Solving $\cos(x)\sin(7x)=\cos(3x)\sin(5x)$
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!
$\endgroup$4 Answers
$\begingroup$Hint:
Use the linearisation formula:$$\sin a\cos b=\frac12\bigl(\sin(a+b)+\sin(a-b)\bigr).$$
$\endgroup$ 3 $\begingroup$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.
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$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).
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