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Trig

Hi quick question.

On D do you rearrange cos squared theta + sin squared theta = 1 to cos squared theta - 1 = sin squared theta to remove the sin squared theta from the question? I'm a little confused with it.

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Reply 1
Avatar for j4ckd4v13z
j4ckd4v13z
2sin⁡2θ−2cos⁡θ−cos⁡2θ=1 2\sin^2 \theta -2\cos\theta - \cos^2\theta = 1

→2(1−cos⁡2θ)−2cos⁡θ−cos⁡2θ=1\rightarrow 2(1- \cos^2\theta) -2\cos\theta -\cos^2\theta = 1

then solve as quadratic.
Reply 2
Avatar for TenOfThem
TenOfThem
18
Original post by ThrashMetal
Hi quick question.

On D do you rearrange cos squared theta + sin squared theta = 1 to cos squared theta - 1 = sin squared theta


That is not correct

cos⁡2θ+sin⁡2θ=1\cos ^2 \theta + \sin ^2 \theta = 1

Subtract cos⁡2θ\cos ^2 \theta from both sides of the equation
Reply 3
Avatar for ThrashMetal
ThrashMetal
OP
2
Original post by TenOfThem
That is not correct

cos⁡2θ+sin⁡2θ=1\cos ^2 \theta + \sin ^2 \theta = 1

Subtract cos⁡2θ\cos ^2 \theta from both sides of the equation
So Sin squared theta = 1-Cos squared theta? Does that mean I can now remove a sin squared theta from the original equation leaving me with 2-2 cos theta - cos squared theta = 1? I'm so confused and I don't know why.
Reply 4
Avatar for TenOfThem
TenOfThem
18
Original post by ThrashMetal
So Sin squared theta = 1-Cos squared theta? Does that mean I can now remove a sin squared theta from the original equation leaving me with 2-2 cos theta - cos squared theta = 1? I'm so confused and I don't know why.


See post 2
Reply 5
Avatar for ThrashMetal
ThrashMetal
OP
2
Original post by TenOfThem
See post 2
Woops my bad. So basically whatever you rearrange cos^2theta+sin^2theta=1 for, you remove the = X and put brackets around what comes before the equals sign, where the = X was?
Reply 6
Avatar for TenOfThem
TenOfThem
18
Original post by ThrashMetal
Woops my bad. So basically whatever you rearrange cos^2theta+sin^2theta=1 for, you remove the = X and put brackets around what comes before the equals sign, where the = X was?


erm

I really have no idea what you mean by this

sin⁡2θ=1−cos⁡2θ\sin ^2 \theta = 1 - \cos ^2 \theta

So

2sin⁡2θ=2(1−cos⁡2θ)2 \sin ^2 \theta = 2(1 - \cos ^2 \theta)
Reply 7
Avatar for ThrashMetal
ThrashMetal
OP
2
Original post by TenOfThem
erm

I really have no idea what you mean by this

sin⁡2θ=1−cos⁡2θ\sin ^2 \theta = 1 - \cos ^2 \theta

So

2sin⁡2θ=2(1−cos⁡2θ)2 \sin ^2 \theta = 2(1 - \cos ^2 \theta)
Yeah I had trouble explaining what I meant. Basically you can remove what the equation equals to and put what ever is left in brackets and solve from there is what I understand.

Thanks for the help guys, I think I fully understand it now :biggrin:

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