Why does cosθ=sin(π2−θ) even when θ>90∘?I get why this relationship holds true in 90 degree...
Why does even when ?
I get why this relationship holds true in 90 degree triangles, but I don't get why this relationship holds true when ? I get that when you find the acute angle that describes , if you see what I mean. But there is no way of showing that in this function! So why would the fact that still hold true here?
Answer & Explanation
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 anticlockwise so its end becomes . You could have got the same result by rotating the radius at (0,1) clockwise through , so
Even a student that has only learned the ‘geometric definition’ of trigonometric functions can derive
Then one would learn the Maclaurin series of and , and then one can show that they have an infinite radius of convergence.
Since and in this interval, by identity theorem this immediately implies the two functions are equal for any