Double Angle Formulas

In the Angle Addition Identities post it was shown that

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

and

$ \displaystyle \cos(x+y) = \cos x \cos y – \sin x \sin y $.

These identities are valid if $\sin \theta \equiv \frac{opp}{hyp}$ and $\cos \theta \equiv \frac{adj}{hyp}$, which implies that $\theta \in [0,\frac{\pi}{2})$ in order for $\theta$ to be an acute angle of a right triangle. These constraints are imposed because the two Angle Addition Identities were derived using these constraints.

The next step, following the steps in reference [1], is setting $y$ equal to $x$. Doing this yields

$ \displaystyle \sin(2x) = \sin x \cos x + \sin x \cos x = 2 \sin x \cos x$.

Hence

$ \boxed{ \displaystyle \sin(2x) = 2 \sin x \cos x } $.

Similarly, following the steps in reference [1], substituting $x$ for $y$ in

$ \displaystyle \cos(x+y) = \cos x \cos y – \sin x \sin y $

yields

$ \displaystyle \cos(2x) = \cos x \cos x – \sin x \sin x = \cos ^2 x – \sin ^2 x $.

Hence

$ \boxed{ \displaystyle \cos(2x) = \cos ^2 x – \sin ^2 x } $.

Using this Identity, the previous equation becomes

$  \displaystyle \cos(2x) = (1 – \sin^2 x) – \sin^2 x = 1 – 2 \sin^2 x$.

Hence

$ \boxed{ \displaystyle \cos(2x) = 1 – 2 \sin^2 x }$.

Solving for $\sin x$ and taking the positive root to ensure that $\sin x$, a ratio of lengths, is not negative yields

$  \displaystyle \sin x = \sqrt{ \frac{1-\cos(2x)}{2} } $.

Finally, define $x \equiv \frac{\alpha}{2}$ and substitute to get the half-angle identity [2],

$  \displaystyle \boxed{ \sin \frac{\alpha}{2} = \sqrt{ \frac{1-\cos \alpha }{2}}  }$.

Therefore,

$  \displaystyle  2\sin^2 \frac{\alpha}{2} =  1-\cos \alpha   $.

 

References

[1] https://www.intmath.com/analytic-trigonometry/3-double-angle-formulas.php

[2] https://www.intmath.com/analytic-trigonometry/4-half-angle-formulas.php

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