Question 1174774: Given sina=5/8, with θ in quadrant I, find cos(2a). (I got 15/32 by using the cos(2θ) identity, but I am not sure if I solved it right.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! sin(a) = 5/8
a is in quadrant 1.
you want to find cos(2a).
first find a.
a = arcsin(5/8) = 38.68218745 degrees.
cos(2a) would be equal to cos(2 * 38.68218745) = .21875.
your answer of 15/32 = .46875.
they're not the same, so something must be wrong.
the formula for cos(2a) is cos(2a) = cos^2(a) - sin^2(a)
in order to use the cos(2a) formula, you need to know cos(a).
you are only given sin(a), so you have to find cos(a).
since sine = opposite / hypotenuse, then sin(a) = 5/8 means the side opposite angle a = 5 and the hypotenuse = 8.
you can use the formula of (hypotenuse)^2 = (side opposite)^2 + (side adjacent)^2.
when side opposite = 5 and hypotenuse = 8, this formula becomes 8^2 = 5^2 + (side adjacent)^2.
solve for (side adjacent)^2 to get (side adjacent)^2 = 8^2 - 5^2 = 39.
this makes side adjacent = sqrt(39).
cos(a) is therefore equal to sqrt(39) / 8.
you already know that sin(a) is equal to 5/8.
cos(2a) = cos^2(a) - sin^2(a) becomes:
cos(2a) = (sqrt(39)/8)^2 - (5/8)^2 = .21875.
that's the same as i got above after finding out what the angle was from the sine and then doubling it and finding out what the cosine was.
this is a good indication that the answer is correct.
in fractional form, the answer will be:
(sqrt(39)/8)^2 - (5/8)^2 = 39/64 - 25/64 = 14/64 = 7/32.
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