Why does cos⁡θ=sin⁡(π2−θ) even when θ>90∘?I get why this relationship holds true in 90 degree...

Aubrey Hendricks

Aubrey Hendricks



Why does cosθ=sin(π2θ) even when θ>90?
I get why this relationship holds true in 90 degree triangles, but I don't get why this relationship holds true when θ>90? I get that when θ>90 you find the acute angle that describes θ, if you see what I mean. But there is no way of showing that in this function! So why would the fact that cosθ=sin(π2θ) still hold true here?

Answer & Explanation

Rosa Nicholson

Rosa Nicholson


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 θ anticlockwise so its end becomes (cosθ,sinθ). You could have got the same result by rotating the radius at (0,1) clockwise through π2θ, so (cosθ,sinθ)=(sin(π2θ),cos(π2θ))
Ronald Alvarez

Ronald Alvarez


2022-01-29Added 11 answers

Even a student that has only learned the ‘geometric definition’ of trigonometric functions can derive
for 0<x<π2.
Then one would learn the Maclaurin series of sin and cos, and then one can show that they have an infinite radius of convergence.
Since (0,π2)C and cosx=sin(π2x) in this interval, by identity theorem this immediately implies the two functions (cosx  and  sin(π2x)) are equal for any xC

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?