Use-the-reduction-formula-to-rewrite-3-sin-x-3-cos-x-in-the-form-K-sin-x-

Question Number 144072 by bobhans last updated on 21/Jun/21

Use the reduction formula to

rewrite - 3 \sin x - 3 \cos x in the

form K \sin (x + α) .

Answered by liberty last updated on 21/Jun/21

because {\begin{cases} a = - 3 \\ b = - 3 \end{cases} we have K = \sqrt{a^{2} + b^{2}} = 3 \sqrt{2}

since the terminal side of α must go

through (- 3, - 3) we have

\cos α = \frac{- 3}{3 \sqrt{2}} = - \frac{\sqrt{2}}{2} . Now \cos^{- 1} (\frac{\sqrt{2}}{2}) = \frac{3 π}{4}

but the terminal side of \frac{3 π}{4} is in

quadrant II, however we also have

\cos (\frac{5 π}{4}) = - \frac{\sqrt{2}}{2} and the terminal side

for \frac{5 π}{4} does pass through (- 3, - 3) in

quadrant III so α = \frac{5 π}{4} so gives

- 3 \sin x - 3 \cos x = 3 \sqrt{2} \sin (x + \frac{5 π}{4}) .

Answered by physicstutes last updated on 21/Jun/21

k \sin (x + α) = k \sin x \cos α + k \cos x \sin α

\Rightarrow k \cos α = - 3

k \sin α = - 3

k = \sqrt{3^{2} + 3^{2}} = 3 \sqrt{2}

\frac{k \sin α}{k \cos α} = 1 \Rightarrow \tan α = 1 or α = \frac{π}{4}

the above function is in quadrant 3, so we use

α = \frac{π}{4} + π = \frac{5 π}{4}

hence - 3 \sin x - 3 \cos x = 3 \sqrt{2} \sin (x + \frac{5 π}{4})