Use the reduction formula to rewrite −3 sin x −3 cos x in the form K sin (x+α) .


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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
	$because { ((a=−3)),((b=−3)) :} we have K=(√(a^2 +b^2 ))=3(√2) since the terminal side of α must go through (−3,−3) we have cos α = ((−3)/(3(√2))) =−((√2)/2) . Now cos^(−1) (((√2)/2))=((3π)/4) but the terminal side of ((3π)/4) is in quadrant II, however we also have cos (((5π)/4))=−((√2)/2) and the terminal side for ((5π)/4) does pass through (−3,−3) in quadrant III so α=((5π)/4) so gives −3sin x−3cos x = 3(√2) sin (x+((5π)/4)).$
	$because {\begin{cases} a = - 3 \\ b = - 3 \end{cases} we have K = \sqrt{a^{2} + b^{2}} = 3 \sqrt{2}$ $since the terminal side of α must go$ $through (- 3, - 3) we have$ $\cos α = \frac{- 3}{3 \sqrt{2}} = - \frac{\sqrt{2}}{2} . Now \cos^{- 1} (\frac{\sqrt{2}}{2}) = \frac{3 π}{4}$ $but the terminal side of \frac{3 π}{4} is in$ $quadrant II, however we also have$ $\cos (\frac{5 π}{4}) = - \frac{\sqrt{2}}{2} and the terminal side$ $for \frac{5 π}{4} does pass through (- 3, - 3) in$ $quadrant III so α = \frac{5 π}{4} so gives$ $- 3 \sin x - 3 \cos x = 3 \sqrt{2} \sin (x + \frac{5 π}{4}) .$
	Answered by physicstutes last updated on 21/Jun/21

	$k \sin (x + α) = k \sin x \cos α + k \cos x \sin α$ $\Rightarrow k \cos α = - 3$ $k \sin α = - 3$ $k = \sqrt{3^{2} + 3^{2}} = 3 \sqrt{2}$ $\frac{k \sin α}{k \cos α} = 1 \Rightarrow \tan α = 1 or α = \frac{π}{4}$ $the above function is in quadrant 3, so we use$ $α = \frac{π}{4} + π = \frac{5 π}{4}$ $hence - 3 \sin x - 3 \cos x = 3 \sqrt{2} \sin (x + \frac{5 π}{4})$
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