Recently, I was trying to solve a trigonometric equation involving the use of sine and cosine:
$$\cos(x)\sin(7x)=\cos(3x)\sin(5x)$$
I attempted to remove the coefficients of $x$ within the trigonometric functions, but I did not understand how to.
Does anyone understand how I'm supposed to solve this equation?
Thank you!
Hint:
Use the linearisation formula: $$\sin a\cos b=\frac12\bigl(\sin(a+b)+\sin(a-b)\bigr).$$
Hint: there is a set of identities called by names such as product rules or product-to-sum formulas. One of them has $\sin(a)\cos(b)$ on the left-hand side.
Multiply both sides by $2,$ and use the identity $$2\cos x\sin y=\sin (x+y)-\sin(x-y).$$ This gives $$\sin 8x-\sin(-6x)=\sin 8x-\sin(-2x),$$ or in other terms $$\sin 6x-\sin 2x=0.$$ Here you may convert this into a product by using $$\sin a-\sin b=2\cos\left(\frac{a+b}{2}\right)\sin\left(\frac{a-b}{2}\right),$$ to get a product that vanishes, then use the fact that if $AB=0,$ then $A=0$ or $B=0$ to finish off the problem.
PS. Note that if $\cos m=0,$ then $m=(1+2j)π/2,$ and if $\sin n=0,$ then $n=kπ,$ where $j,k$ are arbitrary integers.
An alternate path is to use,
$\cos(x) = \frac{1}{2}(e^{ix} + e^{-ix})$ and $\sin(x) = \frac{1}{2i}(e^{ix} - e^{-ix})$. Let $z=e^{ix}$ and keep in mind that $|z|=1$. After some algebra you will get the polynomial,
$$z^{14} - z^{10} + z^6 - z^2 = 0 $$
Which factors as
$$z^2(z^8 + 1)(z^4 - 1) = 0$$
Since $|z| = 1$ we can ignore the $z^2$. Then we have solutions when either $e^{i8x} = -1$ or $e^{i4x} = 1$. The first equation is satisfied when $8x = (1 + 2k)\pi$ (an odd integer times pi) and the second is satisfied when $4x = 2k\pi$ (an even integer times pi).
Using the product-to-sum identity $\sin{A}\cos{B}=\frac{\sin{(A+B)}+\sin{(A-B)}}{2}$ in turn to each side:
LHS: $\sin{(7x)}\cos{(x)}=\frac{\sin{8x}+sin{6x}}{2}$
RHS: $\sin{(5x)}\cos{(3x)}=\frac{\sin{(8x)}+sin{(2x)}}{2}$
Equating the two, the $\frac{1}{2}$ and the $\sin{(8x)}$ both cancel and we are left with $\sin{(6x)}=\sin{(2x)}$.
From here, geometry/symmetry of the unit circle can be used to obtain the solutions $x=0, \pi, 2\pi...$
More generally, $x=n\pi$ where n is any integer. Other solutions exist such as $x=\frac{-3\pi}{8}+n\pi$ but these require diagrams beyond my ability to produce for the moment in this space.
