2 − θ) = tan θ, so cot² = tan²θ. cos(π − θ) = −cos θ, so cos² = cos²θ. cosec(2π − θ) = −cosec θ. Numerator: sin²θ · (−cot θ) = −sin θ cos θ. Denominator: tan²θ · cos²θ · (−cosec θ) = −sin θ. Ratio = cos θ. ✓ (iii) Top: (−sin θ)(sec θ)(−tan θ) = sin θ sec θ tan θ. Bottom: (sec θ)(−sin θ)(tan θ) = −sin θ sec θ tan θ. Ratio = −1. ✓
Work out each factor with the allied-angle device. sec(3π/2 − θ): 3π/2 is an odd multiple, so sec becomes cosec, and the angle lies where sec is negative, giving −cosec θ. sec(5π/2 − θ) = sec(π/2 − θ) after removing one full turn 2π, which is cosec θ. tan(3π/2 − θ) = cot θ, and tan(5π/2 + θ) = tan(π/2 + θ) = −cot θ. So LHS = (−cosec θ)(cosec θ) − (cot θ)(−cot θ) = −cosec²θ + cot²θ = −(cosec²θ − cot²θ) = −1, since cosec²θ − cot²θ = 1. ✓
In any triangle α + β + γ = 180°. (i) α + β = 180° − γ, so sin(α + β) = sin(180° − γ) = sin γ. ✓ (ii) (α + β)/2 = 90° − γ/2, so sec((α + β)/2) = sec(90° − γ/2) = cosec(γ/2). ✓ (iii) α = 180° − (β + γ), so sin α = sin(β + γ), and cosec α = 1/sin α = 1/sin(β + γ). ✓ (iv) tan(α + β) = tan(180° − γ) = −tan γ, so tan(α + β) + tan γ = 0. ✓
(i) sin 15° = sin(45° − 30°) = sin 45° cos 30° − cos 45° sin 30° = (√2/2)(√3/2) − (√2/2)(1/2) = (√6 − √2)/4. (ii) cos 15° = cos(45° − 30°) = cos 45° cos 30° + sin 45° sin 30° = (√6 + √2)/4. (iii) tan 15° = sin 15°/cos 15° = (√6 − √2)/(√6 + √2). Rationalise: multiply top and bottom by (√6 − √2) to get (8 − 4√3)/4 = 2 − √3. (iv) sin 105° = sin(60° + 45°) = sin 60° cos 45° + cos 60° sin 45° = (√3/2)(√2/2) + (1/2)(√2/2) = (√6 + √2)/4. (v) cos 105° = cos(60° + 45°) = cos 60° cos 45° − sin 60° sin 45° = (1/2)(√2/2) − (√3/2)(√2/2) = (√2 − √6)/4. (vi) tan 105° = tan(60° + 45°) = (tan 60° + tan 45°)/(1 − tan 60° tan 45°) = (√3 + 1)/(1 − √3) = −(2 + √3).
(i) sin(45° + α) = sin 45° cos α + cos 45° sin α. Since sin 45° = cos 45° = 1/√2, this is (1/√2) cos α + (1/√2) sin α = (1/√2)(sin α + cos α). ✓ (ii) cos(α + 45°) = cos α cos 45° − sin α sin 45° = (1/√2) cos α − (1/√2) sin α = (1/√2)(cos α − sin α). ✓
(i) tan(45° + A) = (1 + tan A)/(1 − tan A) and tan(45° − A) = (1 − tan A)/(1 + tan A). Multiply: everything cancels, leaving 1. ✓ (ii) 3π/4 + θ = π − (π/4 − θ), and tan(π − x) = −tan x. So tan(3π/4 + θ) = −tan(π/4 − θ), and the sum is 0. ✓ (iii) sin(θ + 30°) = (√3/2) sin θ + (1/2) cos θ; cos(θ + 60°) = (1/2) cos θ − (√3/2) sin θ. Add: the sin θ terms cancel, leaving cos θ. ✓ (iv) Divide numerator and denominator by cos θ: the fraction becomes (tan θ − tan(θ/2))/(1 + tan θ tan(θ/2)), which is tan(θ − θ/2) by the tan difference formula, i.e. tan(θ/2). ✓ (v) Multiply numerator and denominator by cos θ cos φ: the top becomes cos θ cos φ − sin θ sin φ = cos(θ + φ), and the bottom becomes cos θ cos φ + sin θ sin φ = cos(θ − φ). ✓
