How do you find the exact values of sin(theta/2), cos(theta) where cos theta=4/5 0<=theta<=pi/2?

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How do you find the exact values of sin(theta/2), cos(theta) where cos theta=4/5 0<=theta<=pi/2?

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Answer 1 · 2016-09-23T02:47:39

$sin (\frac{t}{2}) = \frac{\sqrt{10}}{10}$ $sin (t/2) = sqrt10/10$
$cos (\frac{t}{2}) = \frac{3 \sqrt{10}}{10}$ $cos (t/2) = (3sqrt10)/10$

Explanation:

$cos t = \frac{4}{5}$ $cos t = 4/5$ . Find $sin (\frac{t}{2}) and cos (\frac{t}{2})$ $sin (t/2) and cos (t/2)$ .
Use the 2 trig identities:
$2 {cos}^{2} (\frac{t}{2}) = 1 + cos t$ $2cos ^2 (t/2) = 1 + cos t$
$2 {sin}^{2} (\frac{t}{2}) = 1 - cos t$ $2sin^2 (t/2) = 1 - cos t$
We have:
$2 {cos}^{2} (\frac{t}{2}) = 1 + cos t = 1 + \frac{4}{5} = \frac{9}{5}$ $2cos^2 (t/2) = 1 + cos t = 1 + 4/5 = 9/5$
${cos}^{2} (\frac{t}{2}) = \frac{9}{10}$ $cos^2 (t/2) = 9/10$
$cos (\frac{t}{2}) = \pm \frac{3}{\sqrt{10}} = \pm \frac{3 \sqrt{10}}{10}$ $cos (t/2) = +- 3/sqrt10 = +- (3sqrt10)/10$
Since t is in Quadrant I, $cos (\frac{t}{2})$ $cos (t/2)$ is positive --> $\frac{cos t}{2} = \frac{3 \sqrt{10}}{10}$ $cos t/2 = (3sqrt10)/10$
$2 {sin}^{2} (\frac{t}{2}) = 1 - cos t = 1 - \frac{4}{5} = \frac{1}{5}$ $2sin^2 (t/2) = 1 - cos t = 1 - 4/5 = 1/5$
${sin}^{2} (\frac{t}{2}) = \frac{1}{10}$ $sin^2 (t/2) = 1/10$
$sin (\frac{t}{2}) = \pm \frac{1}{\sqrt{10}} = \pm \frac{\sqrt{10}}{10}$ $sin (t/2) = +- 1/sqrt10 = +- sqrt10/10$
Since t is in quadrant I, $sin (\frac{t}{2})$ $sin (t/2)$ is positive -->
$sin (\frac{t}{2}) = \frac{\sqrt{10}}{10}$ $sin (t/2) = sqrt10/10$

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How do you find the exact values of sin(theta/2), cos(theta) where cos theta=4/5 0<=theta<=pi/2?

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