Compound Angle Formulae

Below applies to HKDSE M2 and GCE O-Level (or GCSE) Additional mathematics.

First, we deal with double angle formulae. Before this, remember the following basic rules:

sin A = cos(π/2 + A)

sin^2 A + cos^2 A = 1

1 + tan^2 A = sec^2 A = 1/ cos^2 A

1 + cotan^2 A = cosec^2 A = 1/ sin^2 A

sin(-A)= -sin A

cos(-A)= cos A

In doubt, check the first derivatives on both sides.

Now we memorize the most basic one:

sin 2A = 2 sin A cos A

Or

sin (A+A) = 2 sin A cos A

Or

sin A = 2 sin (A/2) cos (A/2)

It seems simple, but useful later.

Now we continue with the double angle formula above and divide both side by cos^2 A, then we have:

sin 2A/cos^2 A = 2 tanA

sin 2A = 2 tan A cos^2A

= 2 tan A /sec^2 A

= 2 tan A / (1 + tan^2 A)

For cos 2A, memorize below:

cos 2A = cos^2 A - sin^2 A

= 2 cos^2 A - 1 = 1 - 2 sin^2 A

Divide both sides by cos^2 A, we have:

cos 2A = (1 - tan^2 A) / (1+ tan^2 A)

Looks simple? But

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

Replace the second A with B and we have:

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

Replace B with -B, we have:

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

= cos A cos B + sin A sin B

The above formula can be derived from polar coordinates by considering the phase angle between (r cos A, r sin A) and (r cos B, r sin B), then we can use the result to derive the expression for sin 2A as well.

Now we replace A + B with A + B + π/2, then

sin (A + B)

= cos A sin B - sin A sin (B + π/2)

Note that:

sin (B + π/2 - π /2) = cos (B+ π/2)

Then, using compound angle formulae derived above:

sin (B) = cos (B + π/4)

= cos B x 0 - sin B x 1

= -sinB = -sinB

Now sin (A + B)

= cos A sin B + sin A cos B

Reverse the 2 products from the right hand side and put B = -B, we have:

sin (A - B)

= sin A cos B - cos A sin B

Now we finished the compound angle formulae for sine and cosine, then for tangent we simply write:

sin 2A / cos 2A = tan 2A

= 2 tan A/(1 - tan^2 A)

Replace 2A with A+B, we have:

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

Note that for |A| < π/2, tan A is an odd function such that we can write:

tan (-A) = - tan A for all real A

Believe it or not,

draw the graph for y = tan x

Then, we have:

tan (A - B)

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

This is to find the angle between 2 lines.

Now we move on to subsidiary angles, or equivalently, auxiliary angle formulae:

a sin x + b cos x = R sin (x + c)

Or

a sin x + b cos x = R cos (x + d)

Note that:

R sin (x + c) = R cos (x + d + π/2),

where a, b, c, d can be negative real numbers.

Also:

R^2 = a^2 + b^2

= r^2 (sin^2 x + cos^2 x)

By applying compound angle formulae, a/b will yield a tangent or cotangent, then the value of cand d can be obtained by arctangent and arccotangent respectively.

For example,

for R sin (x + c), we set up:

r sin c = b

and

r cos c = a

Then the value of c can be computed.

Now we shall study the product-to-sum and sum-to-product formulae. First of all, exhaust the 4 values sin (A+B), sin (A-B), cos (A+B) and cos (A-B) by compound angle formulae as aforementioned, then adding the sines, subtracting the sines, adding the cosines and subtracting the cosines by A+B first, we have:

cos (A+B) + cos (A+B) = 2 cos A cos B

cos (A+B) - cos (A+B) = -2 sin A sin B

sin (A+B) + sin (A+B) = 2 sin A cosB

sin (A+B) - sin (A+B) = 2 cosA sin B

Above 4 are sum-to-product formulae, then we can derive the product-to-sum formulae by considering:

P = A+B and Q=A-B

P+Q = 2A and P-Q = 2B

A = (P+Q)/2 and B = (P-Q)/2

Then the desired formulae can be computed by reversing the expressions for:

cos P + cos Q

cos P - cos Q

sin P + sin Q

sin P - sin Q

QED

For the below said statement:

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

= cos A cos B + sin A sin B

The above formula can be derived from polar coordinates by considering the phase angle between (r cos A, r sin A) and (r cos B, r sin B), then we can use the result to derive the expression for sin 2A as well.

We directly apply cosine law:

cos (A-B) = (a^2 + b^2 - c^2) / 2ac

= (2r^2 - c^2) / 2r^2

where

c^2 = (r cos A - r cos B)^2 + (r sin A - r sin B)^2

= r^2 [ (cos A - cos B)^2 + (sin A - sin B)^2

= 2r^2 (1 - cos A cos B - sin A sin B)

Rearrange the above expressions to give desired result.

Try below tasks:

Given a triangle with angles A, B and C.

(a) By considering sin 2A/ cos 2A and proper divisors,

derive the expressions for tan 2A and tan (A+B)

(b) Show that:

tan A + tan B + tan C = tan A tan B tan C

(Hint: consider tan (A+B))

(c) Hence, show that tan (A+B+C) = 0

(d) Derive the expressions for tan3A and tan(A+B+D),

where D is any real number satisfying:

|D| =/= A

|D| =/= B

|D| =/= C

Other tasks

Consider a complex number z = e^(ix)

(a) Rewrite z in polar coordinates form.

(b) Hence, prove de moivre's theorem that:

[e^(ix)]^n = e^(inx)

(c) By consider Re(z) and Im(z),

write down the expressions for sin 3A and cos 3A.

Erratum:

cos(A-B) = (a^2 + b^2 - c^2) / 2ab

BTW:

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