Expand with the cosine addition and subtraction formulae: LHS = (cos α cos β − sin α sin β)(cos α cos β + sin α sin β) = cos²α cos²β − sin²α sin²β. Now replace cos²β by 1 − sin²β: LHS = cos²α − cos²α sin²β − sin²α sin²β = cos²α − sin²β(cos²α + sin²α) = cos²α − sin²β. For the second form, instead replace cos²α by 1 − sin²α in cos²α cos²β − sin²α sin²β: LHS = cos²β − sin²α cos²β − sin²α sin²β = cos²β − sin²α. ✓
Numerator: sin(α + β) + sin(α − β) = 2 sin α cos β (the cross terms cancel). Denominator: cos(α + β) + cos(α − β) = 2 cos α cos β. Divide: (2 sin α cos β)/(2 cos α cos β) = sin α/cos α = tan α. ✓
(i) Use a² − b² = (a − b)(a + b) with a = sin(α + β/2), b = sin(α − β/2). Then a − b = 2 cos α sin(β/2) by the sin P − sin Q formula, and a + b = 2 sin α cos(β/2) by the sin P + sin Q formula. Multiply: 4 sin α cos α sin(β/2) cos(β/2) = (2 sin α cos α)(2 sin(β/2) cos(β/2)) = sin 2α · sin β. ✓ (ii) Group the last two terms: cos²(α + β) + 2 sin α sin β cos(α + β) = cos(α + β)[cos(α + β) + 2 sin α sin β] = cos(α + β)(cos α cos β + sin α sin β) = cos(α + β) cos(α − β) = cos²α − sin²β (by Q4). Then the whole LHS = sin²α + sin²β + cos²α − sin²β = 1. ✓
(i) RHS = (1 + sin α sin β/(cos α cos β)) · cos α cos β = cos α cos β + sin α sin β = cos(α − β). ✓ (ii) RHS numerator: 1 + (cos α/sin α)(sin β/cos β) = (sin α cos β + cos α sin β)/(sin α cos β). Dividing by cosec α sec β = 1/(sin α cos β) multiplies by sin α cos β, giving sin α cos β + cos α sin β = sin(α + β). ✓ (iii) cot α cot β + 1 = cos(α − β)/(sin α sin β), and cot β − cot α = sin(α − β)/(sin α sin β). Divide: cos(α − β)/sin(α − β) = cot(α − β). ✓ (iv) Multiply top and bottom by cos α cos β: the top becomes sin α cos β + cos α sin β = sin(α + β), the bottom sin α cos β − cos α sin β = sin(α − β). ✓ (v) cot α cot β − 1 = cos(α + β)/(sin α sin β), and cot α + cot β = sin(α + β)/(sin α sin β). Divide: cos(α + β)/sin(α + β) = cot(α + β). ✓
Both angles are acute, so the missing ratios are positive: cos α = √(1 − (24/25)²) = 7/25, and sin β = √(1 − (20/29)²) = 21/29. Then sin(α − β) = sin α cos β − cos α sin β = (24/25)(20/29) − (7/25)(21/29) = (480 − 147)/725 = 333/725. ✓
α is in quadrant IV, so cos α = +15/17. β is in quadrant III, so sin β = −3/5. (i) sin(α + β) = (−8/17)(−4/5) + (15/17)(−3/5) = (32 − 45)/85 = −13/85. (ii) cos(α + β) = (15/17)(−4/5) − (−8/17)(−3/5) = (−60 − 24)/85 = −84/85. (iii) tan(α + β) = (−13/85)/(−84/85) = 13/84. (iv) sin(α − β) = (−8/17)(−4/5) − (15/17)(−3/5) = (32 + 45)/85 = 77/85. (v) cos(α − β) = (15/17)(−4/5) + (−8/17)(−3/5) = (−60 + 24)/85 = −36/85. (vi) tan(α − β) = (77/85)/(−36/85) = −77/36. Quadrants: α + β has sin < 0 and cos < 0, so it lies in quadrant III; α − β has sin > 0 and cos < 0, so it lies in quadrant II.
