7.5 Writing sums of trig functions as a single trig function

Example 7.5.1

Express 3⁢cos⁡(θ)+4⁢sin⁡(θ)3\cos(\theta)+4\sin(\theta) in the form R⁢sin⁡(θ+α)R\sin(\theta+\alpha).

Solution: start by applying the angle addition formula for sin⁡(θ)\sin(\theta) (Equation 7.8).

R⁢sin⁡(θ+α)\displaystyle R\sin(\theta+\alpha) =R⁢cos⁡(α)⁢sin⁡(θ)+R⁢sin⁡(α)⁢cos⁡(θ)\displaystyle=R\cos(\alpha)\sin(\theta)+R\sin(\alpha)\cos(\theta)
=1⁢sin⁡(θ)+3⁢cos⁡(θ)\displaystyle=\hskip 32.0pt1\sin(\theta)+\hskip 32.0pt3\cos(\theta)

From here, comparing coefficients gives

{R⁢cos⁡(α)=1R⁢sin⁡(α)=3\begin{cases}R\cos(\alpha)=1\\ R\sin(\alpha)=3\end{cases}

This means

R2⁢cos2⁡(α)+R2⁢sin2⁡(α)=1+32\displaystyle R^{2}\cos^{2}(\alpha)+R^{2}\sin^{2}(\alpha)=1+3^{2}
R2(cos2(α)+sin2(α)=10\displaystyle R^{2}(\cos^{2}(\alpha)+sin^{2}(\alpha)=10
R2=10\displaystyle R^{2}=10
R=10\displaystyle R=\sqrt{10}

as well as that

R⁢sin⁡(α)R⁢cos⁡(α)=3\displaystyle\frac{R\sin(\alpha)}{R\cos(\alpha)}=3
tan⁡(α)=3\displaystyle\tan(\alpha)=3
α=arctan⁡(3)\displaystyle\alpha=\arctan(3)

So the solution is

s⁢i⁢n⁢(θ)+3⁢c⁢o⁢s⁢(θ)=10⁢sin⁡(x+arctan⁡(3))sin(\theta)+3cos(\theta)=\sqrt{10}\sin\left(x+\arctan(3)\right)

Note that this technique is very useful for solving equations of the form A⁢cos⁡(θ)+B⁢sin⁡(θ)=cA\cos(\theta)+B\sin(\theta)=c, as we just rewrite the left hand side as a single trigonometric function, and then use the method for solving such trig functions66 6 Explored in the section above..