Miscellaneous Exercise answers: Trigonometric Functions

Open NCERT p. 71One way to think about it

1. Turn the product into a sum: 2 cos A cos B = cos(A + B) + cos(A − B), with A = π/13, B = 9π/13.
2. 2 cos(π/13) cos(9π/13) = cos(10π/13) + cos(−8π/13) = cos(10π/13) + cos(8π/13).
3. Now cos(10π/13) = cos(π − 3π/13) = −cos(3π/13), and cos(8π/13) = cos(π − 5π/13) = −cos(5π/13).
4. So the left side becomes −cos(3π/13) − cos(5π/13) + cos(3π/13) + cos(5π/13), which is 0.

In shortProved: the left side equals 0.

Watch this explained “The other direction is not new”, 8:20 into Trading a sum of two ratios for a product, and back again

Open NCERT p. 71One way to think about it

1. Turn sin 3x + sin x into a product: sin 3x + sin x = 2 sin 2x cos x.
2. Turn cos 3x − cos x into a product: cos 3x − cos x = −2 sin 2x sin x.
3. Substitute both: (2 sin 2x cos x)(sin x) + (−2 sin 2x sin x)(cos x).
4. This is 2 sin 2x sin x cos x − 2 sin 2x sin x cos x, which is 0.

In shortProved: the left side equals 0.

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. 71One way to think about it

1. Expand both squares: cos²x + 2 cos x cos y + cos²y + sin²x − 2 sin x sin y + sin²y.
2. Group: (cos²x + sin²x) + (cos²y + sin²y) + 2(cos x cos y − sin x sin y) = 1 + 1 + 2(cos x cos y − sin x sin y).
3. Use the compound-angle formula cos(x + y) = cos x cos y − sin x sin y, so this is 2 + 2 cos(x + y).
4. Use the half-angle version of the double-angle formula: 1 + cos θ = 2 cos²(θ/2) with θ = x + y.
5. 2 + 2 cos(x + y) = 2(1 + cos(x + y)) = 2 · 2 cos²((x+y)/2) = 4 cos²((x+y)/2), as required.

In shortProved: the left side equals 4 cos²((x+y)/2).

Watch this explained “Squaring the first chord”, 5:31 into One distance calculation on the unit circle yields the cosine of a difference

Open NCERT p. 72One way to think about it

1. Expand both squares: cos²x − 2 cos x cos y + cos²y + sin²x − 2 sin x sin y + sin²y.
2. Group: (cos²x + sin²x) + (cos²y + sin²y) − 2(cos x cos y + sin x sin y) = 2 − 2(cos x cos y + sin x sin y).
3. Use the compound-angle formula cos(x − y) = cos x cos y + sin x sin y, so this is 2 − 2 cos(x − y).
4. Use 1 − cos θ = 2 sin²(θ/2) with θ = x − y.
5. 2 − 2 cos(x − y) = 2(1 − cos(x − y)) = 2 · 2 sin²((x−y)/2) = 4 sin²((x−y)/2), as required.

In shortProved: the left side equals 4 sin²((x−y)/2).

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. 72One way to think about it

1. Pair the outermost terms and the innermost terms: (sin x + sin 7x) + (sin 3x + sin 5x).
2. sin x + sin 7x = 2 sin 4x cos 3x, and sin 3x + sin 5x = 2 sin 4x cos x.
3. Add: 2 sin 4x cos 3x + 2 sin 4x cos x = 2 sin 4x (cos 3x + cos x).
4. Now cos 3x + cos x = 2 cos 2x cos x.
5. So the sum is 2 sin 4x · 2 cos 2x cos x = 4 cos x cos 2x sin 4x, as required.

In shortProved: the left side equals 4 cos x cos 2x sin 4x.

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. 72One way to think about it

1. Prove: [(sin 7x + sin 5x) + (sin 9x + sin 3x)] / [(cos 7x + cos 5x) + (cos 9x + cos 3x)] = tan 6x.
2. On top: sin 7x + sin 5x = 2 sin 6x cos x, and sin 9x + sin 3x = 2 sin 6x cos 3x.
3. So the top is 2 sin 6x cos x + 2 sin 6x cos 3x = 2 sin 6x (cos x + cos 3x).
4. On the bottom: cos 7x + cos 5x = 2 cos 6x cos x, and cos 9x + cos 3x = 2 cos 6x cos 3x.
5. So the bottom is 2 cos 6x cos x + 2 cos 6x cos 3x = 2 cos 6x (cos x + cos 3x).
6. Divide: the shared bracket (cos x + cos 3x) cancels, leaving sin 6x / cos 6x = tan 6x.

