Why does cos⁡θ=sin⁡(π2−θ) even when θ&gt;90∘?I get why this relationship holds true in 90 degree triangles, but I don't get why this relationship holds true when θ&gt;90∘? I get that when θ&gt;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 cos⁡θ=sin⁡(π2−θ) still hold true here?

Question

Why does cos⁡θ=sin⁡(π2−θ) even when θ&amp;gt;90∘?I get why this relationship holds true in 90 degree triangles, but I don&#039;t get why this relationship holds true when θ&amp;gt;90∘? I get that when θ&amp;gt;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 cos⁡θ=sin⁡(π2−θ) still hold true here?

Rosa Nicholson · Accepted Answer

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 (cos⁡θ,sin⁡θ). You could have got the same result by rotating the radius at (0,1) clockwise through π2−θ, so (cos⁡θ,sin⁡θ)=(sin⁡(π2−θ),cos⁡(π2−θ))

Ronald Alvarez · Accepted Answer

Even a student that has only learned the ‘geometric definition’ of trigonometric functions can derive  cos⁡x=sin⁡(π2−x)  for 0&amp;lt;x&amp;lt;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 cos⁡x=sin⁡(π2−x) in this interval, by identity theorem this immediately implies the two functions (cos⁡x  and  sin⁡(π2−x)) are equal for any x∈C

Why does \cos\theta=\sin(\frac{\pi}{2}-\theta) even when \theta > 90^{\circ}? I get

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