Exercise 3.3 answers: Trigonometric Functions

Open NCERT p. 67One way to think about it

1. Use the standard values: sin(π/6) = 1/2, cos(π/3) = 1/2, tan(π/4) = 1.
2. Square them: sin²(π/6) = 1/4, cos²(π/3) = 1/4, tan²(π/4) = 1.
3. Add and subtract: 1/4 + 1/4 − 1 = 1/2 − 1 = −1/2.
4. This is the same as the right-hand side, so the statement is true.

In shortLHS = −1/2 = RHS, so the identity holds.

Watch this explained “The table checks itself”, 13:39 into Coordinates on the unit circle extend the ratios to every real number

Open NCERT p. 67One way to think about it

1. sin(π/6) = 1/2, so 2 sin²(π/6) = 2 × (1/4) = 1/2.
2. cos(π/3) = 1/2, so cos²(π/3) = 1/4.
3. 7π/6 is a half turn plus π/6 (π + π/6), and sin(π + θ) = −sin θ.
4. So sin(7π/6) = −sin(π/6) = −1/2, which gives cosec(7π/6) = −2.
5. cosec²(7π/6) = (−2)² = 4.
6. LHS = 1/2 + 4 × 1/4 = 1/2 + 1 = 3/2.

In shortLHS = 1/2 + 1 = 3/2 = RHS. Proved.

Watch this explained “The half turns”, 8:20 into Shifts by a quarter, a half and a whole turn all fall out of the same two results

Open NCERT p. 67One way to think about it

1. cot(π/6) = √3, so cot²(π/6) = 3.
2. tan(π/6) = 1/√3, so tan²(π/6) = 1/3, and 3 tan²(π/6) = 1.
3. 5π/6 is a half turn back from π/6 (π − π/6), and sin(π − θ) = sin θ.
4. So sin(5π/6) = sin(π/6) = 1/2, which gives cosec(5π/6) = 2.
5. LHS = 3 + 2 + 1 = 6.

In shortLHS = 3 + 2 + 1 = 6 = RHS. Proved.

Watch this explained “The half turns”, 8:20 into Shifts by a quarter, a half and a whole turn all fall out of the same two results

Open NCERT p. 67One way to think about it

1. 3π/4 is a half turn back from π/4 (π − π/4), and sin(π − θ) = sin θ, so sin(3π/4) = sin(π/4) = √2/2.
2. sin²(3π/4) = 1/2, so 2 sin²(3π/4) = 1.
3. cos(π/4) = √2/2, so cos²(π/4) = 1/2, and 2 cos²(π/4) = 1.
4. sec(π/3) = 1/cos(π/3) = 2, so sec²(π/3) = 4, and 2 sec²(π/3) = 8.
5. LHS = 1 + 1 + 8 = 10.

In shortLHS = 1 + 1 + 8 = 10 = RHS. Proved.

Watch this explained “The half turns”, 8:20 into Shifts by a quarter, a half and a whole turn all fall out of the same two results

Open NCERT p. 67Matches NCERT’s answer

(i) sin 75°

1. Write 75° as 45° + 30°.
2. Use sin(A + B) = sin A cos B + cos A sin B.
3. sin 75° = sin 45° cos 30° + cos 45° sin 30° = (√2/2)(√3/2) + (√2/2)(1/2).
4. = √6/4 + √2/4 = (√6 + √2)/4.

Answer(√6 + √2)/4

(ii) tan 15°

1. Write 15° as 45° − 30°.
2. Use tan(A − B) = (tan A − tan B)/(1 + tan A tan B).
3. tan 45° = 1 and tan 30° = 1/√3.
4. tan 15° = (1 − 1/√3)/(1 + 1/√3) = (√3 − 1)/(√3 + 1).
5. Multiply top and bottom by (√3 − 1): = (√3 − 1)²/2 = (4 − 2√3)/2 = 2 − √3.

