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Question Number 48526 by Pk1167156@gmail.com last updated on 25/Nov/18
The maximum and minimum values  of   a cos 2θ+ b sin 2θ  are
Themaximumandminimumvaluesofacos2θ+bsin2θare
Answered by tanmay.chaudhury50@gmail.com last updated on 25/Nov/18
a=rsinα   b=rcosα  a^2 +b^2 =r^2   so r=(√(a^2 +b^2 ))   S=acos2θ+bsin2θ    =rsinαcos2θ+rcosαsin2θ  =rsin(α+2θ)  max value of sin(α+2θ)=1  min value of sin(α+2θ)=−1  so max value of acos2θ+bsin2θ  is   =r×1  =(√(a^2 +b^2 ))    min value of acos2θ+bsin2θ  =r×−1  =−(√(a^2 +b^2 ))
a=rsinαb=rcosαa2+b2=r2sor=a2+b2S=acos2θ+bsin2θ=rsinαcos2θ+rcosαsin2θ=rsin(α+2θ)maxvalueofsin(α+2θ)=1minvalueofsin(α+2θ)=1somaxvalueofacos2θ+bsin2θis=r×1=a2+b2minvalueofacos2θ+bsin2θ=r×1=a2+b2
Answered by ajfour last updated on 25/Nov/18
 acos 2θ+bsin 2θ               = (√(a^2 +b^2 ))sin (tan^(−1) (a/b)+2θ)  so maximum value = (√(a^2 +b^2 ))  and minimum value = −(√(a^2 +b^2 )) .
acos2θ+bsin2θ=a2+b2sin(tan1ab+2θ)somaximumvalue=a2+b2andminimumvalue=a2+b2.

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