Chapter 7 of 8

All 8 chapters
  1. 1Radians, readunderstood
  2. 2The unit circle and any angle, readunderstood
  3. 3Trig graphs and their changes, readunderstood
  4. 4Solving trig equations, readunderstood
  5. 5Identities, readunderstood
  6. 6Reciprocal and inverse trig functions, readunderstood
  7. 7Compound and double angles, readunderstood
  8. 8Combining sin and cos, readunderstood

M5.3.7 Chapter 7

Compound and double angles

If I know sin A and sin B, how do I find sin(A + B), and why is it not sin A + sin B?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

  • Use the formula for sin(A + B)
  • Use double angle formulas to solve an equation

We often know the trig ratios of two angles, A and B, and need those of A + B or A − B. A compound angle is a sum or difference of angles like these. The ratio of a sum is not the sum of the ratios, so we need new formulas. They all follow from one fact about the unit circle.

Take two points on the unit circle, P at angle A and Q at angle B. Then P = (cos A, sin A) and Q = (cos B, sin B). The angle POQ at the centre is A − B. We find the distance PQ in two ways.

A circle with centre O and axes named x and y. A point named P (cos A, sin A) at angle A, which is 70 degrees, and a point named Q (cos B, sin B) at angle B, which is 20 degrees, are on the circle and are joined by a chord labelled chord.xychordP (cos A, sin A)Q (cos B, sin B)O
OP makes angle A with the x axis and OQ makes angle B. The angle POQ between them is A − B.

First use the cosine rule in triangle POQ. Here OP = OQ = 1 and the angle at O is A − B. So PQ² = 1² + 1² − 2 × 1 × 1 × cos(A − B) = 2 − 2 cos(A − B). Second, use the distance formula. Then PQ² = (cos A − cos B)² + (sin A − sin B)². Expand this and use sin²x + cos²x = 1 twice. We get PQ² = (cos²A + sin²A) + (cos²B + sin²B) − 2 cos A cos B − 2 sin A sin B = 2 − 2(cos A cos B + sin A sin B).

The two results for PQ² are equal. So 2 − 2 cos(A − B) = 2 − 2(cos A cos B + sin A sin B). This gives the first formula: cos(A − B) ≡ cos A cos B + sin A sin B.

The others follow. Replace B by −B. Since cos(−B) = cos B and sin(−B) = −sin B, we get cos(A + B) ≡ cos A cos B − sin A sin B. For the sine, use the fact that the reflection of the point at angle θ in the line y = x is the point at angle 90° − θ. Reflection swaps the coordinates, so cos(90° − θ) = sin θ and sin(90° − θ) = cos θ.

Then sin(A + B) = cos(90° − (A + B)) = cos((90° − A) − B). Use the formula for cos(X − Y) with X = 90° − A and Y = B. This gives cos(90° − A) cos B + sin(90° − A) sin B, which is sin A cos B + cos A sin B. Replacing B by −B gives sin(A − B). Dividing sin(A ± B) by cos(A ± B) gives tan.

A table with the head formula. The rows read sin(A + B), then = sin A cos B + cos A sin B. sin(A − B), then = sin A cos B − cos A sin B. cos(A + B), then = cos A cos B − sin A sin B. cos(A − B), then = cos A cos B + sin A sin B.formulasin(A + B)= sin A cos B + cos A sin Bsin(A − B)= sin A cos B − cos A sin Bcos(A + B)= cos A cos B − sin A sin Bcos(A − B)= cos A cos B + sin A sin B
The compound angle formulas. The sign in sin matches the sign inside the bracket. In cos it is the opposite sign.

For the tangent, divide the top and the bottom of sin(A + B)/cos(A + B) by cos A cos B. This gives tan(A + B) ≡ (tan A + tan B)/(1 − tan A tan B), and tan(A − B) ≡ (tan A − tan B)/(1 + tan A tan B). They hold where every tangent in them is defined and the bottom is not 0.

