How do you find the exact value of cos2x using the double angle formula given $cos x = - \frac{2}{13}$ $cosx=-2/13$ , $\frac{π}{2} < x < π$ $pi/2<x<pi$ ?

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Answer 1 · 2015-04-29T15:50:11

Use the trig identity: $cos 2 x = 2 {cos}^{2} x - 1$ $cos 2x = 2cos^2 x - 1$

$cos 2 x = 2 {(- \frac{2}{13})}^{2} - 1 = - \frac{161}{169} = - 0.95$ $cos 2x = 2(-2/13)^2 - 1 = -161/169 = -0.95$

Note . $cos x = - \frac{2}{13} - \to x = 98, 85$ $cos x = -2/13 --> x = 98,85$ (Quadrant II) -->
2x = 197.70. (Quadrant III) #

cos 2x = -0.95 gives 2 arcs with same cos value:
2x = 161.80 (Quadrant II) (rejected) and 2x = 198.0 (Quadrant III). OK

How do you find the exact value of cos2x using the double angle formula given cosx=−213cosx=-2/13, π2<x<πpi/2<x<pi?