Prove that sin7A=7sinA-56sin^3A+112sin^5A-64sin^7A

Answer 1

there is a different way, combining Demoivre's theorem and binomial theorem

sin7x=√-1 coeffcient part of (cosx+isinx)⁷

that is sin7x=7*cos⁶x*sinx-35cos⁴x*sin³x+21cos²x*sin⁵x-sin⁷x

use cos²x=1-sin²x, cos⁴x=(1-sin²x)², cos⁶x=(1-sin²x)³

there is still algebra which needs to be carried out carefully.

but you can see the sinx coefficient is +7 and the sin⁷x coefficient is -(7+35+21+1)= -64

so the rest of the algebra concerns coefficints of sin³x and sin⁵x

Answer 2

There's a lot of algebra in this!!! Here goes:

sin(7A) = sin(3A + 4A)

Using the sine angle addition identity: sin(x + y) = sin(x)cos(y) + cos(x)sin(y) we can say:

sin(7A) = sin(3A + 4A) = sin(3A)cos(4A) + cos(3A)sin(4A)

But sin(3A) = sin(A + 2A) and using the same identity that becomes sin(A)cos(2A) + cos(A)sin(2A) giving:

sin(7A) = [sin(A)cos(2A) + cos(A)sin(2A)]cos(4A) + cos(3A)sin(4A)

Using the double angle formulas cos(2A) = 1 - 2sin²(A) and sin(2A) = 2sin(A)cos(A) our equation becomes:

sin(7A) = [sin(A)(1 - 2sin²(A)) + cos(A)(2sin(A)cos(A))]cos(4A) + cos(3A)sin(4A)

and simplifying:

sin(7A) = [sin(A) - 2sin³(A) + 2sin(A)cos²(A)]cos(4A) + cos(3A)sin(4A)

Using the Pythagorean Identity sin²(A) + cos²(A) = 1 or cos²(A) = 1 - sin²(A) it becomes:

sin(7A) = [sin(A) - 2sin³(A) + 2sin(A)(1 - sin²(A))]cos(4A) + cos(3A)sin(4A)

and simplifying:

sin(7A) = [sin(A) - 2sin³(A) + 2sin(A) - 2sin³(A)]cos(4A) + cos(3A)sin(4A)

sin(7A) = [3sin(A) - 4sin³(A)]cos(4A) + cos(3A)sin(4A)

Now, working on cos(4A), we can say cos(4A) = cos(2A + 2A) and using the cosine angle addition identity:

cos(x + y) = cos(x)cos(y) – sin(x)sin(y)

we can say: cos(4A) = cos(2A + 2A) = cos(2A)cos(2A) - sin(2A)sin(2A)

Using the same double angle identities as before this makes:

cos(4A) = (1 - 2sin²(A))(1 - 2sin²(A)) - (2sin(A)cos(A))(2sin(A)cos(A))

cos(4A) = (1 - 4sin²(A) + 4sin⁴(A)) - (4sin²(A)cos²(A))

and again using the Pythagorean Identity cos²(A) = 1 - sin²(A) it becomes:

cos(4A) = (1 - 4sin²(A) + 4sin⁴(A)) - (4sin²(A)(1 - sin²(A)))

and simplifying:

cos(4A) = 1 - 4sin²(A) + 4sin⁴(A) - 4sin²(A) + 4sin⁴(A)

cos(4A) = 1 - 8sin²(A) + 8sin⁴(A)

Then plugging that into our original equation we get:

sin(7A) = [3sin(A) - 4sin³(A)][1 - 8sin²(A) + 8sin⁴(A)] + cos(3A)sin(4A)

Then multiplying out the left side of the equation we get:

sin(7A) = [3sin(A) - 24sin³(A) + 24sin⁵(A) - 4sin³(A) + 32sin⁵(A) - 32sin⁷] + cos(3A)sin(4A)

sin(7A) = [3sin(A) - 28sin³(A) + 56sin⁵(A) - 32sin⁷] + cos(3A)sin(4A)

You can continue the same way on the right side of the equation to convert the cos(3A) and sin(4A) in the same fashion to get the result. If at any point you want to confirm that you have not made an error, you can graph your interim result on desmos and compare it to sin(4A) to make sure you haven't changed anything.