Aubrey Hendricks

Answered

2022-01-27

Why does $\mathrm{cos}\theta =\mathrm{sin}(\frac{\pi}{2}-\theta )$ 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 $\mathrm{cos}\theta =\mathrm{sin}(\frac{\pi}{2}-\theta )$ still hold true here?

I get why this relationship holds true in 90 degree triangles, but I don't get why this relationship holds true when

Answer & Explanation

Rosa Nicholson

Expert

2022-01-28Added 13 answers

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 $(\mathrm{cos}\theta ,\mathrm{sin}\theta )$ . You could have got the same result by rotating the radius at (0,1) clockwise through $\frac{\pi}{2}-\theta$ , so $(\mathrm{cos}\theta ,\mathrm{sin}\theta )=(\mathrm{sin}(\frac{\pi}{2}-\theta ),\mathrm{cos}(\frac{\pi}{2}-\theta ))$

Ronald Alvarez

Expert

2022-01-29Added 11 answers

Even a student that has only learned the ‘geometric definition’ of trigonometric functions can derive

$\mathrm{cos}x=\mathrm{sin}(\frac{\pi}{2}-x)$

for$0<x<\frac{\pi}{2}$ .

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$(0,\frac{\pi}{2})\subset \mathbb{C}$ and $\mathrm{cos}x=\mathrm{sin}(\frac{\pi}{2}-x)$ in this interval, by identity theorem this immediately implies the two functions $\left(\mathrm{cos}x\text{}\text{and}\text{}\mathrm{sin}(\frac{\pi}{2}-x)\right)$ are equal for any $x\in \mathbb{C}$

for

Then one would learn the Maclaurin series of

Since

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