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Question Number 142794 by EDWIN88 last updated on 05/Jun/21

If 2cos (((5π)/4)+3x)cos ((π/4)+3x)=0 and sin (2x−2y)=cos y where (π/4)≤x≤(π/2) and (π/4)≤y≤(π/2) . find the value of { ((sin (2x+y))),((cos (2x+y))),((cos (2x−y))),((sin (2x−y))) :}

$$\mathrm{If}\:\mathrm{2cos}\:\left(\frac{\mathrm{5}\pi}{\mathrm{4}}+\mathrm{3x}\right)\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}+\mathrm{3x}\right)=\mathrm{0}\:\mathrm{and}\: \\ $$$$\mathrm{sin}\:\left(\mathrm{2x}−\mathrm{2y}\right)=\mathrm{cos}\:\mathrm{y}\:\mathrm{where}\:\frac{\pi}{\mathrm{4}}\leqslant\mathrm{x}\leqslant\frac{\pi}{\mathrm{2}}\:\mathrm{and} \\ $$$$\frac{\pi}{\mathrm{4}}\leqslant\mathrm{y}\leqslant\frac{\pi}{\mathrm{2}}\:.\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\begin{cases}{\mathrm{sin}\:\left(\mathrm{2x}+\mathrm{y}\right)}\\{\mathrm{cos}\:\left(\mathrm{2x}+\mathrm{y}\right)}\\{\mathrm{cos}\:\left(\mathrm{2x}−\mathrm{y}\right)}\\{\mathrm{sin}\:\left(\mathrm{2x}−\mathrm{y}\right)}\end{cases} \\ $$

Answered by Rasheed.Sindhi last updated on 05/Jun/21

2cos (((5π)/4)+3x)cos ((π/4)+3x)=0 cos (((5π)/4)+3x)=0 ∣ cos ((π/4)+3x)=0 ((5π)/4)+3x=(π/2)(2n−1) ∣ (π/4)+3x=(π/2)(2n−1) n∈Z 3x=(π/2)(2n−1)−((5π)/4) ∣ 3x=(π/2)(2n−1)− (π/4) x=(π/6)(2n−1)−((5π)/(12)) ∣ x=(π/6)(2n−1)− (π/(12)) x=(({2(2n−1)−5}π)/(12)) ∣ x=(({2(2n−1)−1}π)/(12)) x=(((4n−7)π)/(12)) ∣ x=(((4n−3)π)/(12)) (π/4)≤x≤(π/2) (π/4)≤(((4n−7)π)/(12))≤(π/2) ∣ (π/4)≤(((4n−3)π)/(12))≤(π/2) 3π≤(4n−7)π≤6π ∣ 3π≤(4n−3)π≤6π 3≤(4n−7)≤6 ∣ 3≤(4n−3)≤6 10≤4n≤13 ∣ 6≤4n≤9 (5/2)≤n≤((13)/4) ∣ (3/2)≤n≤(9/4) 2(1/2)≤n≤3(1/4) ∣1 (1/2)≤n≤2(1/4) n=3 ∣ n=2 x=(((4n−7)π)/(12)) ∣ x=(((4n−3)π)/(12)) x=(((4(3)−7)π)/(12)) ∣ x=(((4(2)−3)π)/(12)) x=((5π)/(12)) sin(2x−2y)=cos y sin(2(((5π)/(12)))−2y)=cos y sin(((5π)/6)−2y)=sin ((π/2)−y) ((5π)/6)−2y=(π/2)−y y=((5π)/6)−(π/2)=((5π−3π)/6)⇒y=(π/3)

