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Let $$2\sin{11^{\circ}}\sin{71^{\circ}}\sin{(x^{\circ}+30^{\circ})}=\sin{2013^{\circ}}\sin{210^{\circ}}$$ where $90^{\circ}<x<180^{\circ}$.

My idea: $$2\sin{11^{\circ}}\sin{71^{\circ}}\sin{(x^{\circ}+30^{\circ})}=\sin{33^{\circ}}\sin{30^{\circ}}$$ and $$\Big[\cos{(71^{\circ}-11^{\circ})}-\cos{(71^{\circ}+11^{\circ})}\Big]\sin{(x^{\circ}+30^{\circ})}=\dfrac{1}{2}\sin{33^{\circ}}$$ and $$(1-2\sin{8^{\circ}})\sin{(x^{\circ}+30^{\circ})}=\sin{33^{\circ}}$$

I used Wolfram|Alpha to find $x=19$ or $101$.

My question: we can't use Wolfram|Alpha. How to find this $x$? Thank you.

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    $\begingroup$ Instead of using ^{0} for degrees sign, use ^{\circ}, it looks nicer. $\endgroup$
    – Kaish
    Apr 30, 2013 at 9:25

3 Answers 3

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$$2\sin (x+30^\circ)\sin 11^\circ\sin 71^\circ=\sin 33^\circ\sin 30^\circ=\frac12(3\sin 11^\circ-4\sin^311^\circ)$$ applying $\sin3y=3\sin y -4\sin^3y$

$$\implies 4\sin (x+30^\circ)\sin 71^\circ=3-4\sin^211^\circ=3-2(1-\cos22^\circ)$$ as $\sin11^\circ\ne0$

$$\implies 2\{\cos(x-41^\circ)-\cos(x+101^\circ)\}=2\left(\cos22^\circ+\frac12\right)$$

One way to solve is to put

$(1)\cos(x-41^\circ)=\cos22^\circ,\cos(x+101^\circ)=-\frac12$

or $(2)\cos(x+101^\circ)=-\cos22^\circ=\cos(180-22)^\circ$ as $\cos(\pi-y)=-\cos y$ $=\cos158^\circ,\cos(x-41^\circ)=\frac12$

$(1)\implies x-41^\circ=360^\circ n\pm22^\circ$ where $n$ is any integer.

Taking $'+'$ sign, $x=360^\circ n+41^\circ+22^\circ=360^\circ n+63^\circ$ $\implies \cos(x+101^\circ)=\cos(360^\circ n+63^\circ+101^\circ)=\cos164^\circ \ne-\frac12$ $\implies x\ne360^\circ n+63^\circ$

Taking $'-'$ sign, $x=360^\circ n+41^\circ-22^\circ=360^\circ n+19^\circ$ $\implies \cos(x+101^\circ)=\cos(360^\circ n+19^\circ+101^\circ)=\cos120^\circ=-\frac12$ $\implies x=360^\circ n+19^\circ$ is a solution

$(2)\implies x+101^\circ=360^\circ m\pm158^\circ$ where $m$ is any integer

Taking $'+'$ sign, $x=360^\circ m+158^\circ-101^\circ=360^\circ m+57^\circ$ $\implies \cos(x-41^\circ)=\cos(360^\circ n+57^\circ-41^\circ)=\cos16^\circ \ne\frac12$ $\implies x\ne360^\circ m+57^\circ$

Taking $'-'$ sign, $x=360^\circ m-158^\circ-101^\circ=360^\circ m-259^\circ$ $\implies \cos(x-41^\circ)=\cos(360^\circ m-259^\circ-41^\circ)=\cos(-300^\circ)=\cos(360^\circ-60^\circ)=\cos60^\circ=\frac12$ $\implies x=360^\circ m-259^\circ$ is another solution

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You've over complicated the whole thing wayyy too much. Don't forget, apart from $\sin(x + 30^{\circ})$, the rest are just constants and so treat them like any other number. Therefore you can just do the whole thing simply and like normal:

$$2\sin(11^{\circ})\sin(71^{\circ})\sin{(x+30^{\circ})}=\sin(2013^{\circ})\sin(210^{\circ})$$

$$ = 2\sin({11^{\circ}})\sin({71^{\circ}})\sin{(x^{\circ}+30^{\circ})}=\sin({33^{\circ}})\sin({30^{\circ}})$$

as they're all multiplied, divide through by the $2\sin(11)\sin(71)$ to get

$$\sin(x + 30^{\circ}) = \frac{\sin(33^{\circ})\sin(30^{\circ})}{2\sin(11^{\circ})\sin(71^{\circ})}.$$

