Asked by Samantha King
x + 5 = 9sin theta, 0 < theta < pi/2 , cos(2theta) in terms of x
cos(2 theta)
cos(2 theta)
Answers
Answered by
Samantha
I know i would bring the 5 over to the other side so it would look like
x = 9sin theta - 5
would I distribute the 9 to the sin and the -5?
So x = 9(sin theta - 45)
just confused on how to get to cos 2 theta
x = 9sin theta - 5
would I distribute the 9 to the sin and the -5?
So x = 9(sin theta - 45)
just confused on how to get to cos 2 theta
Answered by
Samantha
or would it be 9(sin theta - 5/9) i believe i did the work wrong
Answered by
Reiny
since you want cos 2Ø in terms of x from
x+5 = 9sinØ
bringing the 5 over would make things worse
... and no, you would not distribute the 9 over
the 9sinØ - 5
I would proceed this way:
sinØ = ( x+5)/9
now you will have to use one of the half-angle formulas
recall that cos 2Ø = 1 - 2sin^2 Ø
= 1 - 2(x+5)^2 /81
= (81 - 2x^2 - 20x - 50)/81
= (31 - 2x^2 - 20x)/81
x+5 = 9sinØ
bringing the 5 over would make things worse
... and no, you would not distribute the 9 over
the 9sinØ - 5
I would proceed this way:
sinØ = ( x+5)/9
now you will have to use one of the half-angle formulas
recall that cos 2Ø = 1 - 2sin^2 Ø
= 1 - 2(x+5)^2 /81
= (81 - 2x^2 - 20x - 50)/81
= (31 - 2x^2 - 20x)/81