Question

express 3cos(theta)+4sin(theta) in the form r sin(theta+alpha).

Answer 1

This is a theorem:
\[\mathrm{\textrm{If } a,\,b \in \mathbb{R}\textrm{, then:}\\a\sin(x)+b\cos(x)=\sqrt{a^2+b^2}\sin(x+\theta)\\\textrm{such that: }\cos(\theta)=\frac{a}{\sqrt{a^2+b^2}}\quad\wedge\quad\sin(\theta)=\frac{b}{\sqrt{a^2+b^2}}}\]

Answer 2

Proof:
\[
\mathrm{
a\sin(x)+b\cos(x)=\sqrt{a^2+b^2}\sin(x+\theta)\\
a\sin(x)+b\cos(x)=\sqrt{a^2+b^2}(\sin(x)\cos(\theta)+\cos(x)\sin(\theta))\\

a\sin(x)+b\cos(x)=\sqrt{a^2+b^2}
(\sin(x)\frac{a}{\sqrt{a^2+b^2}}+\cos(x)\frac{b}{\sqrt{a^2+b^2}})\\

a\sin(x)+b\cos(x)=a\sin(x)+b\cos(x)

}\]

Answer 3

In your question:

\[\mathrm{3\cos(\theta)+4\sin(\theta)=5\sin(\theta+\alpha)}\]
Where:
\[\mathrm{\sin(\alpha)=\frac{3}{5}\quad\wedge\quad\cos(\alpha)=\frac{4}{5}}\]