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Question

Show that for all real values of θ, the expansion asin2θ+bsinθcosθ+ccos2θ lies between 12(a+c)12b2+(ac)2 and 12(a+c)+12b2+(ac)2

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Solution

asin2θ+bsinθcosθ+ccos2θ
We know that sin2θ=2sinθcosθ
asin2θ+b2×2sinθcosθ+ccos2θ
asin2θ+b2sin2θ+ccos2θ+csin2θcsin2θ [add and substract csin2θ]
(ac)sin2θ+b2sin2θ+c(cos2θ+sin2θ)
We know sin2θ=1cos2θ2
(ac)(1cos2θ)2+b2sin2θ+c
(ac)2ac2+c(ac)2cos2θ+b2sin2θ+c
a+c2constant+b2sin2θ(ac2)cos2θmaxandminvalue
We know asinxbcosx
max=a2+b2
min=a+b2
maxa+c2+(b2)2+((ac)2)2
mina+c2(b2)2+((ac)2)2
max value of expansion =a+c2+12b2+(ac)2
min value of expansion =a+c212b2+(ac)2
Hence proved.

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