i was wondering if anyone could help me to simplify cos((sinx)^-1) i remember someone saying in class that it was something like
2sqareroot(9-x^2) but i don't have a clue how to get there. any help would be much apreciated
Let . You can think of this as the angle whose sine is x/1. In a right triangle, sine is opposite over hypotenuse, so draw a right triangle, label one of the acute angles as theta, the opposite side as x and the hypotenuse as 1. Use the Pythagorean theorem to find the missing (adjacent) side:
Solve for b. Label this on the adjacent side.
Now , and cosine is adjacent over hypotenuse. Can you find now?
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