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Can you please explain how to solve this question please, I already have the answer but I do not know the process in achieving it.

Find the value(s) of $\theta$ such that: $\sin(\theta + 30^\circ ) = \cos 50^\circ$.

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Method 1: $$\sin(\theta+30^\circ)=\cos 50^\circ$$ $$\implies \cos (90^\circ-(\theta+30^\circ))=\cos 50^\circ$$ $$\implies \cos (60^\circ-\theta)=\cos 50^\circ$$ $$\implies \cos \left(\frac{\pi}{3}-\theta\right)=\cos \frac{5\pi}{18}$$ Writing the general solution as follows $$\frac{\pi}{3}-\theta=2n\pi\pm \frac{5\pi}{18}$$ $$\implies \color{blue}{\theta=\frac{\pi}{3}-\left(2n\pi\pm \frac{5\pi}{18}\right)}$$ Method 2: , $$\sin(\theta+30^\circ)=\cos 50^\circ$$ $$\implies \sin (\theta+30^\circ)=\sin (90^\circ-50^\circ)$$ $$\implies \sin (\theta+30^\circ)=\sin 40^\circ$$ $$\implies \sin \left(\theta+\frac{\pi}{6}\right)=\sin\frac{2\pi}{9}$$ Writing the general solution as follows $$\theta+\frac{\pi}{6}=2n\pi+ \frac{2\pi}{9}$$ $$\implies \theta=2n\pi+\frac{2\pi}{9}-\frac{\pi}{6}$$ $$\implies \color{blue}{\theta=2n\pi+\frac{\pi}{18}}$$ or
$$\theta+\frac{\pi}{6}=(2n+1)\pi- \frac{2\pi}{9}$$ $$\implies \theta=2n\pi+\pi-\frac{2\pi}{9}-\frac{\pi}{6}$$ $$\implies \color{blue}{\theta=2n\pi+\frac{11\pi}{18}}$$

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    $\begingroup$ Why do you convert to radians ? $\endgroup$ – Yves Daoust Jul 21 '15 at 16:11
  • $\begingroup$ No matter, you may change it in degree as your choice just by substituting $\pi=180^\circ$. Do not worry, it's not a rule $\endgroup$ – Harish Chandra Rajpoot Jul 21 '15 at 16:44
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Hint 1: $\sin(a)=\sin(b)$ iff $a-b=2k\pi$ or $a+b = (2k+1)\pi$ for some $k\in\mathbb{Z}$.

Hint 2: $\cos(40^\circ)=\sin(50^\circ)$.

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Hint:

$$ \sin \theta = \cos (90^\circ-\theta)$$

$$\cos50^\circ = \sin40^\circ$$

can you solve for $\theta$ using the above?

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HINT:

Use $\sin x=\cos(90^\circ-x)$ or $\cos x=\sin(90^\circ\pm x)$

and $\sin y=\sin A\implies y=n180^\circ+(-1)^nA$

or $\cos u=\cos B\implies u= m360^\circ\pm B$ where $m,n$ are integers

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Knowing the relation between $\sin(\theta)$ and $\cos(\theta)$ is quite crucial. One of the major relation is that the sine function and cosine function are fairly similar with $90^{\circ}$ difference so, $$Sin(x+90)=cos(x)$$ We are given $x=50$, so $$x+90=30+\theta$$ $$\theta=110$$ or $$180-140=40$$ This is $\theta+30$ so, $$\theta=10^{\circ}$$

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  • $\begingroup$ Do you think $\sin(70°+30°)=\cos(50°)$? $\endgroup$ – Did Jul 21 '15 at 14:01
  • $\begingroup$ @Did sorry for the mistake! Was in a hurry $\endgroup$ – Socre Jul 21 '15 at 14:38
  • $\begingroup$ And why $180°-140°$ now? $\endgroup$ – Did Jul 21 '15 at 15:06
  • $\begingroup$ @Did I edited my question $\endgroup$ – Socre Jul 21 '15 at 15:41
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$$\sin(\theta+30°)=\cos(50°)=\sin(40°).$$

Two angles have the same sine when they are equal or supplementary (to a multiple of 360°).

Hence $$\theta=40°-30°+k360°\text{ or }\theta=(180°-40°)-30°+k360°.$$

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