Understanding Trigonometric Inverse Functions - Solving Homework Equations

vanmaiden
Messages
101
Reaction score
1

Homework Statement


I understand that y = sin -1 x. However, why is it that when one writes x = sin y, one leaves out the -1


Homework Equations





The Attempt at a Solution


I know that when you invert a trig function, the dependent variable essentially switches from y to x and the independent variable switches from x to y. However, why do you leave out the -1 when you write the inverse sin as x = sin y ?
 
Physics news on Phys.org
not sure if I understand the question, but if you start from
y = sin^{-1} x

take the sin of both sides
sin(y) = sin(sin^{-1} x) = x
 
lanedance said:
not sure if I understand the question, but if you start from
y = sin^{-1} x

take the sin of both sides
sin(y) = sin(sin^{-1} x) = x

I'm sorry I wasn't very clear. That answers my question though! I had no idea you could take the sin of the sin-1(x) and it would cancel them out. Thank you! :smile:
 
Well, that is how inverse functions work.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top