Question

Given sin 0=7/4and angle 0 is in quadrant 1 what is the exact value of cos 0 in simplest form? Simplify all radics if needed

Answer 1

Answer:

Answer: ¾

Step-by-step explanation:

» From the trigonometric identities:

${ \tt{ \sin {}^{2} \theta + \cos {}^{2} \theta = 1 }}$

» Substitute to the formular:

${ \tt {( \frac{ \sqrt{7} }{4} )}^{2} + { \cos }^{2} \theta = 1 } \\ \\ { \tt{ \frac{7}{16} + { \cos}^{2} \theta = 1 }} \\ \\ { \tt{ \cos {}^{2} \theta = 1 - \frac{7}{16} }} \\ \\ { \tt{( \cos\theta) {}^{2} = \frac{9}{16} }} \\ \\ { \tt{ \sqrt{ {( \cos \theta}^{2}) } = \sqrt{( \frac{9}{16}) } }} \\ \\ { \tt{ \cos \theta = \frac{ \sqrt{9} }{ \sqrt{16} } }} \\ \\ { \boxed{ \tt{ \cos \theta = \frac{3}{4} }}}$