# Chapter 6 - Analytic Trigonometry - Section 6.5 Sum and Difference Formulas - 6.5 Assess Your Understanding - Page 508: 2

$\cos \theta=-\frac{3}{5}$

#### Work Step by Step

Given: $\sin \theta=\frac{4}{5}$ We know the Identity : $\sin ^{2} \theta+\cos ^{2} \theta=1$ By Substituting we get $\left(\frac{4}{5}\right)^{2}+\cos ^{2} \theta=1$ $\cos ^{2} \theta=1-\left(\frac{4}{5}\right)^{2}$ $\cos ^{2} \theta=1-\frac{16}{25}$ $\cos ^{2} \theta=\frac{25-16}{25}$ $\cos ^{2} \theta=\frac{9}{25}$ $\cos \theta=\pm\sqrt{\frac{9}{25}}$ $\cos \theta=\pm \frac{3}{5}$ Given theta in quadrant $II$, then cosine must be negative. Therefore $\cos \theta=-\frac{3}{5}$.

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