The formulas are not the obvious guess. Here is why the guess sin(A + B) = sin A + sin B fails. The graph of y = sin(x + π/3) is not the graph of y = sin x raised by the number sin(π/3). The first has height 1 at its peak, and the second has height 1 + √3/2.

A graph with the x axis named x (radians) from 0 to 2π and a y axis from −2 to 2 named y. A solid wave named sin(x + π/3) peaks at 1. A dashed wave named sin x + 0.87 has the same shape as sin x, lifted up by 0.87 so that its peak is above 1.8.0π/2π3π/22π−2−1012x (radians)ysin(x + π/3)sin x + 0.87
sin(x + π/3) (solid) and sin x + sin(π/3) (dashed) are different curves. The second is too high: at its peak it is above 1.

Worked example

Find the exact value of sin 75°.

  1. Write 75° as 45° + 30°, two angles whose exact values we know.
  2. Use sin(A + B) = sin A cos B + cos A sin B with A = 45° and B = 30°.
  3. sin 75° = sin 45° cos 30° + cos 45° sin 30°.
  4. Put in the exact values: (√2/2)(√3/2) + (√2/2)(1/2).
  5. Multiply: √6/4 + √2/4.
  6. Add the fractions: sin 75° = (√6 + √2)/4.
  7. Check with a calculator: (2.4495 + 1.4142) ÷ 4 = 0.9659, and sin 75° = 0.9659.

Answer: sin 75° = (√6 + √2)/4

Often you know sin A and cos B, not both of each. Use sin²x + cos²x = 1 to find the missing ratios. And you must choose the correct sign for each, from the quadrant the problem names.

Worked example

A is acute with sin A = 3/5. B is obtuse with cos B = −12/13. Find sin(A + B).

  1. We need sin A, cos A, sin B and cos B. We know sin A = 3/5 and cos B = −12/13.
  2. A is acute, so cos A is positive: cos²A = 1 − 9/25 = 16/25, so cos A = 4/5.
  3. B is obtuse, so sin B is positive: sin²B = 1 − 144/169 = 25/169, so sin B = 5/13.
  4. sin(A + B) = sin A cos B + cos A sin B = (3/5)(−12/13) + (4/5)(5/13).
  5. The top of the first term is 3 × (−12) = −36. The top of the second term is 4 × 5 = 20.
  6. Both terms have the bottom 5 × 13 = 65. So sin(A + B) = (−36 + 20)/65 = −16/65.

Answer: sin(A + B) = −16/65

Two right angled triangles side by side. The first has sides 3, 4 and 5, with the angle A marked opposite the side 3. The second has sides 5, 12 and 13, with the angle X marked opposite the side 5.345A51213X
The triangles give sin A = 3/5 and cos A = 4/5 for the acute angle A, and cos X = 12/13 and sin X = 5/13. The obtuse angle B = 180° − X has cos B = −12/13 and sin B = 5/13: the same sizes, with the signs of the second quadrant.

Worked example

A and B are acute angles with tan A = 1/2 and tan B = 1/3. Show that A + B = π/4.

  1. Use tan(A + B) = (tan A + tan B)/(1 − tan A tan B).
  2. The top is 1/2 + 1/3 = 5/6.
  3. The bottom is 1 − (1/2)(1/3) = 1 − 1/6 = 5/6.
  4. So tan(A + B) = (5/6) ÷ (5/6) = 1.
  5. A and B are acute, so A + B is between 0 and π. The only angle in that range with tangent 1 is π/4.
  6. So A + B = π/4.

Answer: tan(A + B) = 1, so A + B = π/4.

Now put B = A in the compound formulas. This gives the double angle formulas. sin 2A = sin A cos A + cos A sin A, so sin 2A ≡ 2 sin A cos A. cos 2A = cos²A − sin²A. Using sin²A + cos²A = 1, this also has two other forms: cos 2A ≡ 2cos²A − 1 and cos 2A ≡ 1 − 2 sin²A. And tan 2A ≡ 2 tan A/(1 − tan²A).

