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

Answered

2022-01-27

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

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 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

Expert

2022-01-29Added 11 answers

Even a student that has only learned the ‘geometric definition’ of trigonometric functions can derive
cosx=sin(π2x)
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

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