Expressing in the form $A \sin(x + c)$
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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$\begingroup$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
$\sin (x) + \sqrt{3} \, \cos (x) = 2 \sin (x + \pi / 3)$.
$\endgroup$ $\begingroup$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 $\sin x - \cos x = \sqrt{2} \sin ( x - \frac{\pi}{4})$.
$\endgroup$ $\begingroup$according to copper.hat you will conclude a formula $$ a\sin(x)+b\cos(x)=\sqrt{a^2+b^2}\sin(x+y),\quad \tan(y)=\frac{b}{a},y\in(-\frac{\pi}{2},\frac{\pi}{2}) $$
$\endgroup$ $\begingroup$Solution for item b of the question (this is the item on Problem Set 1 on MITOCW 18.01SC:
$$A\sin(x+c) = A\sin x \cos c + A \sin c \cos x$$
We can see that if $A \cos c = 1$ in the first term and $A \sin c = -1$ in the second term then we will end up with $\sin x - \cos x$, so we will have shown that $f(x)$ can be written in the form $A \sin(x+c)$.
We have two equations in two unknowns (A and c) and so we can solve for these variables. Square both sides of each equation:
$$A^2 \cos^2 c = 1$$$$A^2 \sin^2 c = 1$$Sum the two equations$$A^2(\cos^2 x + \sin^2 x) = 2$$Use $\cos^2 x + \sin^2 x = 1$$$A^2 = 2 \Rightarrow A = \pm \sqrt{2}$$
Solution 1: $A = \sqrt{2}$, $c = -\frac{\pi}{4}$$$\cos c = \frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2} \Rightarrow c = \frac{\pi}{4} \text{ or } c = \frac{7 \pi}{4}$$$$\sin c = -\frac{1}{\sqrt{2}}= -\frac{\sqrt{2}}{2} \Rightarrow c = \frac{5\pi}{4} \text{ or } c = \frac{7 \pi}{4}$$Therefore the value of $c$ that satisfies both equations is $\frac{7\pi}{4}$, which is the same as $-\frac{\pi}{4}$
Solution 2: $A = -\sqrt{2}$, $c = \frac{3\pi}{4}$$$\cos c = -\frac{1}{\sqrt{2}}= -\frac{\sqrt{2}}{2} \Rightarrow c = \frac{3\pi}{4} \text{ or } c = \frac{5 \pi}{4}$$$$\sin c = \frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2} \Rightarrow c = \frac{\pi}{4} \text{ or } c = \frac{3 \pi}{4}$$Therefore the value of $c$ that satisfies both equations is $\frac{3\pi}{4}$.
$\endgroup$ $\begingroup$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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