Aubrey Hendricks

2022-01-27

Why does $\mathrm{cos}\theta =\mathrm{sin}\left(\frac{\pi }{2}-\theta \right)$ even when $\theta >{90}^{\circ }$?
I get why this relationship holds true in 90 degree triangles, but I don't get why this relationship holds true when $\theta >{90}^{\circ }$? I get that when $\theta >{90}^{\circ }$ you find the acute angle that describes $\theta$, if you see what I mean. But there is no way of showing that in this function! So why would the fact that $\mathrm{cos}\theta =\mathrm{sin}\left(\frac{\pi }{2}-\theta \right)$ still hold true here?

Rosa Nicholson

Expert

Draw a centre-O circle of radius 1 and add in the radius with an end at (1,0), and rotate that radius through an angle $\theta$ anticlockwise so its end becomes $\left(\mathrm{cos}\theta ,\mathrm{sin}\theta \right)$. You could have got the same result by rotating the radius at (0,1) clockwise through $\frac{\pi }{2}-\theta$, so $\left(\mathrm{cos}\theta ,\mathrm{sin}\theta \right)=\left(\mathrm{sin}\left(\frac{\pi }{2}-\theta \right),\mathrm{cos}\left(\frac{\pi }{2}-\theta \right)\right)$

Ronald Alvarez

Expert

Even a student that has only learned the ‘geometric definition’ of trigonometric functions can derive
$\mathrm{cos}x=\mathrm{sin}\left(\frac{\pi }{2}-x\right)$
for $0.
Then one would learn the Maclaurin series of $\mathrm{sin}$ and $\mathrm{cos}$, and then one can show that they have an infinite radius of convergence.
Since $\left(0,\frac{\pi }{2}\right)\subset \mathbb{C}$ and $\mathrm{cos}x=\mathrm{sin}\left(\frac{\pi }{2}-x\right)$ in this interval, by identity theorem this immediately implies the two functions are equal for any $x\in \mathbb{C}$

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