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mrenney · Nov 2, 2010, 06:35 AM

and b. cos(sin^-1(x) - cos^-1(y))

mrenney · Nov 2, 2010, 06:42 AM

and 2sin^2(x)cos(x) = cos(x)?

Curlyben · Nov 2, 2010, 06:50 AM

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galactus · Nov 2, 2010, 02:46 PM

Here is the first one. You try the second one.

Using the addition formula for sin:

$sin(A+B)=sinAcosB+cosAsinB$

$sin(sin^{-1}(x))cos(cos^{-1}(y))+cos(sin^{-1}(x))sin(cos^{-1}(y))$

The first two cancel out and we are left with xy.

The second two can be found by looking at a triangle and seeing they are equal to:

$cos(sin^{-1}(x))=\sqrt{1-x^{2}}, \;\ sin(cos^{-1}(y))=\sqrt{1-y^{2}}$

So, we have $xy+\sqrt{1-x^{2}}\cdot \sqrt{1-y^{2}}$

This is not evaluated, but simplified. To evaluate means to plug something in and get an answer.

Were you given values to sub in?

For part b, use the subtraction formula for cos.

Unknown008 · Nov 2, 2010, 10:44 PM

Originally Posted by mrenny

Find all exact solutions of the equations: cos(2x - (Pi/3)) = -(square root(3)/2)
and 2sin^2(x)cos(x) = cos(x)?

What about showing us what you tried?