# Chapter 7 - Section 7.3 - Double-Angle, Half-Angle, and Product-Sum Formulas - 7.3 Exercises - Page 561: 6

$sin(2x) = -\sqrt {15}/8$ $cos(2x) = 7/8$ $tan(2x) -\sqrt{15}/7$

#### Work Step by Step

1. Find cos(x) and sin(x): $$sin(x) = \frac{1}{csc(x)} = \frac 14$$ $$sin^2(x) + cos^2(x) = 1$$ $$cos(x) = \pm \sqrt{1 - sin^2(x)}$$ $$cos(x) = \pm \sqrt{1 - (1/4)^2} = \pm \sqrt{15/16}$$ $$tan(x) = \frac{sin(x)}{cos(x)}$$ Since $tan(x) \lt 0$, $\frac{sin(x)}{cos(x)}$ must be negative. sin(x) = 1/4, which is positive. Thus, cos(x) must be negative. $$cos(x) = -\sqrt{15/16} = \frac{-\sqrt {15}}{4}$$ 2. Calculate cos(2x) and sin(2x) $$sin(2x) = 2sin(x)cos(x) = 2(1/4)(\frac{-\sqrt {15}} 4)$$ $$sin(2x) = \frac{-2\sqrt {15}}{16} = -\frac {\sqrt {15}} 8$$ $$cos(2x)= cos^2(x) - sin^2(x) = 15/16 - 1/16 = 7/8$$ 3. Calculate tan(2x) $$tan(2x) = \frac{sin(2x)}{cos(2x)} = \frac{\frac{-\sqrt {15}}{8}}{7/8} = -\frac{\sqrt {15}}{7}$$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.