(i) tan α > 0 puts α in I or III; not I, so α is in III: sin α = −3/5, cos α = −4/5. cos β > 0 puts β in I or IV; not I, so β is in IV: sin β = −12/13. Then sin(α + β) = (−3/5)(5/13) + (−4/5)(−12/13) = (−15 + 48)/65 = 33/65, and cos(α + β) = (−4/5)(5/13) − (−3/5)(−12/13) = (−20 − 36)/65 = −56/65. (ii) tan α < 0 puts α in II or IV; not IV, so α is in II: sin α = 15/17, cos α = −8/17. sin β < 0 puts β in III or IV; not IV, so β is in III: cos β = −24/25. Then sin(α + β) = (15/17)(−24/25) + (−8/17)(−7/25) = (−360 + 56)/425 = −304/425, and cos(α + β) = (−8/17)(−24/25) − (15/17)(−7/25) = (192 + 105)/425 = 297/425.
Divide top and bottom by cos 19°: the fraction becomes (1 + tan 19°)/(1 − tan 19°). By the tan addition formula, that is tan(45° + 19°) = tan 64°. ✓
The left side is exactly cos(A − B) with A = 60° + θ and B = 60° − θ, since cos(A − B) = cos A cos B + sin A sin B. And A − B = (60° + θ) − (60° − θ) = 2θ. So LHS = cos 2θ. ✓
Since (α + β)/2 = 90° − γ/2, we have cot((α + β)/2) = tan(γ/2) = 1/cot(γ/2). But by the cot addition formula, cot((α + β)/2) = (cot(α/2) cot(β/2) − 1)/(cot(α/2) + cot(β/2)). Equate and cross-multiply: cot(α/2) cot(β/2) cot(γ/2) − cot(γ/2) = cot(α/2) + cot(β/2). Move cot(γ/2) across and the required identity follows. ✓
γ = 180° − (α + β), so cot γ = cot(180° − (α + β)) = −cot(α + β) = −(cot α cot β − 1)/(cot α + cot β). Multiply out: cot γ (cot α + cot β) = −cot α cot β + 1. Rearranged: cot α cot β + cot β cot γ + cot γ cot α = 1. ✓
The method is always the same: r = √(a² + b²), then a = r cos φ and b = r sin φ (with a minus sign giving the θ − φ form). (i) r = √(24² + 7²) = 25, tan φ = 7/24 → 25 sin(θ + φ). (ii) r = √(12² + 5²) = 13, tan φ = 5/12 → 13 sin(θ − φ). (iii) r = √2; √2 sin(θ − 45°) = √2(sin θ · 1/√2 − cos θ · 1/√2) = sin θ − cos θ, so √2 sin(θ − 45°). (iv) r = √(8² + 6²) = 10, tan φ = 6/8 = 3/4 → 10 sin(θ − φ). (v) r = 1, cos φ = 1/2, sin φ = √3/2, so φ = 60° → sin(θ + 60°). (vi) r = √(13² + 84²) = 85, tan φ = 84/13 → 85 sin(θ − φ).
In both parts α is acute, so the missing ratio is positive: for (i), cos α = 4/5; for (ii), sin α = 3/5. So in both cases sin 2α = 2 sin α cos α = 2 × (3/5) × (4/5) = 24/25, cos 2α = cos²α − sin²α = (16/25) − (9/25) = 7/25, and tan 2α = (24/25)/(7/25) = 24/7. The two parts describe the same 3-4-5 triangle, which is why the answers agree.
