SOLUTION: Using the exact value of the sine and the cosine of 1/6pi and 3/4pi and one of the sum and difference formulas, show that the exact value of sin(7/12pi) is 1/4(sqrt2+sqrt6)
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-> SOLUTION: Using the exact value of the sine and the cosine of 1/6pi and 3/4pi and one of the sum and difference formulas, show that the exact value of sin(7/12pi) is 1/4(sqrt2+sqrt6)
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Question 909400: Using the exact value of the sine and the cosine of 1/6pi and 3/4pi and one of the sum and difference formulas, show that the exact value of sin(7/12pi) is 1/4(sqrt2+sqrt6)
My answer is shown below but i get a minus not a plus:
sin(7/12pi) = sin(3/4pi - 1/6pi)
sin(3/4pi)cos(1/6pi)-cos(3/4pi)sin(1/6pi)
(1/2sqrt2)(1/2sqrt3) - (1/2sqrt2)(1/2)
=1/4(sqrt6 - sqrt2) but it needs to be plus not minus can you please tell me where i'm going wrong? Many thanks Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! Using the exact value of the sine and the cosine of 1/6pi and 3/4pi and one of the sum and difference formulas, show that the exact value of sin(7/12pi) is 1/4(sqrt2+sqrt6)
My answer is shown below but i get a minus not a plus:
sin(7/12pi) = sin(3/4pi - 1/6pi)
= sin(3/4pi)cos(1/6pi)-cos(3/4pi)sin(1/6pi)
Note: cos((3/4)pi) = (-1/2)sqrt(2)
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((1/2)(sqrt2))((1/2)(sqrt3)) - ((-1/2)sqrt2))(1/2)
=1/4(sqrt6 + sqrt2)
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Cheers,
Stan H.
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