$$\mathrm{2cos}\:\left(\frac{\mathrm{5}\pi}{\mathrm{4}}+\mathrm{3x}\right)\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}+\mathrm{3x}\right)=\mathrm{0} \\ $$$$\mathrm{cos}\:\left(\frac{\mathrm{5}\pi}{\mathrm{4}}+\mathrm{3x}\right)=\mathrm{0}\:\mid\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}+\mathrm{3x}\right)=\mathrm{0} \\ $$$$\frac{\mathrm{5}\pi}{\mathrm{4}}+\mathrm{3x}=\frac{\pi}{\mathrm{2}}\left(\mathrm{2}{n}−\mathrm{1}\right)\:\mid\:\frac{\pi}{\mathrm{4}}+\mathrm{3x}=\frac{\pi}{\mathrm{2}}\left(\mathrm{2}{n}−\mathrm{1}\right) \\ $$$$\:\:\:\:{n}\in\mathbb{Z} \\ $$$$\:\:\mathrm{3x}=\frac{\pi}{\mathrm{2}}\left(\mathrm{2}{n}−\mathrm{1}\right)−\frac{\mathrm{5}\pi}{\mathrm{4}}\:\mid\:\mathrm{3x}=\frac{\pi}{\mathrm{2}}\left(\mathrm{2}{n}−\mathrm{1}\right)−\:\frac{\pi}{\mathrm{4}} \\ $$$$\:\:\mathrm{x}=\frac{\pi}{\mathrm{6}}\left(\mathrm{2}{n}−\mathrm{1}\right)−\frac{\mathrm{5}\pi}{\mathrm{12}}\:\mid\:\mathrm{x}=\frac{\pi}{\mathrm{6}}\left(\mathrm{2}{n}−\mathrm{1}\right)−\:\frac{\pi}{\mathrm{12}} \\ $$$$\mathrm{x}=\frac{\left\{\mathrm{2}\left(\mathrm{2n}−\mathrm{1}\right)−\mathrm{5}\right\}\pi}{\mathrm{12}}\:\mid\:\mathrm{x}=\frac{\left\{\mathrm{2}\left(\mathrm{2n}−\mathrm{1}\right)−\mathrm{1}\right\}\pi}{\mathrm{12}} \\ $$$$\mathrm{x}=\frac{\left(\mathrm{4n}−\mathrm{7}\right)\pi}{\mathrm{12}}\:\mid\:\mathrm{x}=\frac{\left(\mathrm{4n}−\mathrm{3}\right)\pi}{\mathrm{12}} \\ $$$$\frac{\pi}{\mathrm{4}}\leqslant\mathrm{x}\leqslant\frac{\pi}{\mathrm{2}} \\ $$$$\frac{\pi}{\mathrm{4}}\leqslant\frac{\left(\mathrm{4n}−\mathrm{7}\right)\pi}{\mathrm{12}}\leqslant\frac{\pi}{\mathrm{2}}\:\:\mid\:\frac{\pi}{\mathrm{4}}\leqslant\frac{\left(\mathrm{4n}−\mathrm{3}\right)\pi}{\mathrm{12}}\leqslant\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{3}\pi\leqslant\left(\mathrm{4n}−\mathrm{7}\right)\pi\leqslant\mathrm{6}\pi\:\mid\:\mathrm{3}\pi\leqslant\left(\mathrm{4n}−\mathrm{3}\right)\pi\leqslant\mathrm{6}\pi \\ $$$$\mathrm{3}\leqslant\left(\mathrm{4n}−\mathrm{7}\right)\leqslant\mathrm{6}\:\mid\:\mathrm{3}\leqslant\left(\mathrm{4n}−\mathrm{3}\right)\leqslant\mathrm{6} \\ $$$$\mathrm{10}\leqslant\mathrm{4n}\leqslant\mathrm{13}\:\mid\:\mathrm{6}\leqslant\mathrm{4n}\leqslant\mathrm{9} \\ $$$$\frac{\mathrm{5}}{\mathrm{2}}\leqslant\mathrm{n}\leqslant\frac{\mathrm{13}}{\mathrm{4}}\:\mid\:\frac{\mathrm{3}}{\mathrm{2}}\leqslant\mathrm{n}\leqslant\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\mathrm{2}\frac{\mathrm{1}}{\mathrm{2}}\leqslant\mathrm{n}\leqslant\mathrm{3}\frac{\mathrm{1}}{\mathrm{4}}\:\mid\mathrm{1}\:\frac{\mathrm{1}}{\mathrm{2}}\leqslant\mathrm{n}\leqslant\mathrm{2}\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${n}=\mathrm{3}\:\mid\:{n}=\mathrm{2} \\ $$$$\mathrm{x}=\frac{\left(\mathrm{4n}−\mathrm{7}\right)\pi}{\mathrm{12}}\:\:\mid\:\mathrm{x}=\frac{\left(\mathrm{4n}−\mathrm{3}\right)\pi}{\mathrm{12}} \\ $$$$\mathrm{x}=\frac{\left(\mathrm{4}\left(\mathrm{3}\right)−\mathrm{7}\right)\pi}{\mathrm{12}}\:\:\mid\:\mathrm{x}=\frac{\left(\mathrm{4}\left(\mathrm{2}\right)−\mathrm{3}\right)\pi}{\mathrm{12}} \\ $$$$\mathrm{x}=\frac{\mathrm{5}\pi}{\mathrm{12}}\:\: \\ $$$$\mathrm{sin}\left(\mathrm{2}{x}−\mathrm{2}{y}\right)=\mathrm{cos}\:{y} \\ $$$$\mathrm{sin}\left(\mathrm{2}\left(\frac{\mathrm{5}\pi}{\mathrm{12}}\right)−\mathrm{2}{y}\right)=\mathrm{cos}\:{y} \\ $$$$\mathrm{sin}\left(\frac{\mathrm{5}\pi}{\mathrm{6}}−\mathrm{2}{y}\right)=\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−{y}\right) \\ $$$$\frac{\mathrm{5}\pi}{\mathrm{6}}−\mathrm{2}{y}=\frac{\pi}{\mathrm{2}}−{y} \\ $$$${y}=\frac{\mathrm{5}\pi}{\mathrm{6}}−\frac{\pi}{\mathrm{2}}=\frac{\mathrm{5}\pi−\mathrm{3}\pi}{\mathrm{6}}\Rightarrow{y}=\frac{\pi}{\mathrm{3}} \\ $$