(i) cot α − tan α = cos α/sin α − sin α/cos α = (cos²α − sin²α)/(sin α cos α) = cos 2α/((1/2) sin 2α) = 2 cot 2α. ✓ (ii) 1 + cos 2α = 2cos²α and sin 2α = 2 sin α cos α, so the fraction is (2 sin α cos α)/(2cos²α) = tan α. ✓ (iii) 1 − cos α = 2sin²(α/2) and sin α = 2 sin(α/2) cos(α/2); divide to get tan(α/2). ✓ (iv) Multiply top and bottom by (cos α − sin α): the fraction becomes (cos α − sin α)²/(cos²α − sin²α) = (1 − sin 2α)/cos 2α. And sec 2α − tan 2α = (1 − sin 2α)/cos 2α. ✓ (v) ⚠️ Sign flag (derived step): 1 + sin α = (sin(α/2) + cos(α/2))² and 1 − sin α = (sin(α/2) − cos(α/2))², so the square root is really |sin(α/2) + cos(α/2)| / |sin(α/2) − cos(α/2)|. The identity as printed drops the absolute values, so it holds as written only where sin(α/2) − cos(α/2) > 0 (for instance α = 200° works, α = 60° gives the negative of the printed right side). With that condition understood, the algebra is: √((1 + sin α)/(1 − sin α)) = (sin(α/2) + cos(α/2))/(sin(α/2) − cos(α/2)). (vi) cosec θ + 2 cosec 2θ = 1/sin θ + 1/(sin θ cos θ) = (1 + cos θ)/(sin θ cos θ). Divide by sec θ (i.e. multiply by cos θ): (1 + cos θ)/sin θ = (2cos²(θ/2))/(2 sin(θ/2) cos(θ/2)) = cot(θ/2). ✓ (vii) tan α tan 2α = tan α · 2tan α/(1 − tan²α) = 2tan²α/(1 − tan²α). Add 1: (1 + tan²α)/(1 − tan²α) = sec 2α. ✓ (viii) cos θ + cos 3θ = 2 cos 2θ cos θ, so the fraction is (2 sin θ sin 2θ)/(2 cos 2θ cos θ) = tan θ · tan 2θ. ✓ (ix) Common denominator sin θ cos θ: the numerator becomes sin 3θ cos θ − cos 3θ sin θ = sin 2θ = 2 sin θ cos θ, so the fraction is 2. ✓ (x) Common denominator sin θ cos θ: the numerator becomes cos 3θ sin θ + sin 3θ cos θ = sin 4θ, and sin 4θ/(sin θ cos θ) = 2 sin 4θ/sin 2θ = 4 cos 2θ. ✓ (xi) tan(θ/2) + cot(θ/2) = 2/sin θ, and cot(θ/2) − tan(θ/2) = 2cos θ/sin θ. Divide: 1/cos θ = sec θ. ✓ (xii) Common denominator sin θ cos θ: the numerator becomes sin 3θ sin θ + cos 3θ cos θ = cos 2θ, so the fraction is cos 2θ/((1/2) sin 2θ) = 2 cot 2θ. ✓ (xiii) RHS = (1/2)(sin⁴θ + cos⁴θ)/(sin²θ cos²θ) = (1/2)(1 − 2sin²θ cos²θ)/(sin²θ cos²θ). Writing sin²θ cos²θ = (1/4)sin²2θ gives (2 − sin²2θ)/sin²2θ. LHS = (3 + 1 − 2sin²2θ)/(2sin²2θ) = the same. ✓ (xiv) 1 + sin 2θ = (sin θ + cos θ)² and 1 − sin 2θ = (cos θ − sin θ)², so LHS = ((sin θ + cos θ)/(cos θ − sin θ))². And tan(π/4 + θ) = (1 + tan θ)/(1 − tan θ) = (cos θ + sin θ)/(cos θ − sin θ). Square both sides to match. ✓ (xv) cos²(5π/8) = cos²(3π/8) and cos²(7π/8) = cos²(π/8), since cos(π − x) = −cos x and the square kills the sign. Also cos²(3π/8) = sin²(π/8). So the sum is 2cos²(π/8) + 2sin²(π/8) = 2. ✓
⚠️ Derived/assumed-sign step (flagged): the book's nested radicals drop absolute values. Strictly: 2 + 2cos 4θ = 4cos²2θ, so √(2 + 2cos 4θ) = 2|cos 2θ|. Then 2 + 2|cos 2θ| = 4cos²θ provided cos 2θ ≥ 0, and √(4cos²θ) = 2|cos θ|. So the printed identity 2cos θ = √(2 + √(2 + 2cos 4θ)) holds under the sign assumptions cos 2θ ≥ 0 and cos θ ≥ 0; in full generality the left side is 2|cos θ|. The chain of algebra is: √(2 + √(2 + 2cos 4θ)) = √(2 + 2cos 2θ) = √(4cos²θ) = 2cos θ (with the stated sign assumptions).
