# Write the trigonometric expression as an algebraic expression containing u and v. `cos (cos^(-1)u + sin^(-1)v)`cos(cos^-1u+sin^-1v)

*print*Print*list*Cite

### 1 Answer

`cos(cos^(-1)u + sin^(-1)v)`

Let `A=cos^(-1) u` and one of the angle of a right triangle.

Then, set the hypotenuse of the triangle equal to 1. So, the adjacent side of A is u. Using the Pythagorean formula, the opposite side is `sqrt(1-u^2)` . Base on this triangle,

`cos A=(adj.)/(hyp.)=u/1` and `sin A = (opp.)/(hyp.)=sqrt(1-u^2)/1`

`cosA=u ` `sinA=sqrt(1-u^2)`

Also, let `B=sin^(-1)v` and one of the angle of another right triangle. Then, set the hypotenuse equal to 1. This means that the opposite side of B is v. And its adjacent side is `sqrt (1-v^2)` . So,

`cosB= (adj.)/(hyp.)=sqrt(1-v^2)/1` and `sinB=(opp.)/(hyp.)=v/1`

`cosB=sqrt(1-v^2) ` `sinB=v`

Substituting` A= cos^(-1)u` and `B=sin^(-1)v` to the given expression results to:

`cos(cos^(-1)u + sin^(-1)v)=cos(A + B)`

From here, apply the trigonometric identities of sum of two angles.

`=cosAcosB-sinAsinB`

Then, substitute `cosA=u`, `cosB=sqrt(1-v^2)`, `sinA=sqrt(1-u^2)` and `sinB=v`.

`=usqrt(1-v^2)-vsqrt(1-u^2)`

**Hence, the algebraic expression of `cos(cos^(-1)u+sin^(-1)v)` is `usqrt(1-v^2)` `-` `vsqrt(1-u^2)` .**