Answer2 − √3

Watch this explained “Assembling the sine of a sum”, 5:09 into Shifts by a quarter, a half and a whole turn all fall out of the same two results

Open NCERT p. 67One way to think about it

1. Let A = π/4 − x and B = π/4 − y.
2. The LHS is cos A cos B − sin A sin B, which is exactly cos(A + B).
3. A + B = π/2 − (x + y).
4. So LHS = cos(π/2 − (x + y)).
5. cos(π/2 − θ) = sin θ for any θ, so cos(π/2 − (x + y)) = sin(x + y).

In shortLHS = cos(π/2 − (x+y)) = sin(x+y) = RHS. Proved.

Watch this explained “Equate, cancel, read it off”, 7:28 into One distance calculation on the unit circle yields the cosine of a difference

Open NCERT p. 67One way to think about it

1. Use tan(A + B) = (tan A + tan B)/(1 − tan A tan B) with A = π/4, tan(π/4) = 1: tan(π/4 + x) = (1 + tan x)/(1 − tan x).
2. Use tan(A − B) = (tan A − tan B)/(1 + tan A tan B): tan(π/4 − x) = (1 − tan x)/(1 + tan x).
3. Divide: LHS = [(1 + tan x)/(1 − tan x)] ÷ [(1 − tan x)/(1 + tan x)] = [(1 + tan x)/(1 − tan x)] × [(1 + tan x)/(1 − tan x)].
4. = ((1 + tan x)/(1 − tan x))².

In shortLHS = ((1+tan x)/(1−tan x))² = RHS. Proved.

Watch this explained “What the tidying cost”, 3:08 into The tangent and cotangent versions, and the angles they refuse to cover

Open NCERT p. 67One way to think about it

1. Prove: cos(π + x) cos(−x) / [sin(π − x) cos(π/2 + x)] = cot²x.
2. cos(π + x) = −cos x (half turn forward).
3. cos(−x) = cos x (cosine is an even function).
4. So the numerator = (−cos x)(cos x) = −cos²x.
5. sin(π − x) = sin x (half turn back).
6. cos(π/2 + x) = −sin x (quarter turn forward).
7. So the denominator = (sin x)(−sin x) = −sin²x.
8. LHS = (−cos²x)/(−sin²x) = cos²x/sin²x = cot²x.

In shortLHS = cot²x = RHS. Proved.

Watch this explained “The half turns”, 8:20 into Shifts by a quarter, a half and a whole turn all fall out of the same two results

Open NCERT p. 67One way to think about it

1. cos(2π + x) = cos x and cot(2π + x) = cot x, because a whole turn brings the point back to itself.
2. For cos(3π/2 + x), write 3π/2 + x = π + (π/2 + x). A half turn forward reverses cosine, cos(π + θ) = −cos θ, so cos(3π/2 + x) = −cos(π/2 + x).
3. A quarter turn forward gives cos(π/2 + x) = −sin x. So cos(3π/2 + x) = −(−sin x) = sin x.
4. For cot(3π/2 − x), write 3π/2 − x = π + (π/2 − x). Cotangent repeats after a half turn, cot(π + θ) = cot θ, so cot(3π/2 − x) = cot(π/2 − x) = tan x.
5. LHS = sin x · cos x · [tan x + cot x].
6. tan x + cot x = sin x/cos x + cos x/sin x = (sin²x + cos²x)/(sin x cos x) = 1/(sin x cos x).
7. LHS = sin x cos x × 1/(sin x cos x) = 1.

In shortLHS = sin x cos x × 1/(sin x cos x) = 1 = RHS. Proved.

Watch this explained “The shifts the list does not print”, 12:27 into Shifts by a quarter, a half and a whole turn all fall out of the same two results

Open NCERT p. 67One way to think about it

1. Let A = (n + 1)x and B = (n + 2)x.
2. cos(A − B) = cos A cos B + sin A sin B, so the LHS is exactly cos(A − B).
3. A − B = (n + 1)x − (n + 2)x = −x.
4. cos(−x) = cos x (cosine is an even function).
5. So LHS = cos(−x) = cos x.

In shortLHS = cos(−x) = cos x = RHS. Proved.

