Chapter 7.2, Problem 55E

To determine

To write: the given expression in terms of sine only.

Answer to Problem 55E

  $-\sqrt{3}\sin x+\cos x=2\sin \left(x+150°\right)$

Explanation of Solution

Given information:

Given expression

  $-\sqrt{3}\sin x+\cos x$

Formula used:

For real numbers a and b, $a\sin x+b\cos x=k\sin \left(x+\theta \right)$

Where $\cos \theta =\frac{a}{\sqrt{a^2+b^2}}$ , $\sin \theta =\frac{b}{\sqrt{a^2+b^2}}$ and $k=\sqrt{a^2+b^2}$

Calculation:

Consider given expression.

  $-\sqrt{3}\sin x+\cos x$

Compare the above expression with $a\sin x+b\cos x$

  $a=-\sqrt{3},b=1$

Therefore,

  $\begin{array}{c}k=\sqrt{a^2+b^2}\\ =\sqrt{{\left(-\sqrt{3}\right)}^2+{\left(1\right)}^2}\\ =\sqrt{3+1}\\ =2\end{array}$

And

  $\begin{array}{l}\cos \theta =-\frac{\sqrt{3}}{2}\\ \sin \theta =\frac{1}{2}\end{array}$

So, $\cos \theta$ is negative, and $\sin \theta$ is positive.

Hence, $\theta$ lies in 2^nd coordinate

  $\theta =150°$

Thus, the given expression can be written as

  $-\sqrt{3}\sin x+\cos x=2\sin \left(x+150°\right)$