# Answer to Question #5910 in Trigonometry for Kewal

Question #5910
sin3A+sin2A-sinA = 4sinAcosA/2cos3A/2
1
2012-02-14T12:00:01-0500
we will use &quot;sum to product&quot; formula for sum of sines
sin(x)+sin(y)=2sin((x+y)/2) cos((x-y)/2)

whence sin(3A)-sin(A)=sin(3A)+sin(-A)=2sin(A)cos(2A)

hence we obtain sin3A+sin2A-sinA=2sin(A)cos(2A)+sin(2A)
Recall that sin(2A)=2sin(A)cos(A)

therefore

sin3A+sin2A-sinA=2sin(A)cos(2A)+sin(2A)=
=2sin(A)cos(2A)+2sin(A)cos(A)=2sin(A)(cos(2A)+cos(A))

there is the same &quot;sum to product&quot; formula for sum of cosines
cos(x)+cos(y)=2cos((x+y)/2)cos((x-y)/2)

therefore

2sin(A)(cos(2A)+cos(A))= 2sin(A)*2*cos(3*A/2)*cos(A/2)=
=4sinAcosA/2cos3A/2 and that equals to the right part of the initial equality, so that&#039;s it.

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Assignment Expert
27.06.17, 11:01

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.

Dinesh
27.06.17, 09:00

find value of 8 sin^3 40 - 6 Cos 40