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The Pythagorean Identity states that $(\sin x)^2+(\cos x)^2=1$. Given $\cos\theta=\frac{3}{4}$, find $\sin\theta$.

Can anyone help explain how to do this? Do I just plug in $\frac34$ into $x$ for cosine? And how do I find sine?

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    Please, type the question instead of using a linked picture. – 2017-01-07

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You know that $(\sin \theta)^2 + (\cos \theta)^2 = 1$. Since $\cos \theta = \frac34$, substitute this in. Then the equation is: $$(\sin \theta)^2 + (\frac34)^2 = 1$$ $$(\sin \theta)^2 + \frac9{16} = 1$$ $$(\sin \theta)^2 = \frac7{16}$$ $$\sin \theta = \sqrt{\frac7{16}}$$ (or $\sin \theta = -\sqrt{\frac7{16}}$) Simplifying gives $$\sin \theta = \frac{\sqrt 7}{4}$$ (or $\sin \theta = -\frac{\sqrt 7}{4}$)

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    $(\sin \theta)^2=7/16\iff \sin \theta \in \{\sqrt 7\;/4, -\sqrt 7\;/4\}.$ There are two possible values for $\sin \theta,$ and there does exist a valid $\theta$ for each of them – 2017-01-07
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    okay then. I just added the negative solution – 2017-01-07
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$$ \sin^2(x)+\cos^2(x)=1$$

Given $\cos(x) = \frac{3}{4} $

$$ \sin^2(x)+ (\frac{3}{4})^2 =1 $$

$$ \sin^2(x)= \frac{7}{16}$$

$$ \sin(x) = ? $$