Question

Write the given expression in terms of x and y only. sin(sin^−1 (x) + cos^−1 (y))

Answer 1

Start off by using the sum of sines formula. From there you should be able to use some triangle identities to rewrite everything :)

Answer 2

I should have stated where I was with the problem I basically have this
x=\[\sin(\sin ^{-1}x)\]
y=\[\cos(\cos ^{-1}y)\]
Then I tried substituting and got
\[\sin(\sin ^{-1}x)\cos(\cos ^{-1}y)+\]\[(\sqrt{1-\cos(\cos ^{-1}y)^{2}})(\sqrt{1-\sin(\sin ^{-1}x)^{2}}\]

But then when I tried submitting that answer, it came out incorrect. Did I do something wrong along the way or mess up a step?

Answer 3

Well, you only want x's and y's, not any sines or cosines. So I would just be using the sum of sines formula. So if a = sin^-1(x) and b = cos^-1(y), I have sin(a + b) = sin(a)cos(b) + cos(a)sin(b).
This becomes sin(sin^-1(x))cos(cos^-1(y)) + cos(sin^-1(x))sin(cos^-1(y)).
So sin and sin inverse cancel as well as cos and cos inverse. So you have:

xy + cos(sin^-1(x))sin(cos^-1(y)).

Personally, I would translate the remaining inverse expressions like this:
For cos(sin^-1(x)) - Some triangle has a sine value of x. What's the cosine of that same triangle?

And similarly for the other one
sin(cos^-1(y)) - Some triangle as a cosine value of y. What's the sine of that same triangle?

So with that idea, I draw two triangles and get all their sides





For the first triangle, I need its cosine value, so that ends up being \(\sqrt{1-x^{2}}\). For the 2nd triangle, I need a sine value, and that ends up being \(\sqrt{1-y^{2}}\). So putting everything together, we have

\(xy + \sqrt{1-x^{2}}\) *\(\sqrt{1-y^{2}}\)