Answer to Question #5910 in Trigonometry for Kewal
hence we obtain sin3A+sin2A-sinA=2sin(A)cos(2A)+sin(2A)
Recall that sin(2A)=2sin(A)cos(A)
there is the same "sum to product" formula for sum of cosines
=4sinAcosA/2cos3A/2 and that equals to the right part of the initial equality, so that's it.
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There is an identity sin(3x)=3sin(x)-4sin^3(x), hence sin^3(x)=(3sin(x)-sin(3x))/4 and
8sin^3(x)-6cos(x)=8(3sin(x)-sin(3x))/4-6cos(x)=6sin(x)-2sin(3x)-6cos(x). The value of 8sin^3(40 degrees)-6cos(40 degrees) is equal to -2.47159.
find value of 8 sin^3 40 - 6 Cos 40