Solve $\sin 84^\circ \sin(54^\circ-x)=\sin 126^\circ \sin x$.

Question

Find $x$ in degrees, where $$\sin 84^\circ\cdot \sin(54^\circ-x)=\sin126^\circ\cdot \sin x\,.$$

I tried to use trigonometry identities to transform the product in sums, but I can't simplify moreover. I know the result is 30° since I solved it in a calculator, but there must be an algebraic way.

For context, this equation comes from solving this problem: How can I solve this geometry problem without trigonometry?

Answer 1

Use

$$\cos36 = \frac {\sin108}{2\sin36} = \frac {\sin36+2\sin36\cos72}{2\sin36}=\frac12+\cos72$$

to factorize the equation as follows

$$\begin{align} & \sin 84\sin(54-x)-\sin 54 \sin x \\ & =\cos 6 \cos 36 \cos x - ( \cos 36+\cos 6 \sin 36 )\sin x \\ & =\frac12(\cos30+\cos 42 ) \cos x - \left(\frac12 + \cos 72 + \frac12(\sin 42+ \sin30 )\right)\sin x \\ & =\frac12\left(\frac{\sqrt3}2+\cos 42 \right) \cos x - \left(\frac34 +\cos (42+30) + \sin30\sin 42\right)\sin x \\ & =\frac12\left(\frac{\sqrt3}2+\cos 42 \right) \cos x - \left(\frac34 + \frac{\sqrt3}2\cos42 \right)\sin x \\ & =\frac12\left(\frac{\sqrt3}2+\cos 42 \right) (\cos x -\sqrt3 \sin x) = 0 \end{align}$$

Thus,

$$\tan x = \frac1{\sqrt3}$$

and the angle in the source problem is $30^\circ$.

Answer 2

This solution unfortunately requires that the answer is known beforehand. It only provides a proof that $x\equiv 30^\circ \pmod{180^\circ}$.

Let $\theta:=18^\circ$ and $s:=\sin(\theta)$. We see that $$3\theta=54^\circ=90^\circ-36^\circ=90^\circ-2\theta\,,$$ so $$3s-4s^3=\sin(3\theta)=\cos(2\theta)=1-2s^2\,.$$ This means $$(s-1)(4s^2+2s-1)=4s^3-2s^2-3s+1=0.$$ Because $s\neq 1$ obviously, $4s^2+2s-1=0$ which means $$\sin(\theta)=s=\frac{1}{2}-2s^2=(1-2s^2)-\frac12=\big(1-2\sin^2(\theta)\big)-\frac12\,.$$ Hence, $$\sin(\theta)=\big(1-2\sin^2(\theta)\big)-\frac12=\cos(2\theta)-\frac12\,,$$ or $$\cos(36^\circ)-\sin(18^\circ)=\frac12\,.$$ (From the result above, we see that $s=\dfrac{-1\pm\sqrt{5}}{4}$. As $s>0$. we must have $\sin(18^\circ)=s=\dfrac{-1+\sqrt{5}}{4}$. This shows that $\cos(36^\circ)=\sin(18^\circ)+\dfrac12=\dfrac{1+\sqrt{5}}{4}$.)

This is equivalent to $$\sin(126^\circ)-\cos(72^\circ)=\frac12\,.$$ By writing $72^\circ=30^\circ+42^\circ$, we see that $$\cos(72^\circ)=\cos(30^\circ)\cos(42^\circ)-\sin(30^\circ)\sin(42^\circ)\,.$$ That is, $$\begin{align}\frac12\sin(42^\circ)+\sin(126^\circ)&=\sin(126^\circ)+\sin(30^\circ)\sin(42^\circ)\\&=\frac12+\cos(30^\circ)\cos(42^\circ)\,.\end{align}$$ Because $138^\circ=180^\circ-42^\circ$, we have $$\frac12\sin(138^\circ)+\sin(126^\circ)=\frac12-\cos(30^\circ)\cos(138^\circ)\,.$$ Since $\sin(30^\circ)=\cos(60^\circ)$, we obtain $$\frac12\big(\sin(30^\circ)+\sin(138^\circ)\big)+\sin(126^\circ)=\frac{1+\cos(60^\circ)}{2}-\cos(30^\circ)\cos(138^\circ)\,.$$ Thus, $$\sin(84^\circ)\cos(54^\circ)+\sin(126^\circ)=\cos^2(30^\circ)-\cos(30^\circ)\cos(138^\circ)\,.$$ As $$\cos(30^\circ)-\cos(138^\circ)=2\sin(84^\circ)\sin(54^\circ)=\frac{\sin(84^\circ)\sin(54^\circ)}{\sin(30^\circ)}\,,$$ we conclude that $$\begin{align}\cos^2(30^\circ)-\cos(30^\circ)\cos(138^\circ)&=\cos(30^\circ)\,\big(\cos(30^\circ)-\cos(138^\circ)\big)\\&=\frac{\cos(30^\circ)}{\sin(30^\circ)}\,\sin(84^\circ)\sin(54^\circ)\,,\end{align}$$ whence $$\tan(30^\circ)=\frac{\sin(30^\circ)}{\cos(30^\circ)}=\frac{\sin(84^\circ)\sin(54^\circ)}{\sin(84^\circ)\cos(54^\circ)+\sin(126^\circ)}\,.$$

If $x$ satisfies the given equation, then from $$ \sin(54^\circ -x )=\sin(54^\circ)\cos(x)-\cos(54^\circ)\sin(x)\,,$$ we must have $$\begin{align}\sin(126^\circ)\sin(x)&=\sin(84^\circ)\sin(54^\circ-x)\\&=\sin(84^\circ)\big(\sin(54^\circ)\cos(x)-\cos(54^\circ)\sin(x)\big)\,.\end{align}$$ This shows that $$\tan(x)=\frac{\sin(x)}{\cos(x)}=\frac{\sin(84^\circ)\sin(54^\circ)}{\sin(84^\circ)\cos(54^\circ)+\sin(126^\circ)}=\tan(30^\circ)\,.$$ Ergo, $$x=30^\circ+n\cdot 180^\circ\,,$$ where $n$ is an integer.