double and half
\[ \sin (2t) = \frac {e^{i 2t} - e^{-i 2t}}{2 i} \]
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
\[ \cos (2t) = \cos ^2(t) - \sin ^2(t) = (1 - \sin ^2(t)) - \sin ^2(t) = 1 - 2 \sin ^2(t) \]
\[ \cos (2t) = 2 \cos ^2(t) - 1 \]
\[ \frac {1 + \cos (2t)}{2} = \cos ^2(t) \]
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.
2025-01-12 23:28:58