Find the exact value of sin(pi/12) , cos (11pi/12) , and tan (7pi/12)?

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Find the exact value of sin(pi/12) , cos (11pi/12) , and tan (7pi/12)?

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Answer 1 · 2015-09-25T01:51:28

Find sin (pi/12); cos (pi/12) and tan ((7pi)/12)

Explanation:

Call $\frac{sin π}{12} = sin t .$ $sin pi/12 = sin t.$ Use trig identity: $cos 2 a = 1 - {sin}^{2} a$ $cos 2a = 1 - sin^2 a$
$cos 2 \frac{π}{12} = \frac{cos π}{6} = \frac{\sqrt{3}}{2} = 1 - {sin}^{2} t .$ $cos 2pi/12 = cos pi/6 = sqrt3/2 = 1 - sin^2 t.$
${sin}^{2} t = 1 - \frac{\sqrt{3}}{2} = \frac{2 - \sqrt{3}}{4}$ $sin^2 t = 1 - sqrt3/2 = (2 - sqrt3)/4$
$sin t = \pm \frac{\sqrt{2 - \sqrt{3}}}{2}$ $sin t = +- sqrt(2 - sqrt3)/2$ . Since $\frac{π}{12}$ $pi/12$ has its sin positive, then
$sin (\frac{π}{12}) = sin t = \frac{\sqrt{2 - \sqrt{3}}}{2}$ $sin (pi/12) = sin t = sqrt(2 - sqrt3)/2$

Call $cos (\frac{π}{2}) = cos t$ $cos (pi/2) = cos t$
$cos (\frac{2 π}{12}) = \frac{\sqrt{3}}{2} = 2 {cos}^{2} t - 1$ $cos ((2pi)/12) = sqrt3/2 = 2cos^2 t - 1$
${cos}^{2} t = \frac{2 + \sqrt{3}}{4}$ $cos^2 t = (2 + sqrt3)/4$
$cos t = \pm \frac{\sqrt{2 + \sqrt{3}}}{2}$ $cos t = +- sqrt(2 + sqrt3)/2$ . Since cos $(\frac{π}{12})$ $(pi/12)$ is positive, then
$cos (\frac{π}{12}) = cos t = \frac{\sqrt{2 + \sqrt{3}}}{2}$ $cos (pi/12) = cos t = sqrt(2 + sqrt3)/2$

$tan (\frac{7 π}{12}) = tan (\frac{π}{12} + π) = tan (\frac{π}{12}) = \frac{sin}{cos} =$ $tan ((7pi)/12) = tan (pi/12 + pi) = tan (pi/12) = sin/(cos) =$
$= \frac{\sqrt{2 - \sqrt{3}}}{\sqrt{2 + \sqrt{3}}}$ $= (sqrt(2 - sqrt3))/(sqrt(2 + sqrt3))$

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Find the exact value of sin(pi/12) , cos (11pi/12) , and tan (7pi/12)?

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