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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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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