A table with the head formula. The rows read sin 2A = 2 sin A cos A. cos 2A = cos²A − sin²A. cos 2A = 2cos²A − 1. cos 2A = 1 − 2 sin²A. tan 2A = 2tan A/(1 − tan²A).formulasin 2A = 2 sin A cos Acos 2A = cos²A − sin²Acos 2A = 2cos²A − 1cos 2A = 1 − 2 sin²Atan 2A = 2tan A/(1 − tan²A)
The double angle formulas. Use cos²A − sin²A when both appear, 2cos²A − 1 when only cos A appears, and 1 − 2 sin²A when only sin A appears.

Choose the form of cos 2A that leaves one trig function. If the equation has cos x and cos 2x, use cos 2x = 2cos²x − 1. If it has sin x and cos 2x, use 1 − 2 sin²x. Then solve the quadratic, exactly as in the last two chapters. The double angle formulas also make sin²A = (1 − cos 2A)/2 and cos²A = (1 + cos 2A)/2.

Worked example

Solve 3 cos 2x + cos x + 2 = 0 for 0 ≤ x ≤ 2π. Give the answers to 3 decimal places.

  1. The equation has cos 2x and cos x. Use cos 2x = 2cos²x − 1, which has only cos x.
  2. We get 3(2cos²x − 1) + cos x + 2 = 0, which is 6cos²x − 3 + cos x + 2 = 0.
  3. This is a quadratic in cos x: 6cos²x + cos x − 1 = 0, because −3 + 2 = −1.
  4. Factorise: (3 cos x − 1)(2 cos x + 1) = 0. Check: 6cos²x + 3 cos x − 2 cos x − 1 = 6cos²x + cos x − 1.
  5. So cos x = 1/3 or cos x = −1/2.
  6. cos x = 1/3: the calculator gives 1.2310. cos is positive in the first and fourth quadrants, so x = 1.231 and x = 2π − 1.2310 = 5.052.
  7. cos x = −1/2: the reference angle is π/3. So x = π − π/3 = 2π/3 = 2.094 and x = π + π/3 = 4π/3 = 4.189.
  8. In order: x = 1.231, 2.094, 4.189 or 5.052.

Answer: x = 1.231, 2.094, 4.189 or 5.052

A graph with the x axis named x (radians) from 0 to 2π and a y axis from −4 to 6 named y. A curve named f(x), which is 3 cos 2x + cos x + 2, starts at 6, falls below the x axis and crosses it at four dots, between π/4 and π/2, at 2π/3, at 4π/3 and between 3π/2 and 7π/4.0π/2π3π/22π−4−20246x (radians)yf(x)
The curve y = 3 cos 2x + cos x + 2 (called f(x) here) crosses the x axis at the four solutions.

Worked example

Prove that (sin A + cos A)² ≡ 1 + sin 2A.

  1. Start with the left side and expand: sin²A + 2 sin A cos A + cos²A.
  2. Use sin²A + cos²A ≡ 1 for the squares: 1 + 2 sin A cos A.
  3. Use sin 2A ≡ 2 sin A cos A for the last term: 1 + sin 2A.
  4. This is the right side. So the identity is proved.

Answer: The left side becomes the right side, so the identity is proved.

Adding two compound formulas gives the factor formulas. Write C = P + Q and D = P − Q, so P = (C + D)/2 and Q = (C − D)/2. Then sin(P + Q) + sin(P − Q) = 2 sin P cos Q. In the new letters: sin C + sin D = 2 sin((C + D)/2) cos((C − D)/2). The same steps give three more. Next, sin C − sin D = 2 cos((C + D)/2) sin((C − D)/2). Then cos C + cos D = 2 cos((C + D)/2) cos((C − D)/2). Last, cos C − cos D = −2 sin((C + D)/2) sin((C − D)/2).

Worked example

Show that sin 75° + sin 15° = √6/2.

