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
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
- Cosine of Complement equals Sine
- Tangent of Complement equals Cotangent
- Cotangent of Complement equals Tangent
- Secant of Complement equals Cosecant
- Cosecant of Complement equals Secant
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: Formulae $(5)$
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 5$: Trigonometric Functions: Functions of Angles in All Quadrants in terms of those in Quadrant I
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): Appendix $12$: Trigonometric formulae: Symmetry
- 2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): Appendix $14$: Trigonometric formulae: Symmetry