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Factor the algebraic expression below in terms of a single trigonometric function. cos x - sin ^2x - 1
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you remember the property from the unit circle. \[\large \sin^2(x)+\cos^2(x) = 1\]
Yes I do
Solve that for sin^2(x) \[\sin^2(x)=1-\cos^2(x)\] Put that sin^2(x) into the expression for sin^2(x) \[\cos(x)-(1-\cos^2(x)) -1\]
That gives you \[\large \cos^2(x) + \cos(x) - 2\]
You can factor that out to \[\large (\cos x + 2)*(\cos x - 1)\]
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Oh wow okay thanks!
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