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Write the expression in terms of a single sin

$$ -\sqrt{3}\sin x + \cos x$$

So this was a question on my test today. I used a cofunction identity to solve it. Can someone tell me what the correct answer is. I think mine was something like $1-\sqrt{3}\sin x +1$ or something like that.

Michael Galuza
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TheNewGuy
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4 Answers4

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I would write \begin{align*}-\sqrt{3}\sin x + \cos x &= 2\left(-\frac12\sqrt{3}\sin x+\frac12\cos x\right) \\ &=2\left(-\cos\left(\frac16 \pi\right) \sin x+\sin\left(\frac16 \pi\right) \cos x\right) \\ &= 2\left(\sin\left(\frac16 \pi-x\right)\right)\end{align*}

wythagoras
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\begin{align} -\sqrt{3}\sin x + \cos x &= -2(\sin x\cos(-\tfrac\pi6) +\cos x\sin(-\tfrac\pi6)) =-2\sin(x-\tfrac\pi6) \end{align}

g.kov
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We have: $$ -\sqrt{3}\sin x + \cos x=\\ =2\left(-\frac{\sqrt{3}}{2}\sin x + \frac{1}{2}\cos x\right)=\\ =2\left(\cos\frac{5\pi}{6}\sin x + \sin\frac{5\pi}{6}\cos x\right)=\\ =2\sin\left(\frac{5\pi}{6}+x\right). $$ Notice that: $$ 2\sin\left(x+\frac{5\pi}{6}\right)= -2\sin\left(x-\frac{\pi}{6}\right)= -2\sin\left(x+\arctan\left(-\frac{1}{\sqrt 3}\right)\right). $$

MattAllegro
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    I think the harmonic identity should be \begin{gather} a\sin(x)+b\cos(x)= \sqrt{a^2+b^2}\ \text{sgn}(a)\sin\left(x+\arctan\left(\frac{b}{a}\right)\right) \end{gather} for every $a, b\in\mathbb{R}$ with $a\neq 0.$ – azc Dec 13 '15 at 06:04
  • @azc Hi! Of course, indeed $2\sin(x+\frac{5\pi}{6})=-2\sin(x-\frac{\pi}{6})=-2\sin(x+\arctan(-\frac{1}{√{3}}))$ being $\arctan(x)\in[-\frac{\pi}{2}, \frac{\pi}{2}]$. – MattAllegro Dec 14 '15 at 18:21
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$$ \boxed{ A \sin(x) + B \cos(x) = R \sin \left(x + \varphi \right) }$$

where $R=\sqrt{A^2+B^2}$ and $\tan \varphi = \frac{B}{A}$

Why?

Expand $R \sin(x+\varphi) = R \sin\varphi \cos(x) + R \cos\varphi \sin(x)$ and match terms.

John Alexiou
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