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sin-16-cos-16-without-using-cos-3-algebric-method-please-




Question Number 58521 by malwaan last updated on 24/Apr/19
sin 16^°  =?  cos 16^°  =?  without using cos(3θ)  algebric method please
\boldsymbolsin16°=?\boldsymbolcos16°=?\boldsymbolwithout\boldsymbolusing\boldsymbolcos(3θ)\boldsymbolalgebric\boldsymbolmethod\boldsymbolplease
Answered by tanmay last updated on 24/Apr/19
trying calculus method  y=sinx  (dy/dx)=cosx  (dy/dx)≈((△y)/(△x))  △y=(dy/dx)△x  now x=15^o    x+△x=16^o    so △x=1^o =(π/(180))radian  sin15^o =sin(45−30)=sin45^o cos30^o −cos45^o sin30^o   sin15^o =(1/( (√2)))×((√3)/2)−(1/( (√2)))×(1/2)=(((√3) −1)/(2(√2)))  cos15^o =cos(45^o −30^o )=(1/( (√2)))×((√3)/2)+(1/( (√2)))×(1/2)  cos15^o =(((√3) +1)/(2(√2)))  △y=(dy/dx)×△x  △y=(cosx)△x  △y=cos15^o ×(π/(180))  △y=((((√3) +1)/(2(√2))))×(π/(180))  hence sin16^o =(((√3) −1)/(2(√2)))+(((√3) +1)/(2(√2)))×(π/(180))  sin16^o =0.2726776311  sin16^o ≈0.2726  now if y=cosx  (dy/dx)=((△y)/(△x))=−sinx  △y=(−sinx)△x  =(−(((√3) −1)/(2(√2))))×(π/(180))  =(−0.0045172445)  so cos (16^o )=(((√3) +1)/(2(√2)))−0.0045172445  cos16^o =0.9614085818≈0.9614
tryingcalculusmethody=sinxdydx=cosxdydxyxy=dydxxnowx=15ox+x=16osox=1o=π180radiansin15o=sin(4530)=sin45ocos30ocos45osin30osin15o=12×3212×12=3122cos15o=cos(45o30o)=12×32+12×12cos15o=3+122y=dydx×xy=(cosx)xy=cos15o×π180y=(3+122)×π180hencesin16o=3122+3+122×π180sin16o=0.2726776311sin16o0.2726nowify=cosxdydx=yx=sinxy=(sinx)x=(3122)×π180=(0.0045172445)socos(16o)=3+1220.0045172445cos16o=0.96140858180.9614
Commented by malwaan last updated on 25/Apr/19
thank you so much sir
thankyousomuchsirthankyousomuchsir

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