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

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

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

If we subtract the equations instead, we get

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

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

We can combine those to get

$\displaystyle \tan \theta =\frac{{\sin \theta }}{{\cos \theta }}=\frac{{\left( {{{e}^{{i\theta }}}-{{e}^{{-i\theta }}}} \right)/2i}}{{\left( {{{e}^{{i\theta }}}+{{e}^{{-i\theta }}}} \right)/2i}}=\frac{1}{i}\cdot \frac{{{{e}^{{i\theta }}}-{{e}^{{-i\theta }}}}}{{{{e}^{{i\theta }}}+{{e}^{{-i\theta }}}}}$

We can then multiply both the numerator and the denominator by e^–iθ to get

$\displaystyle \tan \theta =\frac{1}{i}\cdot \frac{{{{e}^{{2i\theta }}}-1}}{{{{e}^{{2i\theta }}}+1}}$

These three formulas, for the sine, the cosine and the tangent, 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 Euler’s identity, so it is a bit of circular reasoning.

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Last modified: May 21, 2025 @ 18:34