Answered by MJS_new last updated on 05/Jun/21

2cos (((5π)/4)+3x) cos ((π/4)+3x)=0 sin 6x −1=0 ⇒ x=((5π)/(12)) ⇒ y=(π/3)

$$\mathrm{2cos}\:\left(\frac{\mathrm{5}\pi}{\mathrm{4}}+\mathrm{3}{x}\right)\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}+\mathrm{3}{x}\right)=\mathrm{0} \\ $$$$\mathrm{sin}\:\mathrm{6}{x}\:−\mathrm{1}=\mathrm{0}\:\Rightarrow\:{x}=\frac{\mathrm{5}\pi}{\mathrm{12}} \\ $$$$\Rightarrow\:{y}=\frac{\pi}{\mathrm{3}} \\ $$

Answered by Rasheed.Sindhi last updated on 06/Jun/21

⟨_• ^• ⟩An Other Way⟨_• ^• ⟩ 2cos (((5π)/4)+3x)cos ((π/4)+3x)_(notice that the diff is π) =0 cos ((π/4)+π+3x)cos ((π/4)+3x)=0 cos(π+θ)=−cosθ −cos ((π/4)+3x)cos ((π/4)+3x)=0 cos^2 ((π/4)+3x)=0 cos ((π/4)+3x)=0 cos ((π/4)+3x)=cos((π/2)(2n−1)) (π/4)+3x=(π/2)(2n−1) x=(1/3)((((2n−1)π)/2)−(π/4))=((2(2n−1)π−π)/(12)) =(((4n−3)π)/(12)) (π/4)≤x≤(π/2) (π/4)≤(((4n−3)π)/(12))≤(π/2) 3≤4n−3≤6 6≤4n≤9 (3/2)≤n≤(9/4) 1(1/2)≤n≤2(1/4) n=2 x=(((4n−3)π)/(12))=(({4(2)−3}π)/(12))=((5π)/(12)) sin(2x−2y)=cos y sin(2x−2y)=sin((π/2)−y) 2x−2y=(π/2)−y y=(π/3)

