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Express in the form $A\sin(x+c)$
a) $\sin x+\sqrt3\cos x$; b) $\sin x-\cos x$

sol: a) $A=\sqrt{1+3}=2$, $\tan c=\frac{\sqrt 3}1$, $c=\frac\pi3$. So $\sin x+\sqrt3\cos x=2\sin(x+\frac\pi3)$
b) $\sqrt 2\sin(x-\frac\pi4)$

Can someone please explain the method used in the provided solution above? (I'm not familiar with this way of solving whatsoever.)

Thanks in advance =]

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    $\begingroup$ There are a few spiritual duplicates of this question. The key is to normalize and use the trigonometric addition formulas. Try and work backwards: expand out $A\sin(x+c)$, and then figure out $A$ and $c$ (do use the formula $\cos^2+\sin^2=1$ to find out $A$). $\endgroup$ – anon Sep 10 '12 at 16:15
  • $\begingroup$ Thanks, I tried expanding the desired form and it all worked out nicely. Though I didn't understand where the identity $cos^2\alpha+sin^2\alpha=1$ comes to use. $\endgroup$ – Py42 Sep 10 '12 at 16:35
  • $\begingroup$ If you know $A\cos(c)$ and $A\sin(c)$, then you can find $A$ via $A^2=(A\cos c)^2+(A\sin c)^2$. $\endgroup$ – anon Sep 10 '12 at 16:39
  • $\begingroup$ Oh, I just compared $Acos(c)$ with its value after finding the angle c. $\endgroup$ – Py42 Sep 10 '12 at 16:43
  • $\begingroup$ A similar question was posted here math.stackexchange.com/questions/877499/… However, it was now autodeleted. $\endgroup$ – Martin Sleziak Aug 5 '14 at 6:50
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The anonymous commenter has already answered your question, but in case you have any remaining doubts, I will provide a detailed answer.

For starters, do note that

$A \, \sin (x + c) = \left(A \, \cos(c) \right) \, \sin(x) + \left( A \, \sin(c)\right) \, \cos(x)$

Since you have $\sin (x) + \sqrt{3} \, \cos (x)$, it follows that $A \, \cos(c) = 1$ and $A \, \sin(c) = \sqrt{3}$. Therefore, since $\sin^2 (x) + \cos^2 (x) = 1$, we have that $A^2 = 4$, which yields $A = 2$, and $2 \cos (c) = 1$, which yields $c = \pi / 3$. Finally, we conclude that

$\sin (x) + \sqrt{3} \, \cos (x) = 2 \sin (x + \pi / 3)$.

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Using the appropriate formula for $\sin$ you have $A \sin(x+c) = A \sin x \cos c + A \cos x \sin c$. You need to determine $A,c$ so the formula holds true for a), b).

Equating $A \sin x \cos c + A \cos x \sin c = \sin x + \sqrt{3} \cos x$ gives $A \cos c = 1$, $A \sin c = \sqrt{3}$. This gives $\tan c = \frac{A \sin c}{A \cos c} = \sqrt{3}$. If $\tan c = \sqrt{3}$, then $\sin c = \frac{\sqrt{3}}{2}$ and $\cos c = \frac{1}{2}$. This gives $A \frac{1}{2} = 1$, so $A = 2$. You can check that $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}, \cos \frac{\pi}{3} = \frac{1}{2}$, from which it follows that $\sin x + \sqrt{3} \cos x = 2 \sin ( x + \frac{\pi}{3})$.

Similarly, $A \sin x \cos c + A \cos x \sin c = \sin x - \cos x$ gives $A \cos c = 1$, $A \sin c = -1$. This gives $\tan c = \frac{A \sin c}{A \cos c} = -1$, which in turn gives $\sin c = -\frac{1}{\sqrt{2}}$, $\cos c = \frac{1}{\sqrt{2}}$. Then $A \cos c = A \frac{1}{\sqrt{2}} = 1$ gives $A = \sqrt{2}$. You can check that $\sin (-\frac{\pi}{4}) = -\frac{1}{\sqrt{2}}, \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$, from which it follows that $\sin x - \cos x = \sqrt{2} \sin ( x - \frac{\pi}{4})$.

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according to copper.hat you will conclude a formula $$ a\sin(x)+b\cos(x)=\sqrt{a^2+b^2}\sin(x+y),\quad \tan(y)=\frac{b}{a},y\in(-\frac{\pi}{2},\frac{\pi}{2}) $$

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I ran into the same problem with the MIT OCW problem set. I found out that A represents the amplitude and c represents the phase shift. I sketched the sine and cosine curve then the result of sin - cos. Where sin = cos, the result is zero at pi/4 and 3pi/4, so that makes c= -pi/4. the max amplitude is at 3pi/4 where sin = (square root of 2)/2 and cos = - (sqr2)/2, making sin-cos square root of 2, so that's the amplitude. so the answer is sinx-cosx = sqrt(2)(x-pi/4)

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