Watch this explained “The difference form, for free”, 9:29 into One distance calculation on the unit circle yields the cosine of a difference

Open NCERT p. 67One way to think about it

1. Use cos C − cos D = −2 sin((C+D)/2) sin((C−D)/2), with C = 3π/4 + x and D = 3π/4 − x.
2. (C + D)/2 = 3π/4 and (C − D)/2 = x.
3. cos C − cos D = −2 sin(3π/4) sin x.
4. sin(3π/4) = √2/2 (a half turn back from π/4: sin(π − π/4) = sin(π/4)).
5. LHS = −2 × (√2/2) × sin x = −√2 sin x.

In shortLHS = −2 sin(3π/4) sin x = −√2 sin x = RHS. Proved.

Watch this explained “The four, in the form you will use”, 5:09 into Trading a sum of two ratios for a product, and back again

Open NCERT p. 67One way to think about it

1. Expand sin(A + B) sin(A − B) using the sum and difference formulas: sin(A+B) = sinA cosB + cosA sinB, sin(A−B) = sinA cosB − cosA sinB.
2. Multiplying these two: sin(A+B) sin(A−B) = sin²A cos²B − cos²A sin²B.
3. Replace cos²B with 1 − sin²B and cos²A with 1 − sin²A: this becomes sin²A − sin²B.
4. So sin(A+B) sin(A−B) = sin²A − sin²B for any A and B.
5. Take A = 6x and B = 4x: sin²6x − sin²4x = sin(6x+4x) sin(6x−4x) = sin10x sin2x.

In shortLHS = sin10x sin2x = sin2x sin10x = RHS. Proved.

Watch this explained “Two more shapes worth keeping”, 13:23 into Trading a sum of two ratios for a product, and back again

Open NCERT p. 67One way to think about it

1. Write cos²2x = 1 − sin²2x and cos²6x = 1 − sin²6x. Then LHS = (1 − sin²2x) − (1 − sin²6x) = sin²6x − sin²2x.
2. For any A and B: sin(A + B) sin(A − B) = (sin A cos B + cos A sin B)(sin A cos B − cos A sin B) = sin²A cos²B − cos²A sin²B.
3. Replace cos²B by 1 − sin²B and cos²A by 1 − sin²A: sin²A(1 − sin²B) − (1 − sin²A) sin²B = sin²A − sin²B.
4. So sin²A − sin²B = sin(A + B) sin(A − B). Take A = 6x and B = 2x: sin²6x − sin²2x = sin 8x sin 4x.

In shortLHS = sin 8x sin 4x = sin 4x sin 8x = RHS. Proved.

Watch this explained “Two more shapes worth keeping”, 13:23 into Trading a sum of two ratios for a product, and back again

Open NCERT p. 67One way to think about it

1. Group sin2x + sin6x and use sin C + sin D = 2 sin((C+D)/2) cos((C−D)/2).
2. With C = 2x, D = 6x: (C+D)/2 = 4x and (C−D)/2 = −2x, and cos(−2x) = cos2x.
3. So sin2x + sin6x = 2 sin4x cos2x.
4. LHS = 2 sin4x cos2x + 2 sin4x = 2 sin4x (cos2x + 1).
5. Use cos2x = 2cos²x − 1, so cos2x + 1 = 2cos²x.
6. LHS = 2 sin4x × 2cos²x = 4cos²x sin4x.

In shortLHS = 4cos²x sin4x = RHS. Proved.

Watch this explained “The four, in the form you will use”, 5:09 into Trading a sum of two ratios for a product, and back again

Open NCERT p. 67One way to think about it

1. Prove: cot4x (sin5x + sin3x) = cotx (sin5x − sin3x).
2. sin5x + sin3x = 2 sin4x cos x (sum-to-product, half-sum 4x, half-difference x).
3. sin5x − sin3x = 2 cos4x sin x (difference-to-product, same half-sum and half-difference).
4. LHS = cot4x × 2 sin4x cos x = (cos4x/sin4x) × 2 sin4x cos x = 2 cos4x cos x.
5. RHS = cot x × 2 cos4x sin x = (cos x/sin x) × 2 cos4x sin x = 2 cos4x cos x.
6. LHS and RHS both equal 2 cos4x cos x.