sin²θ = (1 − cos 2θ)/2, so sin⁴θ = (1 − cos 2θ)²/4 = (1 − 2cos 2θ + cos²2θ)/4. Replace cos²2θ by (1 + cos 4θ)/2: sin⁴θ = (1 − 2cos 2θ + (1 + cos 4θ)/2)/4 = (2 − 4cos 2θ + 1 + cos 4θ)/8 = (3 − 4cos 2θ + cos 4θ)/8. Compare Example 13, which does the same job for cos⁴θ.
(i) θ = 18°: 5θ = 90°, so 3θ = 90° − 2θ and sin 3θ = cos 2θ. With s = sin θ: 3s − 4s³ = 1 − 2s², i.e. 4s³ − 2s² − 3s + 1 = 0, i.e. (s − 1)(4s² + 2s − 1) = 0. Since s ≠ 1, 4s² + 2s − 1 = 0, giving the positive root s = (√5 − 1)/4. So sin 18° = (√5 − 1)/4 and cos 18° = √(10 + 2√5)/4. (ii) θ = 36°: 5θ = 180°, so 3θ = 180° − 2θ and sin 3θ = sin 2θ. Dividing by sin θ: 3 − 4sin²θ = 2cos θ, i.e. 4cos²θ − 2cos θ − 1 = 0, giving cos 36° = (√5 + 1)/4 and sin 36° = √(10 − 2√5)/4. (iii) sin 54° = cos 36° = (√5 + 1)/4, cos 54° = sin 36° = √(10 − 2√5)/4. (iv) sin 72° = cos 18° = √(10 + 2√5)/4, cos 72° = sin 18° = (√5 − 1)/4. Hence: cos 108° = −cos 72° and cos 144° = −cos 36°, so the product is cos²36° · cos²72° = ((√5 + 1)/4)² · ((√5 − 1)/4)² = ((5 − 1)/16)² = (1/4)² = 1/16. ✓
(i) 2 sin 3θ cos θ = sin(3θ + θ) + sin(3θ − θ) = sin 4θ + sin 2θ. (ii) 2 cos 5θ sin 3θ = sin(5θ + 3θ) − sin(5θ − 3θ) = sin 8θ − sin 2θ. (iii) sin 5θ cos 2θ = (1/2)[sin 7θ + sin 3θ]. (iv) 2 sin 7θ sin 2θ = −[cos 9θ − cos 5θ] = cos 5θ − cos 9θ. (v) cos(x + y) sin(x − y) = (1/2)[sin 2x − sin 2y]. (vi) cos(2x + 30°) cos(2x − 30°) = (1/2)[cos 4x + cos 60°] = (1/2) cos 4x + 1/4. (vii) sin 12° sin 46° = (1/2)[cos(12° − 46°) − cos(12° + 46°)] = (1/2)[cos 34° − cos 58°]. (viii) sin(x + 45°) sin(x − 45°) = (1/2)[cos 90° − cos 2x] = −(1/2) cos 2x.
(i) sin 5θ + sin 3θ = 2 sin((5θ + 3θ)/2) cos((5θ − 3θ)/2) = 2 sin 4θ cos θ. (ii) sin 8θ − sin 4θ = 2 cos 6θ sin 2θ. (iii) cos 6θ + cos 3θ = 2 cos(9θ/2) cos(3θ/2). (iv) cos 7θ − cos θ = −2 sin 4θ sin 3θ. (v) cos 12° + cos 48° = 2 cos 30° cos 18° = 2 × (√3/2) × cos 18° = √3 cos 18°. (vi) sin(x + 30°) + sin(x − 30°) = 2 sin x cos 30° = √3 sin x.