  1. Use sin C + sin D = 2 sin((C + D)/2) cos((C − D)/2) with C = 75° and D = 15°.
  2. (C + D)/2 = 45° and (C − D)/2 = 30°.
  3. So sin 75° + sin 15° = 2 sin 45° cos 30° = 2 × (√2/2) × (√3/2).
  4. This is 2√6/4 = √6/2.
  5. Check: sin 75° = 0.9659 and sin 15° = 0.2588. The sum is 1.2247, and √6/2 = 1.2247.

Answer: sin 75° + sin 15° = √6/2

See it three ways

With things

Two turns of a wheel, first by A and then by B, are the same as one turn by A + B. The height after both turns depends on the position after the first turn and the extra turn, so it needs both sines and both cosines.

As a picture

A circle with centre O and axes named x and y. A point named P (cos A, sin A) at angle A, which is 70 degrees, and a point named Q (cos B, sin B) at angle B, which is 20 degrees, are on the circle and are joined by a chord labelled chord.xychordP (cos A, sin A)Q (cos B, sin B)O

In words

The distance between two points on the unit circle depends on the difference of their angles. Working it out two ways gives the cosine of a difference.

In symbols

cos(A − B) = cos A cos B + sin A sin B. Then sin(A + B) = sin A cos B + cos A sin B. Put B = A: sin 2A = 2 sin A cos A.

Try it

Test the formula with a calculator

You need: a calculator, a pencil.

  1. Choose A = 50° and B = 20°. Work out sin 50° and cos 50°, sin 20° and cos 20°.
  2. Work out sin A cos B + cos A sin B using your numbers.
  3. Work out sin 70° directly. Compare.
  4. Work out sin A + sin B and compare it with sin 70°.

What you will see: The formula gives 0.9397, the same as sin 70°. The sum sin 50° + sin 20° = 1.108 is not even a possible sine.

If you cannot try it, this is what happens.
A table with the heads quantity and value. The rows read sin 50° = 0.766, cos 20° = 0.940, cos 50° = 0.643, sin 20° = 0.342, sin 70° = 0.940.quantityvaluesin 50°0.766cos 20°0.940cos 50°0.643sin 20°0.342sin 70°0.940

A wrong idea many people have

sin(A + B) = sin A + sin B, and sin 2A = 2 sin A.

Why it feels right: In algebra, 2(x + y) = 2x + 2y, so a function seems to pass through a bracket in the same way.

The test that breaks it: Put A = B = 45°. Then sin 90° = 1, but sin 45° + sin 45° = 1.414, which is more than 1. And 2 sin 45° = 1.414, while sin 90° = 1.

What is true: sin is a function, not a number, so it does not distribute over a sum. The right formulas are sin(A + B) = sin A cos B + cos A sin B and sin 2A = 2 sin A cos A.

A wrong idea many people have

cos(A + B) = cos A cos B + sin A sin B.

Why it feels right: It looks like the formula for sin(A + B), and a plus sign inside the bracket seems to give a plus sign.

The test that breaks it: Put A = B = 45°. Then cos 90° = 0. But cos 45° cos 45° + sin 45° sin 45° = 1/2 + 1/2 = 1. The right side is wrong.

What is true: In the cosine formulas the sign is the opposite of the sign in the bracket: cos(A + B) = cos A cos B − sin A sin B. Test every formula with A = B = 45° if you are unsure.

A wrong idea many people have

cos 2A = 2 cos A, or cos 2A = cos²A.

Why it feels right: A double angle seems to double the result, and squares are in the formula.

The test that breaks it: Put A = 60°. cos 120° = −1/2. But 2 cos 60° = 1, and cos²60° = 1/4. Neither matches.

What is true: cos 2A has three correct forms: cos²A − sin²A, 2cos²A − 1 and 1 − 2 sin²A. At A = 60°: 1/4 − 3/4 = −1/2, 2/4 − 1 = −1/2, and 1 − 6/4 = −1/2.

