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Question

Prove that (23+3)sinθ+23cosθ lies between -(23+15) and (23+15).


Solution

(23+3)sinθ+23cosθ
assume a=23+3,b=23
a2+b2=12+9+123+12=33+123
Dividing and multiplying the above equation with above value we get,
33+123(23+333+123sinθ+2333+123cosθ)
Assume tanϕ=ab,
We have sinϕ=aa2+b2.cosϕ=ba2+b2
So above expression changes to 33+123(sinϕsinθ+cosϕcostheta)
which is equal to 33+123cos(θϕ)
We know that maximum and minimum value of any cosine term is +1 and -1
33+123=15+12+6+123
We know that 123+6<125 because value of 53 is more than 0.5
so if we replace 123+6 with 125 the above inequality still holds
So range of above expression can be 15+12+125=23+15
(23+15)<33+123cos(θϕ)<23+15

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