Question

Evaluate the expression Sin^-1((sqrt2)/2)

Answer 1

\(\large Sin^{-1}\frac{\sqrt2}{2} =\theta \) is the same as asking \(\large sin\theta=\frac{\sqrt2}{2} \).
What is theta?

Answer 2

Maybe an easier way to try to understand this is to see the question as asking you, "Where is the sin of theta square root of 2 over 2". If you have a unit circle, you could find the answer immmeditely, but if not you have to know which type of special triangle has the square root of 2 in its triple. Do you know that?

Answer 3

The square root of 2 is in the 45-45-90. Actually, the square root was originally in the denominator but the denominator was rationalized. We know this because there isn't a special triangle that has a triple with a 2 and a square root of 2 in it. Next we need to remember what the sin of an angle is given by. The side opposite the angle over the hypotenuse, right? So let's draw a generic angle and figure out where the measures go, then we can find the measure of theta. If sin is opposite over hypotenuse, and the square root of 2 has been rationalized, that means that it was originally in the denominator. Do you know what the other sides are? They are 1's, both of them. For a 45 right triangle, the sides are 1,1,sqrt 2. The angle that is across from the side that measures 1 is the angle that is theta. The 45 degree angle is the angle in question. In radians it is pi / 4.

Answer 4

Do you understand any of this?

Answer 5

im sorry I do! My computer just froze

Answer 6

I was just able to type!

Answer 7

@IMStuck

Answer 8

As answered earlier, \(\Large \theta = \frac \pi 4 ~ radians\ or \ 45 \ degrees.\)

Answer 9

ok thank you

Answer 10

You are welcome.