Chapter 7.2, Problem 56E

To determine

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

Answer to Problem 56E

  $\sin x+\cos x=\sqrt{2}\sin \left(x+45°\right)$

Explanation of Solution

Given information:

Given expression

  $\sin x+\cos x$

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.

  $\sin x+\cos x$

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

  $a=1,b=1$

Therefore,

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

And

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

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

Hence, $\theta$ lies in 1^st Quadrant

  $\theta =45°$

Thus, the given expression can be written as

  $\sin x+\cos x=\sqrt{2}\sin \left(x+45°\right)$