In shortBoth sides equal 2 cos4x cos x. Proved.

Watch this explained “The four, in the form you will use”, 5:09 into Trading a sum of two ratios for a product, and back again

Open NCERT p. 67One way to think about it

1. Prove: (cos9x − cos5x)/(sin17x − sin3x) = −sin2x/cos10x.
2. cos9x − cos5x = −2 sin7x sin2x (half-sum 7x, half-difference 2x; a cosine difference carries a minus sign).
3. sin17x − sin3x = 2 cos10x sin7x (half-sum 10x, half-difference 7x).
4. Ratio = (−2 sin7x sin2x)/(2 cos10x sin7x) = −sin2x/cos10x, cancelling sin7x (valid where sin7x ≠ 0).

In shortLHS = −sin2x/cos10x = RHS. Proved.

Watch this explained “A quotient collapsing”, 9:25 into Trading a sum of two ratios for a product, and back again

Open NCERT p. 67One way to think about it

1. sin5x + sin3x = 2 sin4x cos x (half-sum 4x, half-difference x).
2. cos5x + cos3x = 2 cos4x cos x (half-sum 4x, half-difference x).
3. Ratio = (2 sin4x cos x)/(2 cos4x cos x) = sin4x/cos4x = tan4x, cancelling cos x (valid where cos x ≠ 0).

In shortLHS = tan4x = RHS. Proved.

Watch this explained “A quotient collapsing”, 9:25 into Trading a sum of two ratios for a product, and back again

Open NCERT p. 67One way to think about it

1. Prove: (sinx − siny)/(cosx + cosy) = tan((x − y)/2).
2. sin x − sin y = 2 cos((x+y)/2) sin((x−y)/2).
3. cos x + cos y = 2 cos((x+y)/2) cos((x−y)/2).
4. Ratio = sin((x−y)/2)/cos((x−y)/2) = tan((x−y)/2), cancelling cos((x+y)/2) (valid where it is not zero).

In shortLHS = tan((x−y)/2) = RHS. Proved.

Watch this explained “A quotient collapsing”, 9:25 into Trading a sum of two ratios for a product, and back again

Open NCERT p. 67One way to think about it

1. Prove: (sinx + sin3x)/(cosx + cos3x) = tan2x.
2. sin x + sin3x = 2 sin2x cos x (half-sum 2x, half-difference x).
3. cos x + cos3x = 2 cos2x cos x (half-sum 2x, half-difference x).
4. Ratio = (2 sin2x cos x)/(2 cos2x cos x) = sin2x/cos2x = tan2x, cancelling cos x (valid where cos x ≠ 0).

In shortLHS = tan2x = RHS. Proved.

Watch this explained “A quotient collapsing”, 9:25 into Trading a sum of two ratios for a product, and back again

Open NCERT p. 67One way to think about it

1. Work on the top: sin x − sin 3x. Use sin A − sin B = 2 cos((A+B)/2) sin((A−B)/2) with A = x, B = 3x.
2. sin x − sin 3x = 2 cos(2x) sin(−x) = −2 cos 2x sin x.
3. Work on the bottom: sin²x − cos²x = −(cos²x − sin²x) = −cos 2x.
4. Divide: (−2 cos 2x sin x)/(−cos 2x) = 2 sin x, since cos 2x cancels.
5. This is exactly the right-hand side, so the identity is proved.

In shortLHS simplifies to 2 sin x = RHS, so the identity holds.

Watch this explained “The four, in the form you will use”, 5:09 into Trading a sum of two ratios for a product, and back again

Open NCERT p. 67One way to think about it

1. On top, pair the two outer terms: cos 4x + cos 2x = 2 cos 3x cos x.
2. So the top becomes 2 cos 3x cos x + cos 3x = cos 3x (2 cos x + 1).
3. On the bottom, pair the two outer terms the same way: sin 4x + sin 2x = 2 sin 3x cos x.
4. So the bottom becomes 2 sin 3x cos x + sin 3x = sin 3x (2 cos x + 1).
5. The bracket (2 cos x + 1) is a factor of the bottom, so it is not zero wherever the left side is defined. Cancel it, leaving cos 3x / sin 3x = cot 3x.

