Thread: Trig expressions

(1 pt) Find the exact value (NO DECIMALS) of each expression by sketching a triangle:
A) sin(cos^-1(3/5))
b) cos(sin^-1(4/5))
I have no idea where to start

Originally Posted by RD33

(1 pt) Find the exact value (NO DECIMALS) of each expression by sketching a triangle:
A) sin(cos^-1(3/5))
b) cos(sin^-1(4/5))
I have no idea where to start

A)

Use the fact that

cos^2(x)+sin^2(x) = 1

We have

cos^-1(3/5) <=> cos(x) = 3/5 => cos^2(x) = 9/25

Using the trigonometric one we get

9/25+sin^2(x) = 25/25 <=> sin^2(x) = 16/25 <=> sin(x) = +/- 4/5

But since cos^-1 is defined for 0 to pi, the only solution is sin(x) = 4/5

By this we now have the expression

sin(cos^-1(3/5)) = sin(sin^-1(4/5)) = 4/5

I leave the second one as exercise for you!

Thank you. I got the second one right honestly by guessing.

Cos(sin^-1(4/5))

I got sin(x)=4/5

Sin^2= 16/25

And I'm not sure where to go from there

Originally Posted by RD33

Thank you. I got the second one right honestly by guessing.

Cos(sin^-1(4/5))

I got sin(x)=4/5

Sin^2= 16/25

And I'm not sure where to go from there

sin^2(x)+cos^2(x) = 1 => 16/25+cos^2(x) = 25/25 <=> cos^2(x) = 9/25 <=> cos(x) = +/- 3/5

Since we know that sin^1(x) is defined for -pi<=x<=pi we know that cos^-1(3/5)

Then we have

Cos(sin^-1(4/5)) = cos(cos^-1(3/5))) = 3/5