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March 31, 2015

March 31, 2015

Posted by **anon** on Thursday, March 3, 2011 at 1:35am.

a.)x=y+cos è

b.)cosy=x^2

(can you show how you would solve)

a.)

x= y+ cos è

cos è = x-y

theta = cos^-1(x-y)

b.)

cosy=x^2

cos(y) = x^2

y = Cos^-1(x^2)

- Trigonometry -
**MathMate**, Thursday, March 3, 2011 at 8:31amThey are both correct!

- Trigonometry -
**Reiny**, Thursday, March 3, 2011 at 8:34amin a) is è a variable or a constant.

If it is a variable then we don't find an "inverse"

(the inverse is found by interchanging the x and y variables in a 2 variable relation)

in b)

original : cosy = x^2

inverse: cosx = y^2

solving this for y:

y = ±√cosx,

solving this for x:

x = cos^-1 (y^2) , where -1 < y < 1

- Trigonometry -
**MathMate**, Thursday, March 3, 2011 at 1:17pmReiny, I interpret "inverse" being inverse trigonometric function, since he is doing a trigonometry course. However, I stand to be corrected!

- Trigonometry -
**Reiny**, Thursday, March 3, 2011 at 3:06pmYou are right.

Funny how one's mind can get stuck along one track, and other possibilities get blocked.

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