Solve sin^2ϴ = 1 + cosϴ 0 ≤ ϴ ≤ 2pi I dont even know where to start! any help would be appreciated thank you
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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 $\displaystyle sin^2 \theta = 1-cos^2 \theta$ Sub in, 1 will cancel and solve the quadratic
so am i substituting sin^2(x) with 1 - cos2x / 2 then solving?
No, that wasn't the given indication. Read it again.
i dont understand, which identity are you using to sub in?
Originally Posted by cordelljheckel i dont understand, which identity are you using to sub in? Just like e^(i*pi) said, $\displaystyle sin^2(\theta) = 1 - cos^2(\theta)$ Substitute into the first equation to get: $\displaystyle 1-cos^2\theta = 1 + cos\theta$ Now solve.
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