Key words

compound angle
a sum or difference of two angles, such as A + B or A − B.
double angle
the angle 2A, found by putting B = A in a compound angle formula.

Back to your guess

The answer is C: sin(A + B) = sin A cos B + cos A sin B.

The formula is sin(A + B) = sin A cos B + cos A sin B. Test option A: A = B = 90° gives sin 180° = 0, but sin 90° + sin 90° = 2. Option B gives 1 × 1 = 1. Only option C gives 1 × 0 + 0 × 1 = 0.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. Which expression is equal to sin(A − B)?

    Tap your answer.

    1. Not this one. That is sin(A + B). For a difference, the sign in the middle is a minus. Go to M5.3.7. Try another.

    2. Not this one. That is cos(A − B), the cosine formula. Go to M5.3.7. Try another.

    3. Not this one. sin does not distribute over a difference. The terms are products of a sine and a cosine. Go to M5.3.7. Try another.

    4. Yes, that is right. sin(A − B) = sin A cos B − cos A sin B. The sign in the middle of the sine formula matches the sign inside the bracket.

    See the answer

    sin A cos B − cos A sin B

    sin(A − B) = sin A cos B − cos A sin B. The sign in the middle of the sine formula matches the sign inside the bracket.

    • If you chose A: That is sin(A + B). For a difference, the sign in the middle is a minus. Go to M5.3.7.
    • If you chose B: That is cos(A − B), the cosine formula. Go to M5.3.7.
    • If you chose C: sin does not distribute over a difference. The terms are products of a sine and a cosine. Go to M5.3.7.

  2. Which expression is equal to sin 2A?

    Tap your answer.

    1. Yes, that is right. Put B = A in sin(A + B) = sin A cos B + cos A sin B. Both terms are sin A cos A, so sin 2A = 2 sin A cos A.

    2. Not this one. That is half of sin 2A. Put B = A: sin A cos A + cos A sin A = 2 sin A cos A. Go to M5.3.7. Try another.

    3. Not this one. At A = 45°: sin 90° = 1, but 2 sin 45° = 1.414. A function does not double like that. Go to M5.3.7. Try another.

    4. Not this one. At A = 45°: sin²45° = 1/2, but sin 90° = 1. Go to M5.3.7. Try another.

    See the answer

    2 sin A cos A

    Put B = A in sin(A + B) = sin A cos B + cos A sin B. Both terms are sin A cos A, so sin 2A = 2 sin A cos A.

    • If you chose B: That is half of sin 2A. Put B = A: sin A cos A + cos A sin A = 2 sin A cos A. Go to M5.3.7.
    • If you chose C: At A = 45°: sin 90° = 1, but 2 sin 45° = 1.414. A function does not double like that. Go to M5.3.7.
    • If you chose D: At A = 45°: sin²45° = 1/2, but sin 90° = 1. Go to M5.3.7.

  3. One of these is NOT a correct form of cos 2A. Which one?

    Tap your answer.

    1. Not this one. This is correct. It is the first form, from cos(A + A). Go to M5.3.7. Try another.

    2. Yes, that is right. The correct forms are cos²A − sin²A, 2cos²A − 1 and 1 − 2 sin²A. The form 1 − 2cos²A has the signs the wrong way round: at A = 0 it gives −1, but cos 0 = 1.

    3. Not this one. This is correct. Replace sin²A by 1 − cos²A in the first form. Go to M5.3.7. Try another.

    4. Not this one. This is correct. Replace cos²A by 1 − sin²A in the first form. Go to M5.3.7. Try another.

    See the answer

    1 − 2cos²A

    The correct forms are cos²A − sin²A, 2cos²A − 1 and 1 − 2 sin²A. The form 1 − 2cos²A has the signs the wrong way round: at A = 0 it gives −1, but cos 0 = 1.

