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Mathematics 8 Online
OpenStudy (anonymous):

***Medal AND Fan Rewarded*** Solve for cosθ in sin^2θ+cos^2θ=1

OpenStudy (anonymous):

subtract \(\sin^2(\theta)\) from both sides

OpenStudy (anonymous):

This is an identity.

OpenStudy (anonymous):

can you tell me the answer i should have gotten? @satellite73

OpenStudy (anonymous):

\[\sin^2θ+\cos^2θ=1\]\[\cos^2θ=1-\sin^2θ\]\[\cos\theta=\sqrt{1-\sin^2\theta}\]

OpenStudy (anonymous):

dont be confused by the fact that its "cos \(\theta\)" and "sin\(\theta\)" if it really throws you off that much just look at it with variables instead:\[sin^2\theta+\cos^2\theta=1\]\[~~~\color{blue}{a}^2~~+~~~\color{blue}{b}^2~~=1\]\[~~~~~~~~~~~~~~~~b^2~~=1~-~a^2\]\[~~~~~~~~~~~~~~~~b~~~=\sqrt{~1~-~a^2~}\]now just replace \(a\) and \(b\) with \(\sin\theta\) and \(\cos\theta\):\[b=\sqrt{1-a^2}\]\[cos\theta=\sqrt{1-\sin^2\theta}\]

OpenStudy (anonymous):

Thanks for your response!

OpenStudy (anonymous):

hope it helped :)

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