Ca somone show me how to solve

Sin-1 (- square root 3/2)

Question

Ca somone show me how to solve

Sin-1 (- square root 3/2)

anonymous · Answer

someone*

anonymous · Answer

can't really show how to solve but you need to know your unit circle
the answer is 7pi/3 and 11pi/3

psymon · Answer

sin^-1 and sin are inverse. They basically flip-flop values around. Normally we have:

sinx = some number. 

But what sin^-1 does is say 

sin^-1(some number) = x

You can always move these around in this way. So when your problem says 
sin^-1 (-sqrt(3)/2) = x, I can rewrite that as

sinx = -sqrt(3)/2

So if you dont know the answer, you can always look on the unit circle and check what angle of sin has a value of -sqrt(3)/2 That kind of make sense?

anonymous · Answer

I have the unit circle

 and the book says its - pi over 3 , but how?

psymon · Answer

Well, I can explain that, I just want to make sure you get what I was trying to say above.

anonymous · Answer

ohh okay

anonymous · Answer

sqrt3/2 for sin is pi/3 but because it was negative the answer is in the 3rd and 4th quadrants

psymon · Answer

Kinda make sense? If so, Ill explain the answer.

anonymous · Answer

right costanza, its only in the 3rd and 4th quarters,  but i guess what im trying to find out if its in the 3rd and 4th q, how or where does it = to pi/3

psymon · Answer

I can explain that, but only if you let me. Otherwise ill let costanza do it. Im guessing what I was trying to say didnt make sense then?

anonymous · Answer

no it did not make since sorry.. (lost) , thats fine please explain psymon..

psymon · Answer

Basically, if you ever have sin^-1(y) = x, you can rewrite it as
sinx = y. 

So just examples. 

sin(pi/2) = 1 becomes sin^-1(1) = pi/2
cos(pi) = -1 becomes cos^-1(-1) = pi

As for yours

sin^-1(-sqrt(3)/2) = x
sinx = -sqrt(3)/2

psymon · Answer

Not sure if you catch the pattern or not, though.

anonymous · Answer

o i c, got it!

thanks for explaining!! =)

psymon · Answer

Yep. Now for the answer!

inverse sin MUST MUST MUST be in between -pi/2 and pi/2. So even though sin^-1(-sqrt(3)/2) can be 4pi/3, that is in the 3rd quadrant, which is NOT in between -pi/2 and pi/2. So basically, sin^-1 only has answers in quadrant 1 and 4, nowhere else. Now the two values for which sin is normally -sqrt(3)/2 are 4pi/3 and 5pi/3. Well, 5pi/3 is in the correct quadrant, but it is not in between -pi/2 and pi/2. So instead of calling it 5pi/3, we call it by what the same angleof it is negative, which is -pi/3

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