Compound Angle Formula

Compound angle formulae

The compound angle formula can be used to find the sine, cosine, and tangent of the sum and difference of angles. The basic operations on sums and differences of trigonometric functions can be computed using the concept of compound angles.

Consider the triangle.

We consider a right-angled triangle PRT. Within this triangle, there's a point Q such that angles PQR and QRT are also right angles. Angle RPQ is labeled A, and angle QRT is labeled B. This makes angle PRT (the larger angle) equal to A + B.

Angles A and B are acute such that $A + B < 90^\circ$.

Consider the right-angled triangle PRT:

$RT = RS + ST$

$\sin(A + B) = \frac{RT}{PR} = \frac{RS + ST}{PR}$

Since $ST = QU$, we can rewrite this as:

$\sin(A + B) = \frac{RS + QU}{PR}$

We can express $RS$ and $QU$ in terms of other trigonometric ratios within the smaller triangles:

$RS = QR \cos B$

$QU = PQ \sin B$

Substituting these into the equation for $\sin(A + B)$:

$\sin(A + B) = \frac{QR \cos B + PQ \sin B}{PR}$

We can rewrite this as:

$\sin(A + B) = \frac{QR}{PR} \cos B + \frac{PQ}{PR} \sin B$

Now, observe the following from the diagram:

$\frac{QR}{PR} = \sin A$

$\frac{PQ}{PR} = \cos A$

Substituting these values back into the equation:

$\sin(A + B) = \sin A \cos B + \cos A \sin B$ (7.4)

Therefore, the compound angle formula for finding the sine of the sum of two angles is given by:

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

Suppose B is replaced by –B. Then, substituting it into equation (7.4) (which is not provided in the text but assumed to be sin(A+B) = sinAcosB + cosAsinB), gives:

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

Note that, $\cos(-x) = \cos x$, $\sin(-x) = -\sin x$, and $\tan(-x) = -\tan x$ because cos x is an even function while sin x and tan x are odd functions.

Thus, $\sin(A - B) = \sin A \cos B - \cos A \sin B$.

Therefore, the compound angle formula for finding the sine of the difference of two angles is given by:

$\sin(A - B) = \sin A \cos B - \cos A \sin B$.

From Figure 7.3 (again, missing but the logic is presented), considering the right-angled triangle PRT:

$PT = PU + TU$

$\cos(A+B) = \frac{PT}{PR}$

$PU = PQ \cos A$

$SQ = QR \sin B$

$TU = SQ$

$PT = PU + TU$

$PT = PQ \cos A - QR \sin B$

$\frac{PT}{PR} = \frac{PQ}{PR} \cos A - \frac{QR}{PR} \sin B$

But $\frac{PQ}{PR} = \cos B$, $\frac{QR}{PR} = \sin A$.

Therefore, the compound angle formula for finding the cosine of the sum of two angles is given by:

$\cos(A+B) = \cos A \cos B - \sin A \sin B$ (7.5)

Suppose B is replaced by –B then substituting into equation (7.5), gives:

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

Thus, $\cos(A - B) = \cos A \cos B + \sin A \sin B$.

Therefore, the compound angle formula for finding the cosine of the difference of two angles is given by:

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

The derivation of the compound angle formula for $\tan(A + B)$ can be done as follows:

From trigonometric identities $\tan(A + B) = \frac{\sin(A + B)}{\cos(A + B)}$. Substituting equations (7.4) and (7.5) gives:

$\tan(A + B) = \frac{\sin A \cos B + \cos A \sin B}{\cos A \cos B - \sin A \sin B}$

Divide by $\cos A \cos B$ both the numerator and denominator to get:

$\tan(A + B) = \frac{\frac{\sin A \cos B}{\cos A \cos B} + \frac{\cos A \sin B}{\cos A \cos B}}{\frac{\cos A \cos B}{\cos A \cos B} - \frac{\sin A \sin B}{\cos A \cos B}}$

$\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ (7.6)

Similarly, if B is replaced by –B then substituting it into equation (7.6) results to:

$\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$

In summary, the compound angle formulae are written as:

1. $\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$
2. $\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$
3. $\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$

Example 11

Prove that $\cos(x + y)\cos(x - y) = \cos^2 x - \sin^2 y$.