(i) Top: sin 3x − sin x = 2 cos 2x sin x. Bottom: cos x − cos 3x = 2 sin 2x sin x. Ratio = cos 2x/sin 2x = cot 2x. ✓ (ii) Top: 2 sin 5x cos 3x. Bottom: 2 cos 5x cos 3x. Ratio = tan 5x. ✓ (iii) Top: 2 cos((A + B)/2) sin((A − B)/2). Bottom: 2 sin((A + B)/2) cos((A − B)/2). Ratio = tan((A − B)/2) cot((A + B)/2). ✓ (iv) Top: 2 sin 60° cos 20°. Bottom: 2 cos 60° cos 20°. Ratio = tan 60° = √3. ✓
(i) ⚠️ FLAWED AS PRINTED — read this carefully. Pair the terms: cos 15° + cos 195° = 2 cos 105° cos 90° = 0, and cos 105° + cos 285° = 2 cos 195° cos 90° = 0. Both pairs cancel exactly. What remains is cos 160° alone, which is about −0.94, not 0. So the statement as printed is false; the five terms add up to cos 160°, not 0. No replacement angle is invented here — if the book meant a different fifth angle, it is not stated, so the honest answer is: the sum equals cos 160°. (ii) Pair outer with inner: sin 2θ + sin 8θ = 2 sin 5θ cos 3θ and sin 4θ + sin 6θ = 2 sin 5θ cos θ, so the numerator is 2 sin 5θ (cos 3θ + cos θ). Likewise the denominator is 2 cos 5θ (cos 3θ + cos θ). Ratio = tan 5θ. ✓ (iii) cos²X = (1 + cos 2X)/2, so the difference is (1/2)[cos(π/2 − α) − cos(π/2 + α)] = (1/2)[sin α − (−sin α)] = sin α. ✓ (iv) sin A sin B = (1/2)[cos(A − B) − cos(A + B)] with A = π/4 − θ, B = π/4 + θ: this gives (1/2)[cos(−2θ) − cos(π/2)] = (1/2) cos 2θ. ✓ (v) sin θ + sin 7θ = 2 sin 4θ cos 3θ and sin 3θ + sin 5θ = 2 sin 4θ cos θ, so the numerator is 2 sin 4θ (cos 3θ + cos θ). Likewise the denominator is 2 cos 4θ (cos 3θ + cos θ). Ratio = tan 4θ. ✓
(i) cos 60° = 1/2, and Example 18 gives cos 20° cos 40° cos 80° = 1/8. Multiply: (1/2)(1/8) = 1/16. ✓ (ii) First show sin 20° sin 40° sin 80° = √3/8: sin 40° sin 80° = (1/2)[cos 40° − cos 120°] = (1/2)[cos 40° + 1/2]; multiply by sin 20°: (1/2) sin 20° cos 40° + (1/4) sin 20° = (1/4)[sin 60° + sin(−20°)] + (1/4) sin 20° = (1/4) sin 60° = √3/8. Now sin(π/9) sin(2π/9) sin(4π/9) is sin 20° sin 40° sin 80° = √3/8, and sin(π/3) = sin 60° = √3/2. Product: (√3/8)(√3/2) = 3/16. ✓ (iii) Similarly sin 50° sin 70° = (1/2)[cos 20° − cos 120°] = (1/2)[cos 20° + 1/2]; multiply by sin 10°: (1/2) sin 10° cos 20° + (1/4) sin 10° = (1/4)[sin 30° + sin(−10°)] + (1/4) sin 10° = (1/4)(1/2) = 1/8. Multiply by sin 30° = 1/2: 1/16. ✓
sin 3θ = 3 sin θ − 4 sin³θ = sin θ (3 − 4sin²θ) = sin θ (4cos²θ − 1) = sin θ (2cos θ − 1)(2cos θ + 1). And 1 + 2cos 2θ = 1 + 2(2cos²θ − 1) = 4cos²θ − 1 = (2cos θ − 1)(2cos θ + 1). Cancel the common factor: the fraction is sin θ. ✓ Deduce: put θ = 15°: sin 45°/(1 + 2cos 30°) = sin 15°, i.e. sin 15° = (√2/2)/(1 + √3). Rationalise: multiply top and bottom by (√3 − 1): sin 15° = (√2/2)(√3 − 1)/2 = (√6 − √2)/4, matching Exercise 10.2 Q1(i).
