How do you find the exact functional value sec (-11pi/12) using the cosine sum or difference identity?

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How do you find the exact functional value sec (-11pi/12) using the cosine sum or difference identity?

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Answer 1 · 2015-08-09T15:35:30

Find $sec (\frac{- 11 π}{12})$ $sec ((-11pi)/12)$

Ans: $- \frac{4}{\sqrt{2} + \sqrt{6}}$ $- 4/(sqrt2 + sqrt6)$

Explanation:

sec = 1/cos
$cos (\frac{- 11 π}{12}) = cos (\frac{11 π}{12}) = cos (\frac{3 π}{12} + \frac{8 π}{12})$ $cos ((-11pi)/12) = cos ((11pi)/12) = cos ((3pi)/12 + (8pi)/12)$ =

$= cos (\frac{π}{4} + \frac{2 π}{3})$ $= cos (pi/4 + (2pi)/3)$
Use the trig identity: cos (a + b) = cos a.cos b - sin a.sin b

$cos a = cos (\frac{π}{4}) = \frac{\sqrt{2}}{2}$ $cos a = cos (pi/4) = sqrt2/2$ ; $cos b = cos (\frac{2 π}{3}) = - \frac{1}{2}$ $cos b = cos ((2pi)/3) = -1/2$
$sin a = sin (\frac{π}{4}) = \frac{\sqrt{2}}{2}$ $sin a = sin (pi/4) = sqrt2/2$ ; $sin b = sin (\frac{2 π}{3}) = \frac{\sqrt{3}}{2}$ $sin b = sin ((2pi)/3) = sqrt3/2$
Therefor:
$cos (\frac{- 11 π}{12}) = cos (\frac{π}{4} + \frac{2 π}{3}) =$ $cos ((-11pi)/12) = cos (pi/4 + (2pi)/3) =$

$(\frac{\sqrt{2}}{2}) (- \frac{1}{2}) - (\frac{\sqrt{2}}{2}) (\frac{\sqrt{3}}{2})$ $(sqrt2/2)(-1/2) - ((sqrt2)/2)((sqrt3)/2)$ = $- \frac{\sqrt{2} + \sqrt{6}}{4}$ $- (sqrt2 + sqrt6)/4$
Check by calculator:
$cos (\frac{- 11 π}{12}) = cos (- 165) = - 0.966$ $cos ((-11pi)/12) = cos (-165) = - 0.966$
$- \frac{\sqrt{2} + \sqrt{6}}{4} = - 0.966 .$ $- (sqrt2 + sqrt6)/4 = - 0.966.$ OK

Finally: $sec (\frac{- 11 π}{12}) = - \frac{4}{\sqrt{2} + \sqrt{6}}$ $sec ((-11pi)/12) = - 4/(sqrt2 + sqrt6)$ = $- \frac{1}{0.966} = - 1.035$ $- 1/0.966 = - 1.035$

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How do you find the exact functional value sec (-11pi/12) using the cosine sum or difference identity?

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