If x sin3|+ y cos3|=sin|cos| and xsin|=ycos|, prove x2+y2=1.

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Basanti Dey

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Question is --- X sin3β + Ycos3β =sinβcosβ and, X sinβ =Y cosβ , have to be proved, X2 + Y2 =1 . Or Xsinβ sin2β + Ycos3β = sinβcosβ . Y cosβsin2β + Ycos2β...
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Question is --- X sin3β + Ycos3β =sinβcosβ and, X sinβ =Y cosβ , have to be proved, X2 + Y2 =1 . Or Xsinβ sin2β + Ycos3β = sinβcosβ . Y cosβsin2β + Ycos2β = sinβcosβ ( as X sinβ = Y cosβ) . Y cosβ ( sin2β + cos2β ) = sinβ cosβ . (as sin2β + cos2β = 1). So Y cosβ = sinβ cosβ . or Y= sinβ . now Xsinβ =Ycos β ( put Y=sinβ) . then Xsinβ =sinβcosβ . so X = cosβ. So X2 + Y2 = cos2β + sin2β =1 proved answer Question is X2 + Y2 =1 . So Y cosβ = sinβ cosβ . or Y= sinβ . now Xsinβ =Ycos β ( put Y=sinβ) . then Xsinβ =sinβcosβ . so X = cosβ. So X2 + Y2 = cos2β + sin2β =1 proved answer Question is --- X sin3β + Ycos3β =sinβcosβ and, X sinβ =Y cosβ , have to be proved, X2 + Y2 =1 . Or Xsinβ sin2β + Ycos3β = sinβcosβ . Y cosβsin2β + Ycos2β = sinβcosβ ( as X sinβ = Y cosβ) . Y cosβ ( sin2β + cos2β ) = sinβ cosβ . (as sin2β + cos2β = 1). So Y cosβ = sinβ cosβ . or Y= sinβ . now Xsinβ =Ycos β ( put Y=sinβ) . then Xsinβ =sinβcosβ . so X = cosβ. So X2 + Y2 = cos2β + sin2β =1 proved answer Question is --- X sin3β + Ycos3β =sinβcosβ and, X sinβ =Y cosβ , have to be proved, X2 + Y2 =1 . Or Xsinβ sin2β + Ycos3β = sinβcosβ . Y cosβsin2β + Ycos2β = sinβcosβ ( as X sinβ = Y cosβ) . Y cosβ ( sin2β + cos2β ) = sinβ cosβ . (as sin2β + cos2β = 1). So Y cosβ = sinβ cosβ . or Y= sinβ . now Xsinβ =Ycos β ( put Y=sinβ) . then Xsinβ =sinβcosβ . so X = cosβ. So X2 + Y2 = cos2β + sin2β =1 proved answer Question is --- X2 + Y2 =1 . read less
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