Solution:

Consider the left-hand side:

$\cos(x + y)\cos(x - y) = (\cos x \cos y - \sin x \sin y)(\cos x \cos y + \sin x \sin y)$

$= \cos^2 x \cos^2 y - \sin^2 x \sin^2 y$

$= \cos^2 x (1 - \sin^2 y) - (1 - \cos^2 x) \sin^2 y$

$= \cos^2 x - \cos^2 x \sin^2 y - \sin^2 y + \cos^2 x \sin^2 y$

$= \cos^2 x - \sin^2 y$

Therefore, $\cos(x + y)\cos(x - y) = \cos^2 x - \sin^2 y$.

Example 12

Simplify $\frac{\tan(A + B) - \tan A}{1 + \tan(A + B) \tan A}$.

Solution:

Using the compound angle formula:

$\frac{\tan(A + B) - \tan A}{1 + \tan(A + B) \tan A} = \tan((A + B) - A)$

$= \tan B$

Therefore, $\frac{\tan(A + B) - \tan A}{1 + \tan(A + B) \tan A} = \tan B$.

Example 13

Simplify $(\sin x \cos y - \cos x \sin y)^2 + (\cos x \cos y + \sin x \sin y)^2$.

Solution:

$(\sin x \cos y - \cos x \sin y)^2 + (\cos x \cos y + \sin x \sin y)^2$

$= \sin^2(x - y) + \cos^2(x - y)$

$= 1$

Example 14

If $A + B + C = 180^\circ$, show that $\tan A + \tan B + \tan C = \tan A \tan B \tan C$.

Solution:

Given $A + B + C = 180^\circ \Rightarrow A + B = 180^\circ - C$.

Applying tangent on both sides:

$\tan(A + B) = \tan(180^\circ - C)$

$\frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{\tan 180^\circ - \tan C}{1 + \tan 180^\circ \tan C}$

Since $\tan 180^\circ = 0$,

$\frac{\tan A + \tan B}{1 - \tan A \tan B} = -\tan C$

$\tan A + \tan B = -\tan C (1 - \tan A \tan B)$

$\tan A + \tan B = -\tan C + \tan A \tan B \tan C$

$\tan A + \tan B + \tan C = \tan A \tan B \tan C$

Hence proved.

Double angle formulae

Double angle formulae are trigonometric identities which express trigonometric functions of $2\theta$ in terms of trigonometric functions of $\theta$. They are special cases of compound angle formulae.

The compound angle formulae under addition of two angles can be used to deduce the double angle formula.

Considering the compound angle formula for sine: $\sin(A+B) = \sin A \cos B + \cos A \sin B$. Replacing B by A, we obtain the double angle formula for sine:

$\sin(A + A) = \sin A \cos A + \cos A \sin A$

$\sin 2A = 2\sin A \cos A$

From the compound angle formula for cosine: $\cos(A + B) = \cos A \cos B - \sin A \sin B$. Replacing B by A gives the double angle formula for cosine:

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

$\cos 2A = \cos^2 A - \sin^2 A$

Using the Pythagorean identity $\sin^2 A = 1 - \cos^2 A$, we can rewrite this as:

$\cos 2A = \cos^2 A - (1 - \cos^2 A) = 2\cos^2 A - 1$

Similarly, using $\cos^2 A = 1 - \sin^2 A$, we get:

$\cos 2A = (1 - \sin^2 A) - \sin^2 A = 1 - 2\sin^2 A$

Thus, we have three forms for the double angle formula for cosine:

$\cos 2A = 2\cos^2 A - 1$

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

$\cos 2A = \cos^2 A - \sin^2 A$

Considering the compound angle formula for tangent: $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$. Replacing B by A gives the double angle formula for tangent:

