Solving for theta

**cordelljheckel** · January 19th 2010, 11:37 AM

Solve sin^2ϴ = 1 + cosϴ 0 ≤ ϴ ≤ 2pi

I dont even know where to start! any help would be appreciated thank you

**e^(i*pi)** · January 19th 2010, 11:44 AM

Originally Posted by cordelljheckel

Solve sin^2ϴ = 1 + cosϴ 0 ≤ ϴ ≤ 2pi

I dont even know where to start! any help would be appreciated thank you

$sin^2 \theta = 1-cos^2 \theta$

Sub in, 1 will cancel and solve the quadratic

**cordelljheckel** · January 19th 2010, 11:48 AM

so am i substituting sin^2(x) with 1 - cos2x / 2 then solving?

**Krizalid** · January 19th 2010, 12:05 PM

No, that wasn't the given indication. Read it again.

**cordelljheckel** · January 19th 2010, 12:36 PM

i dont understand, which identity are you using to sub in?

**Defunkt** · January 19th 2010, 02:14 PM

Originally Posted by cordelljheckel

i dont understand, which identity are you using to sub in?

Just like e^(i*pi) said,

$sin^2(\theta) = 1 - cos^2(\theta)$

Substitute into the first equation to get:

$1-cos^2\theta = 1 + cos\theta$

Now solve.

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