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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))) :}

If2cos(5π4+3x)cos(π4+3x)=0andsin(2x2y)=cosywhereπ4xπ2andπ4yπ2.findthevalueof{sin(2x+y)cos(2x+y)cos(2xy)sin(2xy)

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)

2cos(5π4+3x)cos(π4+3x)=0cos(5π4+3x)=0cos(π4+3x)=05π4+3x=π2(2n1)π4+3x=π2(2n1)nZ3x=π2(2n1)5π43x=π2(2n1)π4x=π6(2n1)5π12x=π6(2n1)π12x={2(2n1)5}π12x={2(2n1)1}π12x=(4n7)π12x=(4n3)π12π4xπ2π4(4n7)π12π2π4(4n3)π12π23π(4n7)π6π3π(4n3)π6π3(4n7)63(4n3)6104n1364n952n13432n94212n314112n214n=3n=2x=(4n7)π12x=(4n3)π12x=(4(3)7)π12x=(4(2)3)π12x=5π12sin(2x2y)=cosysin(2(5π12)2y)=cosysin(5π62y)=sin(π2y)5π62y=π2yy=5π6π2=5π3π6y=π3

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)

2cos(5π4+3x)cos(π4+3x)=0sin6x1=0x=5π12y=π3

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)

AnOtherWay2cos(5π4+3x)cos(π4+3x)noticethatthediffisπ=0cos(π4+π+3x)cos(π4+3x)=0cos(π+θ)=cosθcos(π4+3x)cos(π4+3x)=0cos2(π4+3x)=0cos(π4+3x)=0cos(π4+3x)=cos(π2(2n1))π4+3x=π2(2n1)x=13((2n1)π2π4)=2(2n1)ππ12=(4n3)π12π4xπ2π4(4n3)π12π234n3664n932n94112n214n=2x=(4n3)π12={4(2)3}π12=5π12sin(2x2y)=cosysin(2x2y)=sin(π2y)2x2y=π2yy=π3

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