In shortLHS simplifies to cot 3x = RHS, so the identity holds.

Watch this explained “Pair the outer two”, 10:43 into Trading a sum of two ratios for a product, and back again

Open NCERT p. 68One way to think about it

1. Notice that 3x = x + 2x, so cot 3x = cot(x + 2x).
2. Use the compound-angle formula cot(A + B) = (cot A cot B − 1)/(cot A + cot B) with A = x, B = 2x.
3. cot 3x = (cot x cot 2x − 1)/(cot x + cot 2x).
4. Multiply both sides by (cot x + cot 2x): cot 3x cot x + cot 3x cot 2x = cot x cot 2x − 1.
5. Move the two terms on the left to the right: cot x cot 2x − cot 2x cot 3x − cot 3x cot x = 1, which is what we had to prove.

In shortProved using cot(x + 2x) = cot 3x.

Watch this explained “Same move, other divisor”, 7:07 into The tangent and cotangent versions, and the angles they refuse to cover

Open NCERT p. 68One way to think about it

1. Write 4x as 2(2x), so tan 4x = tan(2 · 2x) = 2 tan 2x / (1 − tan²2x).
2. Let t = tan x. Then tan 2x = 2t/(1 − t²), using the double-angle formula for tangent.
3. 1 − tan²2x = 1 − 4t²/(1 − t²)² = [(1 − t²)² − 4t²]/(1 − t²)² = (1 − 6t² + t⁴)/(1 − t²)².
4. So tan 4x = [2 · 2t/(1 − t²)] ÷ [(1 − 6t² + t⁴)/(1 − t²)²] = 4t(1 − t²)/(1 − 6t² + t⁴).
5. Replacing t by tan x gives tan 4x = 4 tan x (1 − tan²x)/(1 − 6 tan²x + tan⁴x), as required.

In shortProved by applying the tangent double-angle formula twice, to 2x and then to 4x.

Watch this explained “It does not stop”, 14:30 into Setting the second angle equal to the first gives the double and triple angle rules

Open NCERT p. 68One way to think about it

1. Write 4x as 2(2x), so cos 4x = cos(2 · 2x).
2. Use the double-angle formula cos 2θ = 1 − 2 sin²θ with θ = 2x: cos 4x = 1 − 2 sin²2x.
3. Now use sin 2x = 2 sin x cos x, so sin²2x = 4 sin²x cos²x.
4. Substitute: cos 4x = 1 − 2(4 sin²x cos²x) = 1 − 8 sin²x cos²x, as required.

In shortProved: cos 4x = 1 − 8 sin²x cos²x.

Watch this explained “It does not stop”, 14:30 into Setting the second angle equal to the first gives the double and triple angle rules

Open NCERT p. 68One way to think about it

1. Write 6x as 3(2x), so cos 6x = cos(3 · 2x).
2. Use the triple-angle formula cos 3θ = 4 cos³θ − 3 cos θ with θ = 2x: cos 6x = 4 cos³2x − 3 cos 2x.
3. Use cos 2x = 2 cos²x − 1. Let c = cos x, so cos 2x = 2c² − 1.
4. (2c² − 1)³ = 8c⁶ − 12c⁴ + 6c² − 1, so 4(2c² − 1)³ = 32c⁶ − 48c⁴ + 24c² − 4.
5. 3(2c² − 1) = 6c² − 3.
6. Subtract: cos 6x = (32c⁶ − 48c⁴ + 24c² − 4) − (6c² − 3) = 32c⁶ − 48c⁴ + 18c² − 1, as required.

In shortProved: cos 6x = 32 cos⁶x − 48 cos⁴x + 18 cos²x − 1.

Watch this explained “It does not stop”, 14:30 into Setting the second angle equal to the first gives the double and triple angle rules