How do find the general solution of sin7a=sina+sin3a

Step-by-step

Solve  

sin(7a) = sin(3a) + sin(a) (1)

Let  

F(a)=sin(7a)-sin(3a)-sin(a)(2)

the equivalent problem to (1) is  

F(a)=0 (3)

F(a)

=sin(7a)-sin(3a)-sin(a)apply trig sum identities

=sin(6a+a) - sin(3a) - sin(a)

=sin(6a)cos(a)+cos(6a)sin(a) -sin(3a) - sin(a)

apply double angle formulas

=(2sin(3a)cos(3a))cos(a) +

(cos^2(3a)-sin^2(3a))sin(a) -sin(3a) - sin(a)

simplify using sin^2(p)+cos^2(p) = 1

= (2sin(3a)cos(3a))cos(a) +

(1-2sin^2(3a))sin(a) - sin(3a) - sin(a)

simplify algebraically, note 1*sin(a) cancels sin(a)

= (2sin(3a)cos(3a))cos(a) -2sin^2(3a)sin(a) - sin(3a)

factor out sin(3a)

= sin(3a)(2cos(3a)cos(a)-2sin(3a)sin(a) - 1)

now use trig sum formula to reduce to cos(4a)

= sin(3a)(2cos(4a)-1)

So

F(a) = 0   if sin(3a) =0 (4)

or

F(a) = 0   if cos(4a) = 1/2 (5)

using the zero product theorem

From (4)

sin(3a) = 0  

3a = sin-1(0) = n*pi

a = n*pi/3  (6)

From (5)

cos(4a) = 1/2

4a = cos^-1(1/2) = 2n*pi +/- pi/3

a = (2n+1/3)pi/4 or (2n-1/3)pi/4(7)

Combine (6) and (7) to give the general solution

a = n*pi/3 or (2n+1/3)pi/4 or (2n-1/3)pi/4 (8)

For checking, use your calculator to substitute every solution given in (8) into the function F(a) in equation (1) to confirm that the result is almost equality.  The small difference will be due to rounding errors of the calculators.  On mine, they work out to be of the order of +/- 10^-15.

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