How do you find the exact value of the following using the unit circle 6 cos (11pi/6) - 2 sin^2 (5pi/4)?

1 Answer
Mar 17, 2018

color(purple)(6 cos ((11pi)/6) - 2 sin ^2 ((5pi)/4) = 2

Explanation:

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From the above diagram,

hat (11pi)/6 is in IV quadrant where cos is positive.

cos ((11pi)/6) = cos (((11pi)/6) - 2pi) = cos -(pi/6) = cos (pi/6)

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But cos (pi/6) = cos 60 = x = 1/2 :. color(red)(cos ((11pi)/6) = 1/2

hat (5pi)/4 is in III Quadrant where sin is negative.

sin ((5pi)/4) = sin (pi + (pi/4)) = - sin (pi/4) = - sin 45

But sin (pi/4) = 1/sqrt2

:. color(green)(sin ((5pi)/4) = y = -1/sqrt2

Returning to the given sum,

6 cos ((11pi)/6) - 2 sin ^2 ((5pi)/4) = 6 * (1/2) - 2 (-(1/sqrt2))^2

=> (6 * (1/2)) -(2* (1/2)) = 3 - 1 = 2