Question

sin(π+x)cos(π/2 + x)tan(3π/2 - x)cot(2π-x)/sin(2π - x) cos(2π + x)cosec(-x)sin(3π/2 + x)​

Answer 1

Answer:

answer

Step-by-step explanation:

sin(π/2−x)=sin(π/2)cosx−cos(π/2)sinx=cosxsin⁡(π/2−x)=sin⁡(π/2)cos⁡x−cos⁡(π/2)sin⁡x=cos⁡x

sin(π/2+x)=sin(π/2)cosx+cos(π/2)sinx=cosxsin⁡(π/2+x)=sin⁡(π/2)cos⁡x+cos⁡(π/2)sin⁡x=cos⁡x

It does!

Is the following equation true?

sin (π/2-x) = sin (x+π/2)

use sin(A +B) = sinAcosB + sinBcosA

[sin(π/2)]cos(-x) + [cos (π/2)]sin (-x) = [sin(π/2)]cos(x) + [cos (π/2)]sin (x)

note cos (π/2) =0

So the equation simplifies to

[sin(π/2)]cos(-x) = [sin(π/2)]cos(x)

divide by sin (π/2)

Does

cos (-x) = cos (x) ?

Yes because cosine is an even function