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Compound Angle Formulae

Compound Angle Formulae

The compound angle formulae are $$ \sin(A\pm B) = \sin(A)\cos(B) \pm \cos(A)\sin(B) $$ here the sign on the LHS is the same as that on the RHS $$ \cos(A\pm B) = \cos(A)\cos(B) \mp \sin(A)\sin(B) $$ here the sign on the LHS is the opposite of that on the RHS $$ \tan(A\pm B) = \frac{\tan(A) \pm \tan(B)}{1\mp \tan(A)\tan(B)} $$

Compound angle formulae 1

By writing $\sin(3\theta) = \sin(2\theta + \theta)$ show that $\sin(3\theta)$ can be expressed as a function of $\sin(\theta)$
solution - press button to display
By the compound angle formula, $$\sin(3\theta) = \sin(\theta + 2\theta) = \sin(\theta)\cos(2\theta) + \sin(2\theta)\cos(\theta)$$ We can now apply the double angle formula giving $$ \sin(3\theta)= \sin(\theta)(1 - 2\sin^2(\theta)) + 2\sin(\theta)\cos^2(\theta) $$
Applying the Pythagorean identity to $\cos^2\theta$ gives
$$\sin(3\theta) = 3\sin(\theta) - 4\sin^3(\theta)$$
 

Compound angle formulae 2

By using the compound angle formula, determine the exact value of $\sin(15)$ and $\cos(15)$
solution - press button to display
$$\sin(15) = \sin(45 - 30) = \sin(45)\cos(30) - \cos(45)\sin(30)$$

Evaluation of this gives $$\sin(15) = \frac{1}{\sqrt{2}}\left(\frac{\sqrt{3}}{2} - \frac{1}{2}\right) = \frac{\sqrt{6} - \sqrt{2}}{4}$$

$\cos(15)$ can be similarly evaluated

$$\cos(15) = \cos(45 - 30) = \cos(45)\cos(30) - \sin(45)\sin(30) = \frac{\sqrt{6} + \sqrt{2}}{4}$$

Compound angle formula 3

By considering an $\sin(3\theta)$ in terms of $\theta$, show that $\sin(10^\circ)$ satisfies the following equation

$$8s^3 - 6s + 1 = 0$$

where $s = \sin(10^\circ)$

solution - press button to display
As we have previously established, $\sin(3\theta) = 3\sin(\theta) - 4\sin^3(\theta)$.

Let $\theta = 10^\circ$. Then we get $$\sin(30^\circ) = 3\sin(10^\circ) - 4\sin^3(10^\circ)$$. Simplification gives

$$\frac{1}{2} = 3\sin(10^\circ) - 4\sin^3(10^\circ)$$

Further rearrangement gives the required equation

$$1 = 6s - 8s^3 \Rightarrow 8s^3 - 6s + 1 =0$$