In shortLHS simplifies to tan 6x = RHS, so the identity holds.

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

Open NCERT p. 72One way to think about it

1. Group sin 3x − sin x first: sin 3x − sin x = 2 cos 2x sin x.
2. So the left side becomes 2 cos 2x sin x + sin 2x.
3. Write sin 2x = 2 sin x cos x, so the left side is 2 cos 2x sin x + 2 sin x cos x = 2 sin x (cos 2x + cos x).
4. Now cos 2x + cos x = 2 cos(3x/2) cos(x/2).
5. So the left side is 2 sin x · 2 cos(3x/2) cos(x/2) = 4 sin x cos(x/2) cos(3x/2), as required.

In shortProved: the left side equals 4 sin x cos(x/2) cos(3x/2).

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. 72Checked by computer

1. In quadrant II, sin x is positive and cos x is negative, and tan x = sin x / cos x = −4/3.
2. So take sin x = 4k, cos x = −3k for some positive k. Then sin²x + cos²x = 25k² = 1, so k = 1/5.
3. This gives sin x = 4/5 and cos x = −3/5.
4. Since x is between 90° and 180°, x/2 is between 45° and 90° — that is, x/2 is in quadrant I, so sin(x/2), cos(x/2) and tan(x/2) are all positive.
5. Use cos(x/2) = √((1+cos x)/2) = √((1 − 3/5)/2) = √(1/5) = √5/5.
6. Use sin(x/2) = √((1−cos x)/2) = √((1 + 3/5)/2) = √(4/5) = 2√5/5.
7. tan(x/2) = sin(x/2)/cos(x/2) = (2√5/5)/(√5/5) = 2.
8. The answer key at the back of the book prints 1/2 for tan(x/2); but tan(x/2) = sin(x/2) ÷ cos(x/2) = (2√5/5) ÷ (√5/5) = 2, so the answer is 2 (1/2 is cot(x/2)).

Answersin(x/2) = 2√5/5, cos(x/2) = √5/5, tan(x/2) = 2.

Watch this explained “A worked halving”, 12:00 into Setting the second angle equal to the first gives the double and triple angle rules

Open NCERT p. 72Matches NCERT’s answer

1. Since x is between 180° and 270°, x/2 is between 90° and 135° — that is, x/2 is in quadrant II.
2. In quadrant II, sin(x/2) is positive, cos(x/2) is negative, and tan(x/2) is negative.
3. Use cos(x/2) = −√((1+cos x)/2) = −√((1 − 1/3)/2) = −√(1/3) = −√3/3 (negative, since x/2 is in quadrant II).
4. Use sin(x/2) = √((1−cos x)/2) = √((1 + 1/3)/2) = √(2/3) = √6/3 (positive, as expected).
5. tan(x/2) = sin(x/2)/cos(x/2) = (√6/3)/(−√3/3) = −√2.

Answersin(x/2) = √6/3, cos(x/2) = −√3/3, tan(x/2) = −√2.

Watch this explained “A worked halving”, 12:00 into Setting the second angle equal to the first gives the double and triple angle rules

Open NCERT p. 72Matches NCERT’s answer

1. x is in quadrant II, so cos x is negative. cos²x = 1 − sin²x = 1 − 1/16 = 15/16, so cos x = −√15/4.
2. x is between 90° and 180°, so x/2 is between 45° and 90°. That is quadrant I, so sin(x/2), cos(x/2) and tan(x/2) are all positive.
3. sin²(x/2) = (1 − cos x)/2 = (1 + √15/4)/2 = (4 + √15)/8 = (8 + 2√15)/16.
4. Notice (√5 + √3)² = 5 + 3 + 2√15 = 8 + 2√15. So sin²(x/2) = (√5 + √3)²/16, and taking the positive root, sin(x/2) = (√5 + √3)/4.
5. cos²(x/2) = (1 + cos x)/2 = (1 − √15/4)/2 = (4 − √15)/8 = (8 − 2√15)/16 = (√5 − √3)²/16. Since √5 > √3, cos(x/2) = (√5 − √3)/4.
6. tan(x/2) = sin(x/2)/cos(x/2) = (√5 + √3)/(√5 − √3). Multiply top and bottom by (√5 + √3): (√5 + √3)²/(5 − 3) = (8 + 2√15)/2 = 4 + √15.

Answersin(x/2) = (√5 + √3)/4, cos(x/2) = (√5 − √3)/4, tan(x/2) = 4 + √15.

Watch this explained “A worked halving”, 12:00 into Setting the second angle equal to the first gives the double and triple angle rules