    • If you chose A: This is correct. It is the first form, from cos(A + A). Go to M5.3.7.
    • If you chose C: This is correct. Replace sin²A by 1 − cos²A in the first form. Go to M5.3.7.
    • If you chose D: This is correct. Replace cos²A by 1 − sin²A in the first form. Go to M5.3.7.

  4. Ravi writes sin 75° = sin 45° + sin 30°. What is wrong?

    Tap your answer.

    1. Not this one. The angles add, but the sines do not. sin 45° + sin 30° = 1.207, which is more than 1, so it cannot be a sine. Go to M5.3.7. Try another.

    2. Not this one. sin is not a multiplication. sin 75° is the sine of the angle 75°. Go to M5.3.7. Try another.

    3. Yes, that is right. sin 45° + sin 30° = 0.7071 + 0.5 = 1.207, which is bigger than 1, so it cannot be the sine of anything. The correct value is sin 75° = sin 45° cos 30° + cos 45° sin 30° = 0.9659.

    See the answer

    sin(A + B) is not sin A + sin B. The correct formula has products of sines and cosines.

    sin 45° + sin 30° = 0.7071 + 0.5 = 1.207, which is bigger than 1, so it cannot be the sine of anything. The correct value is sin 75° = sin 45° cos 30° + cos 45° sin 30° = 0.9659.

    • If you chose A: The angles add, but the sines do not. sin 45° + sin 30° = 1.207, which is more than 1, so it cannot be a sine. Go to M5.3.7.
    • If you chose B: sin is not a multiplication. sin 75° is the sine of the angle 75°. Go to M5.3.7.

  5. A is acute and tan A = 3/4. What is tan 2A?

    Tap your answer.

    1. Not this one. You doubled tan A. But tan 2A = 2 tan A/(1 − tan²A), and the bottom of this fraction is missing. Go to M5.3.7. Try another.

    2. Not this one. That is sin 2A, since sin A = 3/5 and cos A = 4/5. The question asks for tan 2A. Go to M5.3.7. Try another.

    3. Not this one. That is the reciprocal, cot 2A. tan 2A is the top over the bottom the other way up. Go to M5.3.7. Try another.

    4. Yes, that is right. tan 2A = 2 tan A/(1 − tan²A) = 2 × (3/4)/(1 − 9/16) = (3/2)/(7/16) = 24/7. Check with sin 2A = 24/25 and cos 2A = 7/25: 24/25 ÷ 7/25 = 24/7.

    See the answer

    24/7

    tan 2A = 2 tan A/(1 − tan²A) = 2 × (3/4)/(1 − 9/16) = (3/2)/(7/16) = 24/7. Check with sin 2A = 24/25 and cos 2A = 7/25: 24/25 ÷ 7/25 = 24/7.

    • If you chose A: You doubled tan A. But tan 2A = 2 tan A/(1 − tan²A), and the bottom of this fraction is missing. Go to M5.3.7.
    • If you chose B: That is sin 2A, since sin A = 3/5 and cos A = 4/5. The question asks for tan 2A. Go to M5.3.7.
    • If you chose C: That is the reciprocal, cot 2A. tan 2A is the top over the bottom the other way up. Go to M5.3.7.

  6. From an earlier chapter

    Which of these numbers is NOT a value that y = sec x can take?

    Tap your answer.

    1. Yes, that is right. sec x = 1/cos x, and cos x is between −1 and 1. So sec x is at least 1 or at most −1. A value of 0.5 would need cos x = 2, which is impossible.

    2. Not this one. sec x = −3 when cos x = −1/3. This is a value of sec x. Go to M5.3.6. Try another.

    3. Not this one. sec x = −1 when cos x = −1, at x = π. This is a value of sec x. Go to M5.3.6. Try another.

    4. Not this one. sec x = 2 when cos x = 1/2. This is a value of sec x. Go to M5.3.6. Try another.

    See the answer

    0.5

    sec x = 1/cos x, and cos x is between −1 and 1. So sec x is at least 1 or at most −1. A value of 0.5 would need cos x = 2, which is impossible.

