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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) .

$$\mathrm{Use}\:\mathrm{the}\:\mathrm{reduction}\:\mathrm{formula}\:\mathrm{to} \\ $$$$\mathrm{rewrite}\:−\mathrm{3}\:\mathrm{sin}\:\mathrm{x}\:−\mathrm{3}\:\mathrm{cos}\:\mathrm{x}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{form}\:\mathrm{K}\:\mathrm{sin}\:\left(\mathrm{x}+\alpha\right)\:. \\ $$

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)).

$$\mathrm{because}\:\begin{cases}{\mathrm{a}=−\mathrm{3}}\\{\mathrm{b}=−\mathrm{3}}\end{cases}\:\mathrm{we}\:\mathrm{have}\:\mathrm{K}=\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }=\mathrm{3}\sqrt{\mathrm{2}} \\ $$$$\:\mathrm{since}\:\mathrm{the}\:\mathrm{terminal}\:\mathrm{side}\:\mathrm{of}\:\alpha\:\mathrm{must}\:\mathrm{go} \\ $$$$\mathrm{through}\:\left(−\mathrm{3},−\mathrm{3}\right)\:\mathrm{we}\:\mathrm{have}\: \\ $$$$\mathrm{cos}\:\alpha\:=\:\frac{−\mathrm{3}}{\mathrm{3}\sqrt{\mathrm{2}}}\:=−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:.\:\mathrm{Now}\:\mathrm{cos}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)=\frac{\mathrm{3}\pi}{\mathrm{4}} \\ $$$$\mathrm{but}\:\mathrm{the}\:\mathrm{terminal}\:\mathrm{side}\:\mathrm{of}\:\frac{\mathrm{3}\pi}{\mathrm{4}}\:\mathrm{is}\:\mathrm{in}\: \\ $$$$\mathrm{quadrant}\:\mathrm{II},\:\mathrm{however}\:\mathrm{we}\:\mathrm{also}\:\mathrm{have}\: \\ $$$$\mathrm{cos}\:\left(\frac{\mathrm{5}\pi}{\mathrm{4}}\right)=−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\mathrm{and}\:\mathrm{the}\:\mathrm{terminal}\:\mathrm{side} \\ $$$$\mathrm{for}\:\frac{\mathrm{5}\pi}{\mathrm{4}}\:\mathrm{does}\:\mathrm{pass}\:\mathrm{through}\:\left(−\mathrm{3},−\mathrm{3}\right)\:\mathrm{in} \\ $$$$\mathrm{quadrant}\:\mathrm{III}\:\mathrm{so}\:\alpha=\frac{\mathrm{5}\pi}{\mathrm{4}}\:\mathrm{so}\:\mathrm{gives}\: \\ $$$$−\mathrm{3sin}\:\mathrm{x}−\mathrm{3cos}\:\mathrm{x}\:=\:\mathrm{3}\sqrt{\mathrm{2}}\:\mathrm{sin}\:\left(\mathrm{x}+\frac{\mathrm{5}\pi}{\mathrm{4}}\right). \\ $$

Answered by physicstutes last updated on 21/Jun/21

k sin(x + α) = k sin x cos α + k cos x sin α ⇒ k cos α = −3 k sin α = −3 k = (√(3^2 +3^2 )) = 3(√2) ((ksin α)/(k cos α)) = 1 ⇒ tan α = 1 or α = (π/4) the above function is in quadrant 3 , so we use α = (π/4)+π = ((5π)/4) hence −3 sin x − 3 cos x = 3(√2) sin (x+((5π)/4))

$${k}\:\mathrm{sin}\left({x}\:+\:\alpha\right)\:=\:{k}\:\mathrm{sin}\:{x}\:\mathrm{cos}\:\alpha\:+\:{k}\:\mathrm{cos}\:{x}\:\mathrm{sin}\:\alpha\: \\ $$$$\Rightarrow\:{k}\:\mathrm{cos}\:\alpha\:=\:−\mathrm{3} \\ $$$$\:\:\:\:{k}\:\mathrm{sin}\:\alpha\:=\:−\mathrm{3} \\ $$$${k}\:=\:\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} }\:=\:\mathrm{3}\sqrt{\mathrm{2}} \\ $$$$\:\frac{{k}\mathrm{sin}\:\alpha}{{k}\:\mathrm{cos}\:\alpha}\:=\:\mathrm{1}\:\Rightarrow\:\:\mathrm{tan}\:\alpha\:=\:\mathrm{1}\:\mathrm{or}\:\alpha\:=\:\frac{\pi}{\mathrm{4}} \\ $$$$\mathrm{the}\:\mathrm{above}\:\mathrm{function}\:\mathrm{is}\:\mathrm{in}\:\mathrm{quadrant}\:\mathrm{3}\:,\:\mathrm{so}\:\mathrm{we}\:\mathrm{use} \\ $$$$\alpha\:=\:\frac{\pi}{\mathrm{4}}+\pi\:=\:\frac{\mathrm{5}\pi}{\mathrm{4}} \\ $$$$\mathrm{hence}\:\:−\mathrm{3}\:\mathrm{sin}\:{x}\:−\:\mathrm{3}\:\mathrm{cos}\:{x}\:=\:\mathrm{3}\sqrt{\mathrm{2}}\:\mathrm{sin}\:\left({x}+\frac{\mathrm{5}\pi}{\mathrm{4}}\right) \\ $$

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