# Answer on Trigonometry Question for Kewal

Question #5910

sin3A+sin2A-sinA = 4sinAcosA/2cos3A/2

Expert's answer

we will use "sum to product" 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 "sum to product" 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's it.

Dear Kewal

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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 "sum to product" 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's it.

Dear Kewal

For you and other our visitors we created this video. Please take a look!

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## Comments

Assignment Expert27.06.2017 04:01There 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.

Dinesh27.06.2017 02:00find value of 8 sin^3 40 - 6 Cos 40

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