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Find $\theta\leq90°$ if $$\sin( 2 \theta) = \cos( 3)$$

I know that $\sin 2\theta = 2\sin\theta\cos \theta$, or alternatively, $\theta = \dfrac{\sin^{-1}(\cos 3)}{2}$.

Can somebody help me?

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  • $\begingroup$ 2 sin theta cos theta=cos 3 $\endgroup$ – burm1 Aug 25 '14 at 13:45
  • $\begingroup$ Do you mean $\cos (3\theta)$, or simply $\cos(3)$? $\endgroup$ – Namaste Aug 25 '14 at 13:46
  • $\begingroup$ @cjferes i dont know edit here. $\endgroup$ – burm1 Aug 25 '14 at 13:46
  • $\begingroup$ theta = sin^-1 (cos 3 )/2 $\endgroup$ – burm1 Aug 25 '14 at 13:46
  • $\begingroup$ Hint: wolframalpha.com/input/?i=sin%282x%29%3Dcos%283%29 $\endgroup$ – user153012 Aug 25 '14 at 13:47
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$$\sin2\theta=\cos3^\circ=\sin87^\circ$$

$$\implies2\theta=180^\circ m+(-1)^m87^\circ\text{ where } m \text{ is any integer}$$

$$\implies\theta=90^\circ m+(-1)^m43.5^\circ$$

Check for even $m=2r$(say) and for odd $=2r+1$(say)

Find $m$ such that $\displaystyle\theta\le90^\circ$


Alternatively, $$\cos3^\circ=\sin2\theta=\cos\left(90^\circ-2\theta\right)=\cos\left(2\theta-90^\circ\right)$$

$$\iff2\theta-90^\circ= 360^\circ n\pm3^\circ\text{ where } n \text{ is any integer}$$

$\displaystyle'+'\implies 2\theta= 360^\circ n+93^\circ\iff\theta=180^\circ n+46.5^\circ$

$\displaystyle'-'\implies 2\theta= 360^\circ n+87^\circ\iff\theta=180^\circ n+43.5^\circ$

Find $n$ such that $\displaystyle\theta\le90^\circ$

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  • $\begingroup$ @burm1, Also find the edited answer $\endgroup$ – lab bhattacharjee Aug 26 '14 at 18:12
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Another way to solve the problem: note that

$$\sin(2\theta)=\cos(3)=\sin(90-3)$$

But we know $\theta\leq90°$, so $2\theta\leq180°$. Then, two solutions are possible:

$$2\theta=\left\{\begin{array}{ccc}90-3=87°&\Rightarrow&\theta_1=43.5°\\90+3=93°&\Rightarrow&\theta_2=46.5°\end{array}\right.$$

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  • $\begingroup$ $\theta \le 90^{\circ} \implies 2\theta \le 180^{\circ}$, guess there can be two possible solutions ? $\endgroup$ – ganeshie8 Aug 25 '14 at 14:01
  • $\begingroup$ Didn't notice that, you're right! Editing answer now... $\endgroup$ – cjferes Aug 25 '14 at 14:04
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$$\sin{( 2 \theta)} = \cos{(3^{\circ})} \Rightarrow \sin{(2 \theta)}=\sin{ \left (90^{\circ} \pm 3^{\circ} \right )} \Rightarrow 2 \theta=90^{\circ} \pm 3^{\circ}+360^{\circ}k, k \in \mathbb{Z} \\ \Rightarrow \theta=\left ( 45\pm\frac{3}{2}+180k \right )^{\circ}, k \in \mathbb{Z}$$

$$\Rightarrow \theta=\left ( \frac{93}{2}+180k \right )^{\circ} \text{ or } \theta=\left ( \frac{87}{2}+180k \right )^{\circ}, \ \ \ k\in \mathbb{Z}$$

Then you have to find a $\theta$ such that $\theta \leq 90^{\circ}$.

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  • $\begingroup$ I understand that the problem means $3°$, as the assumption is given in degrees... $\endgroup$ – cjferes Aug 25 '14 at 14:09
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    $\begingroup$ I edited my answer.. $\endgroup$ – Mary Star Aug 25 '14 at 14:36
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Assuming you mean $3^{\circ}$, here are the steps $$ \sin 2\theta=\cos 3^{\circ} $$ $$ \arcsin(\sin 2\theta)=\arcsin(\cos 3^{\circ}) $$ $$ 2\theta=\arcsin(\cos 3^{\circ}) $$ $$ \theta=\frac{1}{2}\arcsin(\cos 3^{\circ})=\frac{87^{\circ}}{2}=43.5^{\circ} $$

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    $\begingroup$ The OP has already posted that answer, in the comments, 5 minutes before your post. $\endgroup$ – Namaste Aug 25 '14 at 13:53
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    $\begingroup$ =1/2arc sin(cos 3) how to proceed after that i dont know the method without using calculator $\endgroup$ – burm1 Aug 25 '14 at 13:58
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    $\begingroup$ @amWhy, oops. Didn't see it. Typing latex on a phone is a bit time consuming. $\endgroup$ – k170 Aug 25 '14 at 13:59
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As you've noted, $\theta = \dfrac{\sin^{-1}(\cos 3)}{2}$. Cosine of $3$ is in the second quadrant and is quite close to $-1$, and sine of something close to $-1$ is going to be in the fourth quadrant, hence this sine is going to be negative. Dividing by two doesn't change the sign, so the aswer is:

$$\theta = -\dfrac{180\sin^{-1}(\cos 3)}{2\pi}=-\dfrac{90\sin^{-1}(\cos 3)}{\pi}$$

(The factor of $\frac{180}{\pi}$ comes from the fact that we need to convert radians to degrees.)

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