Expert: Scott A Wilson - 9/10/2009

Question
prove that (sinA.sin2A+sin3A.sinA6A)/(sinA.cos2A+sin3A.cos6A)=5tanA

(sinA.sin2A+sin3A.sinA6A)/(sinA.cos2A+sin3A.cos6A)
=(2sinA.sin2A+2sin3A.sinA6A)/(sinA.cos2A+2sin3A.cos6A)
=(cos(2A-A)-cos(2A+A)+cos(6A+3A)-cos(6A+3A))/(sin(2A+A)-sin(2A-A)+sin(6A+3A)-sin(6A-3A))
=(cosA-cos3A+cos3A-cos9A)/(sin3A-sinA+sin9A-sin3A)
=(cosA-cos9A)/(sin9A-sinA)
=(2sin5A.sin4A)/(2cos5A.sin4A)=tan5A

sir please explain how this statement (sinA.sin2A+sin3A.sinA6A)/(sinA.cos2A+sin3A.cos6A)

became (2sinA.sin2A+2sin3A.sinA6A)/(sinA.cos2A+2sin3A.cos6A).please reply

Answer
I'm not sure what sinA6A is.  Is it suppose to be sin6A?
I don't see how that is possible.

It looks like the question should be
"prove that (sinA•sin2A + sin3A•sin6A)/(sinA•cos2A + sin3A.cos6A) = 5tanA".

I don't follow where the equations came from, so I'll redo it all.

We need to find sin2A, cos2A, cos3A, sin3A, sin6A, and cos6A.

It is known that sin2A = 2sinA•cosA.
It is also known that cos2A = 2cos²A-1.

We can then say sin3A = sin(2A+A) = sin2A•cosA + sinA•cos2A.
Since we already gave sin2A and cos2A, we can say
sin3A = (2sinA•cosA)cosA + sinA(cos²A-1) = 2sinA•cos²A + sinA•cos²A - sinA.
That's the same as sin3A = 3sinA•cos²A.

For cos3A, we know that is cos(2A+A).
It is then known that cos(2A+A) = cos2A•cosA - sin2A•sinA.
Since the equation for cos2A and sin2A were just given,
we can say that cos2A•cosA - sin2A•sinA changes into
(2cos²A-1)cosA - (2sinA•cosA)sinA. This gives us cos3A = 2cos³A - cosA - 2sin²AcosA.
Now since sin²A + cos²A = 1, we can say that sin²A = 1 - cos²A.
Putting this in gives cos3A = 2cos³A - cosA - 2(1-cos²A)cosA.
Multiplying this out gives cos3A = 2cos³A - cosA - 2cosA + 2cos³A.
This reduces to cos3A = 4cos³A - 3cosA.

For sin6A, we know that it is sin(2(3A))=2sin3A•cos3A.
Since sin3A = 3sinA•cos²A and cos3A = 4cos³A - 3cosA, we can say this is
sin6A = 2(3sinA•cos²A)(4cos³A - 3cosA).

For cos6A, we can say that this is cos(2(3A)) = 2cos²3A - 1.
Now since we know that cos3A = 4cos³A - 3cosA, as we have to do is square it,
and that can be put in our previous equation.
It can now be said the cos6A = 2(4cos³A - 3cosA)² - 1,
which works out to be 2(16cos^6(A) - 24cos^4(A) + 9cos²A) - 1, so
cos6A = 32cos^6(A) - 48cos^4(A) + 18cos²A - 1.

So we have
sin2A = 2sinA•cosA,
cos2A = 2cos²A - 1,
sin3A = 3sinA•cos²A,
cos3A = 4cos³A - 3cosA,
sin6A = 2(3sinA•cos²A)(4cos³A - 3cosA), and
cos6A =  32cos^6(A) - 48cos^4(A) + 18cos²A - 1.

These can be put back in the original and it should reduce.  That is,
(sinA(2sinAcosA) + 3sinA•cos²A(2)(3sinA•cos²A)(4cos³A - 3cosA))
(sinA(2cos²A-1) + (3sinA•cos²A)(32cos^6(A) - 48cos^4(A) + 18cos²A - 1)

is suppose to reduce to 5tanA.  I know 5tanA is 5sinA/5cosA, so what I'll do is print this much out and check what I've done so far, then I'll try and do the rest.