69 584
Assignments Done
Successfully Done
In February 2019

Answer to Question #5910 in Trigonometry 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)



there is the same "sum to product" formula for sum of cosines


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!

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!


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.

27.06.17, 09:00

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

Leave a comment

Ask Your question

Privacy policy Terms and Conditions