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Question Number 142794 by EDWIN88 last updated on 05/Jun/21
If2cos(5π4+3x)cos(π4+3x)=0andsin(2x−2y)=cosywhereπ4⩽x⩽π2andπ4⩽y⩽π2.findthevalueof{sin(2x+y)cos(2x+y)cos(2x−y)sin(2x−y)
Answered by Rasheed.Sindhi last updated on 05/Jun/21
2cos(5π4+3x)cos(π4+3x)=0cos(5π4+3x)=0∣cos(π4+3x)=05π4+3x=π2(2n−1)∣π4+3x=π2(2n−1)n∈Z3x=π2(2n−1)−5π4∣3x=π2(2n−1)−π4x=π6(2n−1)−5π12∣x=π6(2n−1)−π12x={2(2n−1)−5}π12∣x={2(2n−1)−1}π12x=(4n−7)π12∣x=(4n−3)π12π4⩽x⩽π2π4⩽(4n−7)π12⩽π2∣π4⩽(4n−3)π12⩽π23π⩽(4n−7)π⩽6π∣3π⩽(4n−3)π⩽6π3⩽(4n−7)⩽6∣3⩽(4n−3)⩽610⩽4n⩽13∣6⩽4n⩽952⩽n⩽134∣32⩽n⩽94212⩽n⩽314∣112⩽n⩽214n=3∣n=2x=(4n−7)π12∣x=(4n−3)π12x=(4(3)−7)π12∣x=(4(2)−3)π12x=5π12sin(2x−2y)=cosysin(2(5π12)−2y)=cosysin(5π6−2y)=sin(π2−y)5π6−2y=π2−yy=5π6−π2=5π−3π6⇒y=π3
Answered by MJS_new last updated on 05/Jun/21
2cos(5π4+3x)cos(π4+3x)=0sin6x−1=0⇒x=5π12⇒y=π3
Answered by Rasheed.Sindhi last updated on 06/Jun/21
⟨∙∙⟩AnOtherWay⟨∙∙⟩2cos(5π4+3x)cos(π4+3x)⏟noticethatthediffisπ=0cos(π4+π+3x)cos(π4+3x)=0cos(π+θ)=−cosθ−cos(π4+3x)cos(π4+3x)=0cos2(π4+3x)=0cos(π4+3x)=0cos(π4+3x)=cos(π2(2n−1))π4+3x=π2(2n−1)x=13((2n−1)π2−π4)=2(2n−1)π−π12=(4n−3)π12π4⩽x⩽π2π4⩽(4n−3)π12⩽π23⩽4n−3⩽66⩽4n⩽932⩽n⩽94112⩽n⩽214n=2x=(4n−3)π12={4(2)−3}π12=5π12sin(2x−2y)=cosysin(2x−2y)=sin(π2−y)2x−2y=π2−yy=π3
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