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Simplify

 

\(4 \sin x \sin (60^\circ - x) \sin (60^\circ + x)\)


The answer will be a trigonometric function of some simple function of x like "2sinx" or "4cosx".

 

I think the way to solve this is by using Sum to Product and Product to Sum formulas but whenever i try i get a wrong answer.sad

 Jul 4, 2020
 #1
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I'm going to use these identities:

1)  sin(A - B)  =  sin(A)cos(B) - cos(A)sin(B)

2)  sin(A + B)  =  sin(A)cos(B) + cos(A)sin(B)  

3)  sin(3A)  =  3sin(A) - 4sin2(A)

 

4 · sin(x) · sin(60 -x) · sin(60+x)

=  4 · sin(x) · [ sin(60 - x) ] · [ sin(60 + x) ]

=  4 · sin(x) · [ sin(60)cos(x) - cos(60)sin(x) ] · [ sin(60)cos(x) + cos(60)sin(x) ]

=  4 · sin(x) · [ sqrt(3)cos(x)/2 - sin(x)/2 ] · [ sqrt(3)cos(x)/2 + sin(x)/2 ]

=  4 · sin(x) · [ 3cos2(x)/4 - sin2(x)/4 ]

=  sin(x) · [ 3cos2(x) - sin2(x) ]

=  sin(x) · [ 3( 1 - sin2(x) ) - sin2(x) ]

=  sin(x) · [ 3 - 3sin2(x) - sin2(x) ]

=  sin(x) · [ 3 - 4sin2(x) ]

=  3sin(x) - 4sin3(x)

=  sin(3x) 

 Jul 4, 2020
 #2
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Sorry but I don't understand how you got identity 3. The first two i reconize but the third one looks new to me, can you explain it or derive it please?

 

Oh but thanks for answering my question smiley

Guest Jul 5, 2020

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