Sine of Complement equals Cosine

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Theorem

$\map \sin {\dfrac \pi 2 - \theta} = \cos \theta$

where $\sin$ and $\cos$ are sine and cosine respectively.

That is, the cosine of an angle is the sine of its complement.


Proof 1

\(\ds \map \sin {\frac \pi 2 - \theta}\) \(=\) \(\ds \sin \frac \pi 2 \cos \theta - \cos \frac \pi 2 \sin \theta\) Sine of Difference
\(\ds \) \(=\) \(\ds 1 \times \cos \theta - 0 \times \sin \theta\) Sine of Right Angle and Cosine of Right Angle
\(\ds \) \(=\) \(\ds \cos \theta\)

$\blacksquare$


Proof 2

\(\ds \map \sin {\frac \pi 2 - \theta}\) \(=\) \(\ds -\map \sin {\theta - \frac \pi 2}\) Sine Function is Odd
\(\ds \) \(=\) \(\ds \map \cos {\theta - \frac \pi 2 + \frac \pi 2}\) Cosine of Angle plus Right Angle
\(\ds \) \(=\) \(\ds \cos \theta\)

$\blacksquare$


Proof 3

\(\ds \map \sin {\dfrac \pi 2 - \theta}\) \(=\) \(\ds \map \Im {e^{i \paren {\frac \pi 2 - \theta} } }\) Euler's Formula
\(\ds \) \(=\) \(\ds \map \Im {e^{i \frac \pi 2} e^{-i \theta} }\) Exponent Combination Laws
\(\ds \) \(=\) \(\ds \map \Im {\paren {\cos \dfrac \pi 2+i \sin \dfrac \pi 2} e^{-i \theta} }\) Euler's Formula
\(\ds \) \(=\) \(\ds \map \Im {i e^{-i \theta} }\) Cosine of Right Angle, Sine of Right Angle
\(\ds \) \(=\) \(\ds \map \Re {e^{-i \theta} }\)
\(\ds \) \(=\) \(\ds \map \cos {-\theta}\) Euler's Formula
\(\ds \) \(=\) \(\ds \cos \theta\) Cosine Function is Even

$\blacksquare$


Proof 4

ComplementaryAngles.png

Let $\angle xOP$ and $\angle QOy$ be complementary.

Then:

$\angle xOP = \angle QOy$

Hence:

the projection of $OP$ on the $x$-axis

equals:

the projection of $OQ$ on the $y$-axis.

Hence the result.

$\blacksquare$


Examples

Sine of $2 \theta - 90 \degrees$

$\map \sin {2 \theta - 90 \degrees} = -\cos 2 \theta$


Also see


Sources

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