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Solve $\sqrt{2}\sin(x)+\sqrt{6}\cos(x) = \sqrt{3} +1$ for $x$

I started by multiplying both sides of the equation by $\frac{1}{2\sqrt{2}}$ to obtain

$$\displaystyle\frac{\sin(x)}{2}+\frac{\sqrt{3}\cos(x)}{2} = \frac{\sqrt{3} +1}{2\sqrt{2}}$$ $$\iff \sin(60+x) = \frac{\sqrt{3} +1}{2\sqrt{2}}$$

I am stuck here. Any hints on solving the R.H.S will be appreciated.

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  • $\begingroup$ You could also start by squaring both sides. You can then use double angle formulas to reduce to something that requires only the $30^\circ/60^\circ$ exact ratios. Bear in mind that squaring both sides is not a step that can be "undone" logically, so you will probably get some erroneous solutions. Just check them at the end. $\endgroup$ May 27, 2019 at 8:37

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$\sin 75^0 = \frac{\sqrt{3}+1}{2\sqrt2}$

Can you find now?

PROOF

$\sin (x+y) = \sin x\cos y +\cos x \sin y $
$\sin(75^o) = \sin(30^o+45^o) = \sin 30\cos 45 +\cos 30 \sin 45 = \frac{1}{2\sqrt2}+\frac{\sqrt3}{2\sqrt2} = \frac{\sqrt3+1}{2\sqrt2} $

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Try to find what is $$\sin\frac{5\pi}{12}$$ If you get it $$\frac{\sqrt3+1}{2\sqrt2}$$ then you did correctly and you know what to do next using the general definition of $$\sin x=\sin y $$ yields what you know it !

you can find $$\sin\frac{5\pi}{12}$$ using the formula for $$\sin \frac{x}{2}$$ by taking $x=\frac{5\pi}{6}$ in degrees which is equivalent to 75 degrees

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  • $\begingroup$ Did you arrive at $\frac{5\pi}{12}$ using a calculator? $\endgroup$ May 27, 2019 at 8:14
  • $\begingroup$ nope.. firest square it and notice what you get it after squaring .... you will get the idea. I know the formulas for sin 15 and sin 75 that's why i got it quickly.. i know half angle formulas... like for 22.5 also $\endgroup$ May 27, 2019 at 8:16
  • $\begingroup$ if you think my solution wokrs for u .. please vote i will be obliged... $\endgroup$ May 27, 2019 at 8:17
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    $\begingroup$ Cheers friend, an upvote from me (+1) $\endgroup$
    – 19aksh
    May 27, 2019 at 8:32
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    $\begingroup$ You're welcome🙂 $\endgroup$
    – 19aksh
    May 27, 2019 at 8:35
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square both sides, we get $$2 \sin^2(x) + 6 \cos^2(x) + 2\sqrt{12}\cos(x)\sin(x) = (1 + \sqrt{3})^2$$ that is $$2 + 4 \cos^2(x) + 2\sqrt{12}\cos(x)\sin(x) = (1 + \sqrt{3})^2$$ that is $$2 + 4 \cos^2(x) + \sqrt{12}\sin(2x) = (1 + \sqrt{3})^2$$ that is $$2 + 4 \frac{1}{1+\tan^2(x)} + \sqrt{12}\frac{2\tan x}{1+\tan^2(x)} = (1 + \sqrt{3})^2$$ Multiply everything by $1 + \tan^2(x)$ $$2(1+\tan^2(x)) + 4 +2 \sqrt{12}\tan x = (1 + \sqrt{3})^2(1+\tan^2(x)) $$ This is quadratic in $y = \tan(x)$

$$ay^2 + by + c = 0$$ where \begin{align} a &= 2- (1 + \sqrt{3})^2\\ b &=2\sqrt{12}\\ c &= 2 + 4 - (1 + \sqrt{3})^2 \end{align} Solving you get something like \begin{align} y_1 &= 1.0000 \\ y_2 &= 0.2679 \end{align}

Now taking the inverse tangent of that you get \begin{align} x_1 &= \tan^{-1} y_1 =\tan^{-1} 1.0000 = 45^\circ\\ x_2 &= \tan^{-1} y_2 =\tan^{-1} 0.2679= 15^\circ \end{align}

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    $\begingroup$ I think you missed multiplying 1+tanx on the R.H.S $\endgroup$ May 27, 2019 at 8:24
  • $\begingroup$ right .. ill edit $\endgroup$ May 27, 2019 at 8:25
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$$\frac{\sqrt{3} +1}{2\sqrt{2}}=\frac{\sqrt{3}\sqrt{2} +\sqrt{2}}{4}=\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\cdot\frac{1}{2}=\sin{45^\circ}\cos{30^\circ}+\cos{45^\circ}\sin{30^\circ}=\sin{(45^\circ+30^\circ)}=\sin{75^\circ}$$

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