Cosine of Complement equals Sine
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Theorem
- $\map \cos {\dfrac \pi 2 - \theta} = \sin \theta$
where $\cos$ and $\sin$ are cosine and sine respectively.
That is, the sine of an angle is the cosine of its complement.
Proof 1
\(\ds \map \cos {\frac \pi 2 - \theta}\) | \(=\) | \(\ds \cos \frac \pi 2 \cos \theta + \sin \frac \pi 2 \sin \theta\) | Cosine of Difference | |||||||||||
\(\ds \) | \(=\) | \(\ds 0 \times \cos \theta + 1 \times \sin \theta\) | Cosine of Right Angle and Sine of Right Angle | |||||||||||
\(\ds \) | \(=\) | \(\ds \sin \theta\) |
$\blacksquare$
Proof 2
\(\ds \map \cos {\frac \pi 2 - \theta}\) | \(=\) | \(\ds \map \cos {\theta - \frac \pi 2}\) | Cosine Function is Even | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \sin {\theta - \frac \pi 2 + \frac \pi 2}\) | Sine of Angle plus Right Angle | |||||||||||
\(\ds \) | \(=\) | \(\ds \sin \theta\) |
$\blacksquare$
Proof 3
\(\ds \map \cos {\dfrac \pi 2 - \theta}\) | \(=\) | \(\ds \frac 1 2 \paren {e^{i \paren {\frac \pi 2 - \theta} } + e^{-i \paren {\frac \pi 2 - \theta} } }\) | Euler's Cosine Identity | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 \paren {e^{i \frac \pi 2} e^{-i \theta} + e^{-i \frac \pi 2} e^{i \theta} }\) | Exponential of Sum: Complex Numbers | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 \paren {\paren {\map \cos {\frac \pi 2} + i \, \map \sin {\frac \pi 2} } e^{-i \theta} + \paren {\map \cos {-\frac \pi 2} + i \, \map \sin {-\frac \pi 2} } e^{i \theta} }\) | Euler's Formula | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 \paren {i e^{-i \theta} - i e^{i \theta} }\) | Cosine of Right Angle, Sine of Right Angle, Cosine Function is Even, Sine Function is Odd | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {2 i} \paren {e^{i \theta} - e^{-i \theta} }\) | Definition of Imaginary Unit | |||||||||||
\(\ds \) | \(=\) | \(\ds \sin \theta\) | Euler's Sine Identity |
$\blacksquare$
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.
$\blacksquare$
Also see
- Sine of Complement equals Cosine
- Tangent of Complement equals Cotangent
- Cotangent of Complement equals Tangent
- Secant of Complement equals Cosecant
- Cosecant of Complement equals Secant
Historical Note
The result Cosine of Complement equals Sine was discovered and documented by Varahamihira in the $6$th century CE.
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: Formulae $(4)$
- 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