# Write 3cosx-4sinx in the form k(sinx-β)?

May 21, 2018

#### Explanation:

We need

$\sin \left(a - b\right) = \sin a \cos b - \sin b \cos a$

${\cos}^{2} x + {\sin}^{2} x = 1$

Therefore,

$3 \cos x - 4 \sin x = k \sin \left(x - \beta\right)$

$3 \cos x - 4 \sin x = k \left(\sin x \cos \beta - \sin \beta \cos x\right)$

So,

$\left\{\begin{matrix}- 4 = k \cos \beta \\ 3 = - k \sin \beta\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}\cos \beta = - \frac{4}{k} \\ \sin \beta = - \frac{3}{k}\end{matrix}\right.$

${\cos}^{2} \beta + {\sin}^{2} \beta = \frac{16}{k} ^ 2 + \frac{9}{k} ^ 2 = \frac{25}{k} ^ 2 = 1$

$\implies$, ${k}^{2} = 25$

$k = \pm 5$

$k = 5$,$\implies$, $\left\{\begin{matrix}\cos \beta = - \frac{4}{k} = - \frac{4}{5} \\ \sin \beta = - \frac{3}{k} = - \frac{3}{5}\end{matrix}\right.$

$\implies$, $\tan \beta = \frac{3}{4}$

$\implies$, $\beta = {216.9}^{\circ}$

$= 5 \sin \left(x - {216.9}^{\circ}\right)$

$k = - 5$,$\implies$, $\left\{\begin{matrix}\cos \beta = - \frac{4}{k} = \frac{4}{5} \\ \sin \beta = - \frac{3}{k} = \frac{3}{5}\end{matrix}\right.$

$\implies$, $\tan \beta = \frac{3}{4}$

$\implies$, $\beta = {36.9}^{\circ}$

$= - 5 \sin \left(x - {36.9}^{\circ}\right)$

May 21, 2018

$f \left(x\right) = \left(1.8\right) \sin \left(x - {36}^{\circ} 87\right)$

#### Explanation:

$f \left(x\right) = 3 \cos x - 4 \sin x = 3 \left(\cos x - \left(\frac{4}{3}\right) \sin x\right)$ (1)
Call $\cot b = \cos \frac{b}{\sin b} = \frac{4}{3}$ --> $\tan b = \frac{3}{4}$
Calculator gives -->
$b = {36}^{\circ} 87$ --> $\sin b = 0.6$
f(x) becomes:
$f \left(x\right) = 3 \left(\cos x - \left(\cos \frac{b}{\sin b}\right) \sin x\right)$
$f \left(x\right) = 3 \sin b \left(\cos x . \sin b - \cos b . \sin x\right)$
$f \left(x\right) = 3 \left(0.6\right) \sin \left(x - b\right) = \left(1.8\right) \sin \left(x - {36}^{\circ} 87\right)$