    • If you chose B: sec x = −3 when cos x = −1/3. This is a value of sec x. Go to M5.3.6.
    • If you chose C: sec x = −1 when cos x = −1, at x = π. This is a value of sec x. Go to M5.3.6.
    • If you chose D: sec x = 2 when cos x = 1/2. This is a value of sec x. Go to M5.3.6.

  7. Prove that sin 3A ≡ 3 sin A − 4 sin³A.

    See the answer

    The left side becomes the right side, so the identity is proved.

    Write 3A as 2A + A. Then sin 3A = sin 2A cos A + cos 2A sin A. Use sin 2A = 2 sin A cos A and cos 2A = 1 − 2 sin²A. This gives 2 sin A cos²A + (1 − 2 sin²A) sin A. Replace cos²A by 1 − sin²A: 2 sin A (1 − sin²A) + sin A − 2 sin³A. Expand: 2 sin A − 2 sin³A + sin A − 2 sin³A. Collect: 3 sin A − 4 sin³A, the right side.

  8. Use tan 15° = tan(45° − 30°) to find the exact value of tan 15°.

    See the answer

    2 − √3

    tan(A − B) = (tan A − tan B)/(1 + tan A tan B). With tan 45° = 1 and tan 30° = 1/√3: (1 − 1/√3)/(1 + 1/√3) = (√3 − 1)/(√3 + 1). Multiply top and bottom by (√3 − 1): (√3 − 1)²/2 = (4 − 2√3)/2 = 2 − √3. Check: 2 − 1.732 = 0.268 = tan 15°.

  9. Solve cos 2x + sin x = 0 for 0° ≤ x ≤ 360°.

    See the answer

    x = 90°, 210° or 330°

    Use cos 2x = 1 − 2 sin²x, because the equation has sin x. Then 1 − 2 sin²x + sin x = 0, so 2 sin²x − sin x − 1 = 0. Factorise: (2 sin x + 1)(sin x − 1) = 0. sin x = 1 gives x = 90°. sin x = −1/2 gives x = 180° + 30° = 210° and x = 360° − 30° = 330°.

  10. A is acute with sin A = 5/13. B is obtuse with cos B = −3/5. Find sin(A + B) as a fraction.

    See the answer

    33/65

    A is acute, so cos A = 12/13. B is obtuse, so sin B = 4/5. Then sin(A + B) = sin A cos B + cos A sin B = (5/13)(−3/5) + (12/13)(4/5) = −15/65 + 48/65 = 33/65.

  11. Show that cos 105° = (√2 − √6)/4.

    See the answer

    cos 105° = (√2 − √6)/4

    Write 105° as 60° + 45° and use cos(A + B) = cos A cos B − sin A sin B. Then cos 105° = cos 60° cos 45° − sin 60° sin 45°. This is (1/2)(√2/2) − (√3/2)(√2/2) = √2/4 − √6/4 = (√2 − √6)/4. Check: (1.414 − 2.449)/4 = −0.2588, which is cos 105°.

  12. Use sin C + sin D = 2 sin((C + D)/2) cos((C − D)/2) to find the exact value of sin 105° + sin 15°.

    See the answer

    sin 105° + sin 15° = √6/2

    With C = 105° and D = 15°: (C + D)/2 = 60° and (C − D)/2 = 45°. So sin 105° + sin 15° = 2 sin 60° cos 45° = 2 × (√3/2) × (√2/2) = 2√6/4 = √6/2. Check: sin 105° = 0.9659 and sin 15° = 0.2588, and the sum is 1.2247, which is √6/2.

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sin(A ± B) = sin A cos B ± cos A sin B. cos(A + B) = cos A cos B − sin A sin B. For A − B, use plus. sin 2A = 2 sin A cos A. cos 2A = 2cos²A − 1 = 1 − 2 sin²A.

Draw: Two points on a unit circle at angles A and B joined by a chord.

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