Reduce √3 sin x − cos x to the Sine and Cosine of a Single Expression

Solution

We use the standard form:

\[ a\sin x+b\cos x=R\sin(x-\alpha) \]

where

\[ R=\sqrt{a^2+b^2} \]

Given expression:

\[ \sqrt{3}\sin x-\cos x \]

Here,

\[ a=\sqrt{3}, \qquad b=-1 \]

Now,

\[ R=\sqrt{(\sqrt{3})^2+(-1)^2} \]

\[ =\sqrt{3+1} \]

\[ =\sqrt{4}=2 \]

Let

\[ \sqrt{3}\sin x-\cos x = 2\sin(x-\alpha) \]

Using the identity:

\[ 2\sin(x-\alpha) = 2(\sin x\cos\alpha-\cos x\sin\alpha) \]

\[ = 2\cos\alpha\sin x-2\sin\alpha\cos x \]

Comparing coefficients,

\[ 2\cos\alpha=\sqrt{3} \]

\[ \cos\alpha=\frac{\sqrt{3}}{2} \]

and

\[ 2\sin\alpha=1 \]

\[ \sin\alpha=\frac{1}{2} \]

Therefore,

\[ \alpha=\frac{\pi}{6} \]

Hence,

\[ \boxed{ \sqrt{3}\sin x-\cos x = 2\sin\left(x-\frac{\pi}{6}\right) } \]

Also, using cosine form:

\[ \boxed{ \sqrt{3}\sin x-\cos x = 2\cos\left(x-\frac{2\pi}{3}\right) } \]

Final Answer

\[ \boxed{ \sqrt{3}\sin x-\cos x = 2\sin\left(x-\frac{\pi}{6}\right) } \]

or

\[ \boxed{ \sqrt{3}\sin x-\cos x = 2\cos\left(x-\frac{2\pi}{3}\right) } \]