$$\langle_{\bullet} ^{\bullet} \rangle\mathbb{A}\mathrm{n}\:\mathbb{O}\mathrm{ther}\:\mathbb{W}\mathrm{ay}\langle_{\bullet} ^{\bullet} \rangle \\ $$$$\mathrm{2cos}\:\underset{{notice}\:{that}\:{the}\:{diff}\:{is}\:\pi} {\underbrace{\left(\frac{\mathrm{5}\pi}{\mathrm{4}}+\mathrm{3x}\right)\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}+\mathrm{3x}\right)}}=\mathrm{0} \\ $$$$\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}+\pi+\mathrm{3x}\right)\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}+\mathrm{3x}\right)=\mathrm{0} \\ $$$$\mathrm{cos}\left(\pi+\theta\right)=−\mathrm{cos}\theta \\ $$$$−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}+\mathrm{3x}\right)\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}+\mathrm{3x}\right)=\mathrm{0} \\ $$$$\mathrm{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}+\mathrm{3x}\right)=\mathrm{0} \\ $$$$\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}+\mathrm{3x}\right)=\mathrm{0} \\ $$$$\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}+\mathrm{3}{x}\right)=\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}\left(\mathrm{2}{n}−\mathrm{1}\right)\right) \\ $$$$\:\:\:\:\:\:\frac{\pi}{\mathrm{4}}+\mathrm{3}{x}=\frac{\pi}{\mathrm{2}}\left(\mathrm{2}{n}−\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:{x}=\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\right)=\frac{\mathrm{2}\left(\mathrm{2}{n}−\mathrm{1}\right)\pi−\pi}{\mathrm{12}} \\ $$$$\:\:\:\:=\frac{\left(\mathrm{4}{n}−\mathrm{3}\right)\pi}{\mathrm{12}} \\ $$$$\frac{\pi}{\mathrm{4}}\leqslant{x}\leqslant\frac{\pi}{\mathrm{2}} \\ $$$$\frac{\pi}{\mathrm{4}}\leqslant\frac{\left(\mathrm{4}{n}−\mathrm{3}\right)\pi}{\mathrm{12}}\leqslant\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{3}\leqslant\mathrm{4}{n}−\mathrm{3}\leqslant\mathrm{6} \\ $$$$\mathrm{6}\leqslant\mathrm{4}{n}\leqslant\mathrm{9} \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}\leqslant{n}\leqslant\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\mathrm{1}\frac{\mathrm{1}}{\mathrm{2}}\leqslant{n}\leqslant\mathrm{2}\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${n}=\mathrm{2} \\ $$$${x}=\frac{\left(\mathrm{4}{n}−\mathrm{3}\right)\pi}{\mathrm{12}}=\frac{\left\{\mathrm{4}\left(\mathrm{2}\right)−\mathrm{3}\right\}\pi}{\mathrm{12}}=\frac{\mathrm{5}\pi}{\mathrm{12}} \\ $$$$\mathrm{sin}\left(\mathrm{2}{x}−\mathrm{2}{y}\right)=\mathrm{cos}\:{y} \\ $$$$\mathrm{sin}\left(\mathrm{2}{x}−\mathrm{2}{y}\right)=\mathrm{sin}\left(\frac{\pi}{\mathrm{2}}−{y}\right) \\ $$$$\mathrm{2}{x}−\mathrm{2}{y}=\frac{\pi}{\mathrm{2}}−{y} \\ $$$$\:\:{y}=\frac{\pi}{\mathrm{3}} \\ $$

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