1. (tan^-1)x = -1

2. sin^-1(sinx) = π/10

Thanks for your help!

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- Mar 2nd 2009, 08:41 AMlive_laugh_luv27Trig Function Equations
1. (tan^-1)x = -1

2. sin^-1(sinx) = π/10

Thanks for your help! - Mar 2nd 2009, 09:16 AMJameson
You need to understand inverse trig functions to solve these. If the tangent of a ratio of (opposite)/(adjacent) legs of a right triangle give you an angle, then the inverse tangent intuitively takes an angle and gives a ratio. So in your problem, x, is an angle. So using this info, can you draw the right triangle that it is describing? Once you do, you will know the angle because it's a famous right triangle, one of those you have to memorize.

How do you think sine and inverse sine are related?

edit: Ok it seems this isn't enough to grab your attention. For #1) you can also think of it as $\displaystyle x=\tan^{-1}(-1)$. So if x is an angle, -1 is a ratio. That is obviously simplified though, since the ratio is of (opposite)/(adjacent). Thus must be that either the opposite leg is 1 or 1 and that the adjacent leg is 1 or -1, but they can't both be the same sign, or the ratio would be positive.