I need help solving inverse trigonometric problem.

[Eashan1337]

I need to simplify cos⁻¹[(2x - 3√(1 - x²))/(√13)]

 

Any ideas to how to do so?

[daniton]

try this

Cos(z+y)=Cos(z)Cos(y)-Sin(z)Sin(y)

=>

z+y= Cos⁻¹ [Cos(z)Cos(y)-Sin(z)Sin(y)]

using this:

the expression becomes this by little rearranging

 

Cos⁻¹[(2/ √13 )*x - (3/ (√13)) *√(1 -x²)]

Cos(z)=?

Cos(y)=?

try the rest. I think that helps!