Relation between Trigonometry & Inverse Trigonometry Functions

Suppose we are given

θ = sin ^–1 x

We need to find sin θ , cos θ and tan θ

Now,

  θ = sin ^–1 x

So,

  sin θ = x

  sin θ = x/1

And we know that

  sin θ =  (Side opposite to angle θ)/Hypotenuse

  x/1 = AB/AC 

So, AB = x, AC = 1

Now, in right triangle Δ ABC

By Pythagoras Theorem

  AC ² = AB ² + BC ²

  1 ² = x ² + BC ²

  1 – x ² = BC ²

  BC = √(1-x ² )

Value of sin θ , cos θ , tan θ when θ is given as inverse