0

Show that:

\begin{align*} a \cos \theta + b \sin \theta = \sqrt{a^2+b^2} \cos (\theta - \varphi) \end{align*}

fyi, the context is in studying the differential equation:

\begin{align*} y''(t) + c^2 y(t) &= 0 \\ \end{align*}

which has general solution:

\begin{align*} y(t) &= a \cos (ct) + b \sin (ct) \\ \end{align*}

My textbook says that this can easily be transformed into, where $\varphi \in \mathbb{R}$:

\begin{align*} y(t) &= \sqrt{a^2 + b^2} \cos (ct - \varphi) \\ \end{align*}

clay
  • 2,713
  • 23
  • 34
  • https://en.wikipedia.org/wiki/Phasor – Connor Harris Sep 03 '19 at 18:20
  • 1
    Or less glibly, if $a \cos \theta = \Re(a e^{i \theta})$ and $b \sin \theta = \Re(a e^{i \theta - i \pi/2})$, then $a \cos \theta + b \sin \theta = \Re( (a + b e^{-i \pi/2}) e^{i \theta})$; play with this until you get something of the correct form. – Connor Harris Sep 03 '19 at 18:23
  • 1
    Hint: $\cos\theta$ and $\sin\theta$ are always 90° out of phase. Think about what this might mean in terms of a rotating right triangle. – amd Sep 03 '19 at 20:48

2 Answers2

4

Use the compound-angle formula $$\cos(\theta-\varphi)=\cos\varphi\cos\theta+\sin\varphi\sin\theta.$$

To solve $$\cos\varphi=\frac{a}{\sqrt{a^2+b^2}},\,\sin\varphi=\frac{b}{\sqrt{a^2+b^2}}$$ take$$\varphi=\operatorname{atan2}\left(b,\,a\right).$$

clay
  • 2,713
  • 23
  • 34
J.G.
  • 115,835
1

Consider the vector $v = (a, b)$. The inner product of $v$ with the unit vector $u = (\cos \theta, \sin \theta)$ satisfies: $$ a \cos \theta + b \sin \theta = v \cdot u = ||v|| \cdot ||u|| \cdot \cos \beta, $$ where $\beta$ is the angle between $v$ and $u$. Define $\varphi := \theta - \beta$.

D. Ungaretti
  • 2,668