The compound angle formulae

The compound angle formulae can be used to find the sums and differences of sine, cosine, and tangent of compound angles.

From triangle OSM:

∠OMS + A + 90° = 180°
∠OMS = 90° - A

For triangle PQM:

90° + A + ∠PMQ = 180°
∠PQM = 90° - A

Thus, ∠OMS and ∠PQM are opposite angles. Let OP = r

sin B = PQ/OP
PQ = OP sin B = r sin B

cos B = OQ/OP
OQ = OP cos B = r cos B

sin A = QT/OQ
QT = OQ sin A = r cos B sin A

cos A = OT/OQ
OT = OQ cos A = r cos B cos A

Using triangle PQR:

sin A = RQ/PQ
RQ = PQ sin A = r sin B sin A

cos A = PR/PQ
PR = PQ cos A = r sin B cos A

Now,

sin(A + B) = PS/OP = (PR + RS)/r = (PR + QT)/r
= (r sin B cos A + r cos B sin A) / r
= sin B cos A + cos B sin A

Therefore,

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

Replacing B with -B:

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

From Figure 8.28:

cos(A + B) = OS/OP = (OT - ST)/r = (OT - RQ)/r
= (r cos A cos B - r sin A sin B) / r
= cos A cos B - sin A sin B

Replacing B with -B:

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

From the definition tan θ = sin θ / cos θ:

tan(A + B) = sin(A + B) / cos(A + B)
= (sin A cos B + cos A sin B) / (cos A cos B - sin A sin B)

Dividing the numerator and denominator by cos A cos B:

tan(A + B) = (tan A + tan B) / (1 - tan A tan B)

Replacing B with -B:

tan(A - B) = (tan A - tan B) / (1 + tan A tan B)

Summary of compound angle formulae

1. sin(A + B) = sin A cos B + cos A sin B
2. sin(A - B) = sin A cos B - cos A sin B
3. cos(A + B) = cos A cos B - sin A sin B
4. cos(A - B) = cos A cos B + sin A sin B
5. tan(A + B) = (tan A + tan B) / (1 - tan A tan B)
6. tan(A - B) = (tan A - tan B) / (1 + tan A tan B)

Example 1

Find, in surd form, the values of: a) sin(30° + 45°) b) cos(30° + 45°) c) tan(30° + 45°)

Solution:

a) sin(30° + 45°) = sin 30° cos 45° + cos 30° sin 45°
= (1/2)(√2/2) + (√3/2)(√2/2)
= (√2 + √6)/4

b) cos(30° + 45°) = cos 30° cos 45° - sin 30° sin 45°
= (√3/2)(√2/2) - (1/2)(√2/2)
= (√6 - √2)/4

c) tan(30° + 45°) = (tan 30° + tan 45°) / (1 - tan 30° tan 45°)
= (√3/3 + 1) / (1 - (√3/3)(1))
= (√3 + 3) / (3 - √3)
= (√3 + 3)(3 + √3) / (9 - 3)
= (3√3 + 9 + 3 + 3√3) / 6
= (12 + 6√3) / 6
= 2 + √3

Example 2

If tan 105° = -2 - √3, without using mathematical tables or calculators, find the value of tan 165° in the form a + b√3, where a and b are integers.

Solution:

tan 165° = tan(105° + 60°)
= (tan 105° + tan 60°) / (1 - tan 105° tan 60°)
= (-2 - √3 + √3) / (1 - (-2 - √3)(√3))
= -2 / (1 + 2√3 + 3)
= -2 / (4 + 2√3)
= -1 / (2 + √3)
= -1(2 - √3) / (4 - 3)
= -2 + √3