Trig Identities

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double and half

1 Double Angle Formulas

\[ \sin (2t) = \frac {e^{i 2t} - e^{-i 2t}}{2 i} \]

This is a difference of two squares. It factors.

\[ \sin (2t) = \frac {(e^{i t} - e^{-i t})(e^{i t} + e^{-i t})}{2 i} \]

\[ \sin (2t) = 2 \cdot \frac {(e^{i t} - e^{-i t})}{2i} \cdot \frac {(e^{i t} + e^{-i t})}{2} \]

\[ \sin (2t) = 2 \sin (t) \cos (t) \]

This is the double angle formla for sine. How about cosine? Let’s begin with $\cos ^2(t) - \sin ^2(t)$ and rearrange it.

\[ \cos ^2(t) - \sin ^2(t) \]

\[ \cos ^2(t) - \sin ^2(t) = \left ( \frac {e^{i t} + e^{-i t}}{2} \right )^2 - \left ( \frac {e^{i t} - e^{-i t}}{2i} \right )^2 \]

\[ \cos ^2(t) - \sin ^2(t) = \left ( \frac {e^{2i t} + 2 + e^{-2i t}}{4} \right ) - \left ( \frac {e^{2 i t} - 2 + e^{-2i t}}{-4} \right ) \]

\[ \cos ^2(t) - \sin ^2(t) = \frac {2 e^{2i t} + 2 e^{-2i t}}{4} = \frac {e^{2i t} + e^{-2i t}}{2} = \cos (2t) \]

\[ \cos ^2(t) - \sin ^2(t) = \cos (2t) \]

This is the double angle formula for cosine.

Additionally, we have

\[ \cos (2t) = \cos ^2(t) - \sin ^2(t) = \cos ^2(t) - (1 - \cos ^2(t)) = 2 \cos ^2(t) - 1 \]

and

\[ \cos (2t) = \cos ^2(t) - \sin ^2(t) = (1 - \sin ^2(t)) - \sin ^2(t) = 1 - 2 \sin ^2(t) \]

2 Half Angle Formulas

\[ \cos (2t) = 2 \cos ^2(t) - 1 \]

\[ \frac {1 + \cos (2t)}{2} = \cos ^2(t) \]

and we know that $\cos ^2(t) = 1- \sin ^2(t)$. Applying this to the formula above gives us

\[ \frac {1 - \cos (2t)}{2} = \sin ^2(t) \]

These are half-angle formulas. We can see the half better by replacing $t$ with $\frac {x}{2}$

\[ \frac {1 + \cos (x)}{2} = \cos ^2\left ( \tfrac {x}{2} \right ) \]

\[ \frac {1 - \cos (x)}{2} = \sin ^2\left ( \tfrac {x}{2} \right ) \]

Obtain an expression for $\cos \left ( \frac {\pi }{8} \right )$ that only uses square roots. Let $t = \frac {\pi }{8}$ in the equation above.

$\frac {\pi }{8}$ is half of $\frac {\pi }{4}$

\[ \frac {1 + \cos ( \frac {\pi }{4} )}{2} = \cos ^2\left ( \frac {\pi }{8} \right ) \]

\[ \frac {1 + \frac {1}{\sqrt {2}}}{2} = \cos ^2\left ( \frac {\pi }{8} \right ) \]

$\frac {\pi }{8}$ is in quadrant I, therefore cosine is positive.

\[ \sqrt { \frac {1 + \frac {1}{\sqrt {2}}}{2} } = \cos \left ( \frac {\pi }{8} \right ) \]

The double angle formulas help you reduce the argument inside the trigonometric function.

You replace $\sin (2 \theta )$ with $2 \sin (\theta ) \cos (\theta )$. You end up with more functions, but they are easier to work with.

The half angle formulas get rid of squaring.

$\sin ^2\left ( \tfrac {x}{2} \right )$ is replaced with $\frac {1 - \cos (x)}{2}$. Much easier to work with.

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2025-01-07 03:54:21