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Double Angle Formula

Double Angle Formula

The double angle can be derived easiest by considering the properties of triangles inscribed inside circles. By applying the circle theorem that the angle subtended at the centre of a circle is twice that at the circumference, we get the diagram shown.

double angle formula 1

Applying Pythagoras to the large right angled triangle gives

$$ \begin{eqnarray} (2\cos(x))^2 &=& (1 + \cos(2x))^2 + (\sin(2x))^2 \\4\cos^2(x) &=& 1 + 2\cos(2x) + \cos^2(2x) + \sin^2(2x) \\4\cos^2(x) &=&2 + 2\cos(2x) \\ \cos(2x)&=& 2\cos^2(x) - 1\end{eqnarray} $$

Application of the Pythagorean identities then yield equivalent results for $\cos(2x)$: 

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

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

Double Angle Formula 2

In the second image we can quickly derive the value of $\sin(2x)$ by applying the sine rule:

$$ \frac{\sin(2x)}{2\sin(x)} = \frac{\cos(x)}{1} $$ giving the result $$ \sin(2x) = 2\sin(x)\cos(x) $$

The Double Angle Formula 1

Suppose $\sin(10^\circ) = p$. Express $\sin(20^\circ)$ in terms of $p$

solution - press button to display

The double angle formula gives us $\sin(2x) = 2\sin(x)\cos(x)$ Now recall that $\cos^2(x) = 1 - \sin^2(x)$ and hence $\cos(x) = \pm \sqrt{1 - \sin^2(x)}$. As both $\sin(x) \gt 0$ and $\cos(x) \gt 0$ for both $x=10^\circ$ and $x=20^\circ$ then we get $$ \sin(2x) = 2\sin(x)\sqrt{1 - \sin^2(x)} $$ Consequently, $\sin(20^\circ) = 2p\sqrt{1-p^2}$

Double Angle Formula 2

Given that $\cos(30^\circ) = \frac{\sqrt{3}}{2}$, determine the value of $\cos(15^\circ)$
solution - press button to display
The double angle formula says

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

So, $$ \cos(30^\circ) = 2\cos^2(15^\circ)-1 $$

Substitution of the known exact value gives $$\frac{\sqrt{3}}{2} = 2\cos^2(15^\circ)-1 $$

Rearranging this equation yields $$\cos(15^\circ) =\sqrt{ \frac{1 + \frac{\sqrt{3}}{2}}{2} }$$

Double Angle Formula 3

One useful application of the double angle formulae is in integrating powers of trig functions. For example,

Evaluate $$\int \sin^2 x dx$$

solution - press button to display
By the DAF, $\cos(2x) = 1 - 2\sin^2(x)$, which can be rewritten as $\sin^2(x) = \frac{1}{2} - \frac{1}{2}\cos(2x)$. The right hand side of this expression is now straightforward to integrate: $$ \int \sin^2(x)dx = \int \frac{1}{2} - \frac{1}{2}\cos(2x)dx = \frac{1}{2}x - \frac{1}{4}\sin(2x) + c $$