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))) :}


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

$If 2 \cos (\frac{5 π}{4} + 3 x) \cos (\frac{π}{4} + 3 x) = 0 and$ $\sin (2 x - 2 y) = \cos y where \frac{π}{4} ⩽ x ⩽ \frac{π}{2} and$ $\frac{π}{4} ⩽ y ⩽ \frac{π}{2} . find the value of {\begin{cases} \sin (2 x + y) \\ \cos (2 x + y) \\ \cos (2 x - y) \\ \sin (2 x - y) \end{cases}$
	Answered by Rasheed.Sindhi last updated on 05/Jun/21

	$2 \cos (\frac{5 π}{4} + 3 x) \cos (\frac{π}{4} + 3 x) = 0$ $\cos (\frac{5 π}{4} + 3 x) = 0 ∣ \cos (\frac{π}{4} + 3 x) = 0$ $\frac{5 π}{4} + 3 x = \frac{π}{2} (2 n - 1) ∣ \frac{π}{4} + 3 x = \frac{π}{2} (2 n - 1)$ $n \in Z$ $3 x = \frac{π}{2} (2 n - 1) - \frac{5 π}{4} ∣ 3 x = \frac{π}{2} (2 n - 1) - \frac{π}{4}$ $x = \frac{π}{6} (2 n - 1) - \frac{5 π}{12} ∣ x = \frac{π}{6} (2 n - 1) - \frac{π}{12}$ $x = \frac{{2 (2 n - 1) - 5} π}{12} ∣ x = \frac{{2 (2 n - 1) - 1} π}{12}$ $x = \frac{(4 n - 7) π}{12} ∣ x = \frac{(4 n - 3) π}{12}$ $\frac{π}{4} ⩽ x ⩽ \frac{π}{2}$ $\frac{π}{4} ⩽ \frac{(4 n - 7) π}{12} ⩽ \frac{π}{2} ∣ \frac{π}{4} ⩽ \frac{(4 n - 3) π}{12} ⩽ \frac{π}{2}$ $3 π ⩽ (4 n - 7) π ⩽ 6 π ∣ 3 π ⩽ (4 n - 3) π ⩽ 6 π$ $3 ⩽ (4 n - 7) ⩽ 6 ∣ 3 ⩽ (4 n - 3) ⩽ 6$ $10 ⩽ 4 n ⩽ 13 ∣ 6 ⩽ 4 n ⩽ 9$ $\frac{5}{2} ⩽ n ⩽ \frac{13}{4} ∣ \frac{3}{2} ⩽ n ⩽ \frac{9}{4}$ $2 \frac{1}{2} ⩽ n ⩽ 3 \frac{1}{4} ∣ 1 \frac{1}{2} ⩽ n ⩽ 2 \frac{1}{4}$ $n = 3 ∣ n = 2$ $x = \frac{(4 n - 7) π}{12} ∣ x = \frac{(4 n - 3) π}{12}$ $x = \frac{(4 (3) - 7) π}{12} ∣ x = \frac{(4 (2) - 3) π}{12}$ $x = \frac{5 π}{12}$ $\sin (2 x - 2 y) = \cos y$ $\sin (2 (\frac{5 π}{12}) - 2 y) = \cos y$ $\sin (\frac{5 π}{6} - 2 y) = \sin (\frac{π}{2} - y)$ $\frac{5 π}{6} - 2 y = \frac{π}{2} - y$ $y = \frac{5 π}{6} - \frac{π}{2} = \frac{5 π - 3 π}{6} \Rightarrow y = \frac{π}{3}$
	Answered by MJS_new last updated on 05/Jun/21

	$2 \cos (\frac{5 π}{4} + 3 x) \cos (\frac{π}{4} + 3 x) = 0$ $\sin 6 x - 1 = 0 \Rightarrow x = \frac{5 π}{12}$ $\Rightarrow y = \frac{π}{3}$
	Answered by Rasheed.Sindhi last updated on 06/Jun/21
	$⟨_• ^• ⟩An Other Way⟨_• ^• ⟩ 2cos (((5π)/4)+3x)cos ((π/4)+3x)_(notice that the diff is π) =0 cos ((π/4)+π+3x)cos ((π/4)+3x)=0 cos(π+θ)=−cosθ −cos ((π/4)+3x)cos ((π/4)+3x)=0 cos^2 ((π/4)+3x)=0 cos ((π/4)+3x)=0 cos ((π/4)+3x)=cos((π/2)(2n−1)) (π/4)+3x=(π/2)(2n−1) x=(1/3)((((2n−1)π)/2)−(π/4))=((2(2n−1)π−π)/(12)) =(((4n−3)π)/(12)) (π/4)≤x≤(π/2) (π/4)≤(((4n−3)π)/(12))≤(π/2) 3≤4n−3≤6 6≤4n≤9 (3/2)≤n≤(9/4) 1(1/2)≤n≤2(1/4) n=2 x=(((4n−3)π)/(12))=(({4(2)−3}π)/(12))=((5π)/(12)) sin(2x−2y)=cos y sin(2x−2y)=sin((π/2)−y) 2x−2y=(π/2)−y y=(π/3)$
	$⟨_{∙}^{∙} ⟩ A n O ther W ay ⟨_{∙}^{∙} ⟩$ $2 \cos \underset{n o t i c e t h a t t h e d i f f i s π}{\underset{⏟}{(\frac{5 π}{4} + 3 x) \cos (\frac{π}{4} + 3 x)}} = 0$ $\cos (\frac{π}{4} + π + 3 x) \cos (\frac{π}{4} + 3 x) = 0$ $\cos (π + θ) = - \cos θ$ $- \cos (\frac{π}{4} + 3 x) \cos (\frac{π}{4} + 3 x) = 0$ $\cos^{2} (\frac{π}{4} + 3 x) = 0$ $\cos (\frac{π}{4} + 3 x) = 0$ $\cos (\frac{π}{4} + 3 x) = \cos (\frac{π}{2} (2 n - 1))$ $\frac{π}{4} + 3 x = \frac{π}{2} (2 n - 1)$ $x = \frac{1}{3} (\frac{(2 n - 1) π}{2} - \frac{π}{4}) = \frac{2 (2 n - 1) π - π}{12}$ $= \frac{(4 n - 3) π}{12}$ $\frac{π}{4} ⩽ x ⩽ \frac{π}{2}$ $\frac{π}{4} ⩽ \frac{(4 n - 3) π}{12} ⩽ \frac{π}{2}$ $3 ⩽ 4 n - 3 ⩽ 6$ $6 ⩽ 4 n ⩽ 9$ $\frac{3}{2} ⩽ n ⩽ \frac{9}{4}$ $1 \frac{1}{2} ⩽ n ⩽ 2 \frac{1}{4}$ $n = 2$ $x = \frac{(4 n - 3) π}{12} = \frac{{4 (2) - 3} π}{12} = \frac{5 π}{12}$ $\sin (2 x - 2 y) = \cos y$ $\sin (2 x - 2 y) = \sin (\frac{π}{2} - y)$ $2 x - 2 y = \frac{π}{2} - y$ $y = \frac{π}{3}$
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