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Question Number 130054 by EDWIN88 last updated on 22/Jan/21

The maximum value of f(x)=sin^2 (((2π)/3)+x)+sin^2 (((2π)/3)−x) is ?

Themaximumvalueoff(x)=sin2(2π3+x)+sin2(2π3x)is?

Answered by liberty last updated on 22/Jan/21

Commented by MJS_new last updated on 22/Jan/21

0≤sin^2  α ≤1  ⇒ 0≤sin^2  α +sin^2  β ≤2  answer has to be within this interval

0sin2α10sin2α+sin2β2answerhastobewithinthisinterval

Commented by MJS_new last updated on 22/Jan/21

error in line 3, it musr be −cos 2x

errorinline3,itmusrbecos2x

Commented by liberty last updated on 22/Jan/21

thank you sir

thankyousir

Answered by MJS_new last updated on 22/Jan/21

sin^2  (a−x) +sin^2  (a+x) =  =(sin a cos x −cos a sin x)^2 +(sin a cos x +cos a sin x)^2 =  =2(cos^2  a sin^2  x +2sin^2  a cos^2  x)=  =2((1−sin^2  a)sin^2  x +2sin^2  a (1−sin^2  x))=  =2(sin^2  a +(1−2sin^2  a)sin^2  x)=       [a=((2π)/3)]  =(3/2)−sin^2  x  (1/2)≤(3/2)−sin^2  x ≤(3/2)

sin2(ax)+sin2(a+x)==(sinacosxcosasinx)2+(sinacosx+cosasinx)2==2(cos2asin2x+2sin2acos2x)==2((1sin2a)sin2x+2sin2a(1sin2x))==2(sin2a+(12sin2a)sin2x)=[a=2π3]=32sin2x1232sin2x32

Commented by liberty last updated on 22/Jan/21

haha..what wrong my solution

haha..whatwrongmysolution

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