tan 75° = tan(45° + 30°) = (1 + tan 30°)/(1 − tan 30°) = (1 + 1/√3)/(1 − 1/√3) = (√3 + 1)/(√3 − 1). Multiply top and bottom by (√3 + 1): (4 + 2√3)/2 = 2 + √3. And tan 15° = 2 − √3 from Exercise 10.2 Q1(iii). Subtract: (2 + √3) − (2 − √3) = 2√3. ✓
Square the left side: (cos 15° − sin 15°)² = cos²15° + sin²15° − 2 sin 15° cos 15° = 1 − sin 30° = 1 − 1/2 = 1/2. Since cos 15° > sin 15°, the difference is positive, so cos 15° − sin 15° = √(1/2) = 1/√2. ✓
Numerator: sin²α − sin²β = sin(α + β) sin(α − β) by Example 5. Denominator: sin α cos α − sin β cos β = (1/2)(sin 2α − sin 2β) = (1/2) · 2 cos(α + β) sin(α − β) = cos(α + β) sin(α − β). Divide: sin(α + β)/cos(α + β) = tan(α + β). ✓
Group as (sin α + sin β) − [sin(α + β + γ) − sin γ]. The first bracket is 2 sin((α + β)/2) cos((α − β)/2) by the sum formula. The second bracket is 2 cos((α + β + 2γ)/2) sin((α + β)/2) by the difference formula. Factor 2 sin((α + β)/2): LHS = 2 sin((α + β)/2) [cos((α − β)/2) − cos((α + β + 2γ)/2)]. Now cos C − cos D = −2 sin((C + D)/2) sin((C − D)/2) with (C + D)/2 = (α + γ)/2 and (C − D)/2 = −(β + γ)/2. So the bracket is −2 sin((α + γ)/2) · (−sin((β + γ)/2)) = 2 sin((α + γ)/2) sin((β + γ)/2). Multiply by the outside factor: 4 sin((α + β)/2) sin((β + γ)/2) sin((γ + α)/2). ✓
اس باب میں کیا ہے؟ Trigonometric identities کی پوری عمارت, ایک بنیاد پر کھڑی: distance formula سے cos(α − β) کا ثبوت, اس سے دس deductions, پھر allied angles کا odd/even اصول, double-angle, half-angle اور triple-angle formulae, اور آخر میں product ↔ sum کی تبدیلی. ہر formula اگلے کا ذریعہ بنتا ہے — اس لیے اس chapter کو ترتیب سے پڑھنا سب سے فائدہ-مند ہے.
Revision guidance: پہلے fundamental law کا ثبوت سمجھو (دو طریقوں سے ایک فاصلہ), پھر allied angles کا device زبانی یاد کرو — امتحان میں سب سے زیادہ سوال اسی سے آتے ہیں. Double-angle کی تینوں cos صورتیں ایک ساتھ لکھ کر فرق سمجھو, اور product ↔ sum میں cos P − cos Q والی minus علامت کو غور کر یاد کرو. Exercise: 📝 Official Exercise Questions — with Answers (36 questions, 123 parts; no MCQs, no review exercise in this unit). Flags to remember: Example 3(iii) misprint (−1, not 1); Exercise 10.4 Q4(i) is false as printed (sum = cos 160°, not 0); Example 10 prints “tan⁻¹φ” for tan φ; Exercise 10.3 Q3 drops absolute values in its nested radicals.
اس باب میں مثلثیاتی شناختوں کی مکمل عمارت ہے: فاصلے کے فارمولے سے cos(α − β) کا ثبوت، اس سے دس نتائج، ملحقہ زاویوں کا اصول، دگنے، نصف اور تگنے زاویے کے فارمولے، اور آخر میں حاصل ضرب سے مجموعے کی تبدیلی۔
امتحان کے لیے تین چیزیں سب سے اہم ہیں: بنیادی قانون کا ثبوت، ملحقہ زاویوں کا طاق/جفت اصول مع ربع کی علامت، اور cos 2α کی تین صورتیں۔ ان تینوں پر عبور حاصل کر لیں تو اس باب کی مشقیں آسان لگیں گی۔ کتاب کی دو غلطیوں (مثال ۳(iii) اور مشق ۱۰.۴ سوال ۴(i)) کو بھی ذہن میں رکھیں تاکہ امتحان میں غلط جواب نہ لکھیں۔
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