# Some important trig identities

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We have that

${e^{i\theta }} = \cos \theta + i\sin \theta$

If we instead have minus θ as the input we get

${e^{ - i\theta }} = \cos ( - \theta ) + i\sin ( - \theta ) = \cos (\theta ) - i\sin ( - \theta )$

If we now add the two equations we get

${e^{i\theta }} + {e^{ - i\theta }} = 2\cos (\theta )$

or

$\cos (\theta ) = \frac{{{e^{i\theta }} + {e^{ - i\theta }}}}{2}$

If we subtract the equations instead we get

${e^{i\theta }} - {e^{ - i\theta }} = 2isin(\theta )$

or

$sin(\theta ) = \frac{{{e^{i\theta }} - {e^{ - i\theta }}}}{{2i}}$

These two formulas for the sine and the cosine will be helpful in some calculus problems. We may, for example, take the derivative of say sin(θ) to get

$(sin(\theta ))' = (\frac{{{e^{i\theta }} - {e^{ - i\theta }}}}{{2i}})' = \frac{{i{e^{i\theta }} - ( - i){e^{ - i\theta }}}}{{2i}} = \frac{{{e^{i\theta }} + {e^{ - i\theta }}}}{2} = \cos (\theta )$

Just as we know it should be. It is not a derivation of the derivative of sin(θ), since we used the derivative of sin(θ) to find the Euler’s identity – so it is a bit of circular reasoning.

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