Then $\arcsin$ both sides to get

$$x + 30^{\circ} = \sin^{-1} \left(\frac{\sin(33^{\circ})\sin(30^{\circ})}{2\sin(11^{\circ})\sin(71^{\circ})} \right) = 49^{\circ}.$$

Subtracting $30^{\circ}$ then gives you $x = 19^{\circ}$ and as you want it in the interval $90^{\circ} < x <180^{\circ}$, you simply do $90^{\circ} + 19^{\circ} = 109^{\circ}$.

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  • $\begingroup$ oh,I have solution, I have forget,$\sin{(3x)}=4\sin{x}\sin{(60+x)}\sin{(60-x)}$, $\endgroup$
    – math110
    Apr 30, 2013 at 9:27
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    $\begingroup$ How do you get $\sin^{-1}$ of that not obvious expression? $\endgroup$
    – Ma Ming
    Apr 30, 2013 at 9:41
  • $\begingroup$ @math110, Putting $x=11^\circ$ in your identity, $\sin(x+30^\circ)$ in the question will be $\sin(60^\circ-11^\circ)\implies x+30^\circ=49^\circ$ or $180^\circ-49^\circ$ as $\sin(180^\circ-y)=\sin y$ So, $x+30^\circ=49^\circ$ or $131^\circ$ $\endgroup$
    – lab bhattacharjee
    Apr 30, 2013 at 16:19
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    $\begingroup$ @Kaish : I upvoted your answer because someone had downvoted it but I liked your simple approach. However, you didn't explain why $\sin^{-1}(stuff) = 49^\circ$. Also, you claimed that $\sin(x + 30^\circ) = stuff$ implies $x + 30^\circ = \sin^{-1}stuff$, and we know that the inverse trig functions don't always work that nicely. $\endgroup$
    – Stefan Smith
    Apr 30, 2013 at 22:31
  • $\begingroup$ I genuinely thought, when OP said they used Wolfram, they meant they put the whole thing into Wolfram and it gave them a value for $x$. Not the fact that they meant they couldn't use any kind of calculator and so you simply just type in $sin^{-1} (stuff)$ into your calculator and it gives you the answer. $\endgroup$
    – Kaish
    May 1, 2013 at 10:12
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$$2\sin 11^\circ*\sin 71^\circ*\sin (x+30)^\circ=\sin 2013^\circ*\sin 210^\circ$$

$$2\sin 71^\circ*\sin 11^\circ*\sin (x+30)^\circ=\sin 213^\circ*-\sin 30^\circ$$

$$2\sin 71^\circ*\sin 11^\circ*\sin (x+30)^\circ=-\sin 30^\circ*(\sin 213^\circ-\sin 71^\circ+\sin 71^\circ)$$

$$2\sin 71^\circ*\sin 11^\circ*\sin (x+30)^\circ=-\sin 30^\circ*(2\cos 142^\circ*\sin 71^\circ+2\cos 60^\circ*\sin 71^\circ)$$

$$2\sin 71^\circ*\sin 11^\circ*\sin (x+30)^\circ=-\sin 30^\circ*2sin 71^\circ*(\cos 142^\circ+\cos 60^\circ)$$

$$2\sin 71^\circ*\sin 11^\circ*\sin (x+30)^\circ=-\sin 71^\circ*(\cos 142^\circ+\cos 60^\circ)$$

$$2\sin 11^\circ*\sin (x+30)^\circ=(\cos 142^\circ+cos 60^\circ)$$

$$-2\cos 101^\circ*\sin (x+30)^\circ=-2\cos 101^\circ*cos 41^\circ$$

$$\sin (x+30)^\circ=\cos 41^\circ$$

$$\sin (x+30)^\circ=\sin 49^\circ$$

For $$(x+30)^\circ=49^\circ$$ or $$x^\circ=19^\circ$$

Also $$(x+30)^\circ=131^\circ$$ or $$x^\circ=101^\circ$$

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