Question

Prove that $(2\sqrt{3}+3)sin\theta +2\sqrt{3}cos\theta$ lies between -$(2\sqrt{3}+\sqrt{15})$ and $(2\sqrt{3}+\sqrt{15})$.

Answer 1

$(2\sqrt{3}+3)sin\theta +2\sqrt{3}cos\theta$
assume $a=2\sqrt{3}+3,b=2\sqrt{3}$
$\sqrt{a^2+b^2}=\sqrt{12+9+12\sqrt{3}+12}=\sqrt{33+12\sqrt{3}}$
Dividing and multiplying the above equation with above value we get,
$\sqrt{33+12\sqrt{3}}(\frac{2\sqrt{3}+3}{\sqrt{33}+12\sqrt{3}}sin\theta +\frac{2\sqrt{3}}{\sqrt{33}+12\sqrt{3}}cos\theta )$
Assume $tan\varphi =\frac{a}{b},$
We have $sin\varphi =\frac{a}{\sqrt{a^2+b^2}}.cos\varphi =\frac{b}{\sqrt{a^2+b^2}}$
So above expression changes to $\sqrt{33+12\sqrt{3}}(sin\varphi sin\theta +cos\varphi costheta)$
which is equal to $\sqrt{33+12\sqrt{3}}cos(\theta -\varphi )$
We know that maximum and minimum value of any cosine term is +1 and -1
$\sqrt{33+12\sqrt{3}}=\sqrt{15+12+6+12\sqrt{3}}$
We know that $12\sqrt{3}+6<12\sqrt{5}$ because value of $\sqrt{5}-\sqrt{3}$ is more than 0.5
so if we replace $12\sqrt{3}+6$ with $12\sqrt{5}$ the above inequality still holds
So range of above expression can be $\sqrt{15+12+12\sqrt{5}}=2\sqrt{3}+\sqrt{15}$
$-(2\sqrt{3}+\sqrt{15})<\sqrt{33+12\sqrt{3}}cos(\theta -\varphi )<2\sqrt{3}+\sqrt{15}$