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 · 2015-08-19T15:08:19

$sin (\frac{π}{12}) = (\frac{\sqrt{3} - 1}{2 \sqrt{2}})$ $sin (pi/12)=((sqrt(3)-1)/(2sqrt(2)))$
$cos (\frac{11 π}{12}) = - (\frac{1 + \sqrt{3}}{2 \sqrt{2}})$ $cos((11pi)/12)=-((1+sqrt(3))/(2sqrt(2)))$
$tan (\frac{7 π}{12}) = - 2 - \sqrt{3}$ $tan ((7pi)/12)=-2-sqrt(3)$

Explanation:

$sin (\frac{π}{12})$ $sin (pi/12)$
$= sin (\frac{π}{3} - \frac{π}{4})$ $=sin(pi/3 - pi/4)$
$= sin (\frac{π}{3}) cos (\frac{π}{4}) - cos (\frac{π}{3}) sin (\frac{π}{4})$ $=sin(pi/3)cos(pi/4)-cos(pi/3)sin(pi/4)$
$= (\frac{\sqrt{3}}{2}) (\frac{1}{\sqrt{2}}) - (\frac{1}{2}) (\frac{1}{\sqrt{2}})$ $=(sqrt(3)/2)(1/sqrt(2))-(1/2)(1/sqrt(2))$
$= (\frac{\sqrt{3} - 1}{2 \sqrt{2}})$ $=((sqrt(3)-1)/(2sqrt(2)))$

$cos (\frac{11 π}{12})$ $cos((11pi)/12)$
$= cos (\frac{2 π}{3} + \frac{π}{4})$ $=cos((2pi)/3+pi/4)$
$= cos (\frac{2 π}{3}) cos (\frac{π}{4}) - sin (\frac{2 π}{3}) sin (\frac{π}{4})$ $=cos((2pi)/3)cos(pi/4)-sin((2pi)/3)sin(pi/4)$
$= cos (π - \frac{π}{3}) cos (\frac{π}{4}) - sin (π - \frac{π}{3}) sin (\frac{π}{4})$ $=cos(pi-pi/3)cos(pi/4)-sin(pi-pi/3)sin(pi/4)$
$= - cos (\frac{π}{3}) cos (\frac{π}{4}) - sin (\frac{π}{3}) sin (\frac{π}{4})$ $=-cos(pi/3)cos(pi/4)-sin(pi/3)sin(pi/4)$
$= - (\frac{1}{2}) (\frac{1}{\sqrt{2}}) - (\frac{\sqrt{3}}{2}) (\frac{1}{\sqrt{2}})$ $=-(1/2)(1/sqrt(2))-(sqrt(3)/2)(1/sqrt(2))$
$= - (\frac{1 + \sqrt{3}}{2 \sqrt{2}})$ $=-((1+sqrt(3))/(2sqrt(2)))$

$tan (\frac{7 π}{12})$ $tan ((7pi)/12)$
$= tan (\frac{π}{3} + \frac{π}{4})$ $=tan(pi/3 + pi/4)$
$= \frac{\sqrt{3} + 1}{1 - \sqrt{3}}$ $=(sqrt(3) + 1)/(1 - sqrt(3))$
$= \frac{(\sqrt{3} + 1) (1 + \sqrt{3})}{1 - 3}$ $=((sqrt(3) + 1)(1+sqrt(3)))/(1 - 3)$
$= - 2 - \sqrt{3}$ $=-2-sqrt(3)$

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

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