Question

What is `Cos(arcsin(-5/13)+arccos(12/13))`?

Answer 1

`=1`

Explanation:

First you want to let `alpha=arcsin(-5/13)` and `beta=arccos(12/13)`

So now we are looking for `color(red)cos(alpha+beta)!`

`=>sin(alpha)=-5/13" "` and `" "cos(beta)=12/13`

Recall : `cos^2(alpha)=1-sin^2(alpha)=>cos(alpha)=sqrt(1-sin^2(alpha))`

`=>cos(alpha)=sqrt(1-(-5/13)^2)=sqrt((169-25)/169)=sqrt(144/169)=12/13`

Similarly, `cos(beta)=12/13`

`=>sin(beta)=sqrt(1-cos^2(beta))=sqrt(1-(12/13)^2)=sqrt((169-144)/169)=sqrt(25/169)=5/13`

`=>cos(alpha+beta)=cos(alpha)cos(beta)-sin(alpha)sin(beta)`

Then substitue all the values obtained ealier.

`=>cos(alpha+beta)=12/13*12/13-(-5/13)*5/13=144/169+25/169=169/169=color(blue)1`