Answered

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if cos theta is -4/7 and in the third quadrant, find the exact value of sin theta

Sagot :

In the third quadrant, the value of the cosine function is negative and so is the sine function.

Remember the identity

[tex]\sin ^2\theta+\cos ^2\theta=1[/tex]

Then,

[tex]\begin{gathered} \sin ^2\theta=1-\cos ^2\theta=1-(-\frac{4}{7})^2=1-\frac{16}{49}=\frac{33}{49} \\ \Rightarrow\sin ^2\theta=\frac{33}{49} \\ \Rightarrow\sin \theta=\pm\sqrt[]{\frac{33}{49}}=\pm\frac{\sqrt[]{33}}{7} \end{gathered}[/tex]

But remember that we are in the third quadrant; therefore, the value of sine has to be negative

[tex]\Rightarrow\sin \theta=-\frac{\sqrt[]{33}}{7}[/tex]

The answer is -sqrt(33)/7

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