x + 5 = 9sin theta, 0 < theta < pi/2 , cos(2theta) in terms of x
cos(2 theta)
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
or would it be 9(sin theta - 5/9) i believe i did the work wrong
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
To find cos(2θ) in terms of x, we'll need to use an identity known as the double-angle formula for cosine.
The double-angle formula for cosine states that cos(2θ) = cos^2(θ) - sin^2(θ).
In order to find cos(2θ), we need to write cos^2(θ) and sin^2(θ) in terms of x using the given equation: x + 5 = 9sin(θ).
Let's start by solving for sin(θ) using algebra. Rearrange the equation:
9sin(θ) = x + 5
Divide both sides by 9:
sin(θ) = (x + 5)/9
Now, using the Pythagorean identity sin^2(θ) + cos^2(θ) = 1, we can find cos^2(θ) in terms of x:
cos^2(θ) = 1 - sin^2(θ)
cos^2(θ) = 1 - ((x + 5)/9)^2
cos^2(θ) = 1 - (x^2 + 10x + 25)/81
cos^2(θ) = (81 - x^2 - 10x - 25)/81
cos^2(θ) = (56 - x^2 - 10x)/81
Finally, substituting cos^2(θ) back into the double-angle formula:
cos(2θ) = cos^2(θ) - sin^2(θ)
cos(2θ) = (56 - x^2 - 10x)/81 - ((x + 5)/9)^2
Therefore, cos(2θ) in terms of x is given by (56 - x^2 - 10x)/81 - ((x + 5)/9)^2.