$\tan(A + A) = \frac{\tan A + \tan A}{1 - \tan A \tan A}$

$\tan 2A = \frac{2\tan A}{1 - \tan^2 A}$

The double angle formulae can be summarized as follows:

1. $\sin 2A = 2\sin A \cos A$
2. $\cos 2A = 1 - 2\sin^2 A = 2\cos^2 A - 1 = \cos^2 A - \sin^2 A$
3. $\tan 2A = \frac{2\tan A}{1 - \tan^2 A}$

Example 17

Express $\sin 3\alpha$ in terms of $\sin \alpha$.

Solution:

$\sin 3\alpha = \sin(2\alpha + \alpha)$

$= \sin 2\alpha \cos \alpha + \cos 2\alpha \sin \alpha$

$= (2\sin \alpha \cos \alpha)\cos \alpha + (1 - 2\sin^2 \alpha)\sin \alpha$

$= 2\sin \alpha \cos^2 \alpha + \sin \alpha - 2\sin^3 \alpha$

$= 2\sin \alpha (1 - \sin^2 \alpha) + \sin \alpha - 2\sin^3 \alpha$

$= 2\sin \alpha - 2\sin^3 \alpha + \sin \alpha - 2\sin^3 \alpha$

$= 3\sin \alpha - 4\sin^3 \alpha$

Therefore, $\sin 3\alpha = 3\sin \alpha - 4\sin^3 \alpha$.

Example 18

Prove that $\cos^4 x - \sin^4 x = \cos 2x$.

Solution:

$\cos^4 x - \sin^4 x = (\cos^2 x - \sin^2 x)(\cos^2 x + \sin^2 x)$

Since $\cos^2 x + \sin^2 x = 1$,

$= (\cos^2 x - \sin^2 x)(1)$

$= \cos 2x$

Therefore, $\cos^4 x - \sin^4 x = \cos 2x$.

Example 19

Prove that $\frac{\cos x + \sin x}{\cos x - \sin x} - \frac{\cos x - \sin x}{\cos x + \sin x} = 2\tan 2x$.

Solution:

$\frac{\cos x + \sin x}{\cos x - \sin x} - \frac{\cos x - \sin x}{\cos x + \sin x} = \frac{(\cos x + \sin x)^2 - (\cos x - \sin x)^2}{(\cos x - \sin x)(\cos x + \sin x)}$

$= \frac{(\cos^2 x + 2\cos x \sin x + \sin^2 x) - (\cos^2 x - 2\cos x \sin x + \sin^2 x)}{\cos^2 x - \sin^2 x}$

$= \frac{4\cos x \sin x}{\cos 2x}$

$= \frac{2(2\sin x \cos x)}{\cos 2x}$

$= \frac{2\sin 2x}{\cos 2x}$

$= 2\tan 2x$

Example 20

Eliminate $\lambda$ from the following equations: $x = 3 + 4\sin \lambda$ and $y = 5\cos 4\lambda$.

Solution:

Given $x = 3 + 4\sin \lambda$, make $\sin \lambda$ the subject:

$\sin \lambda = \frac{x - 3}{4}$

From $y = 5\cos 4\lambda$, write $\cos 4\lambda$ in terms of $\sin \lambda$:

$\cos 4\lambda = 1 - 2\sin^2 2\lambda$

$= 1 - 2(2\sin \lambda \cos \lambda)^2$

$= 1 - 8\sin^2 \lambda \cos^2 \lambda$

$= 1 - 8\sin^2 \lambda (1 - \sin^2 \lambda)$

Substitute $\sin \lambda = \frac{x-3}{4}$:

$y = 5\left(1 - 8\left(\frac{x-3}{4}\right)^2\left(1 - \left(\frac{x-3}{4}\right)^2\right)\right)$

This is the equation with $\lambda$ eliminated. It can be further simplified, but this is the core of the elimination process.