Question

Prove that $co{s}^{2}X+co{s}^2(X+\frac{\pi }{3})+co{s}^2(X-\frac{\pi }{3})+co{s}^2=\frac{3}{2}.$

Answer 1

We have, LHS $co{s}^{2}X+co{s}^2(X+\frac{\pi }{3})+co{s}^2(X-\frac{\pi }{3})+co{s}^2=\frac{3}{2}$

$=co{s}^{2}X+{\left[cos(x+\frac{\pi }{3})\right]}^2+{\left[cos(x-\frac{\pi }{3})\right]}^2$

$=co{s}^{2}x+{(cosxcos\frac{\pi }{3}-sinxsin\frac{\pi }{3})}^2+{(cosxcos\frac{\pi }{3}+sinxsin\frac{\pi }{3})}^2$

$[\because cos(A+B)=cosAcosB-sinAsinBandcos(A-B)=cosAcosB+sinAsinB]$

$=co{s}^{2}x+{(\frac{1}{2}cosx-\frac{\sqrt{3}}{2}sinx)}^2+co{s}^{2}x+{(\frac{1}{2}cosx+\frac{\sqrt{3}}{2}sinx)}^2$

$=ci{s}^{2}x+[\frac{1}{4}co{s}^{2}x+\frac{3}{4}si{n}^{2}x-\frac{\sqrt{3}}{2}sinxcosx+\frac{1}{4}co{s}^{2}x+\frac{3}{4}si{n}^{2}x+\frac{\sqrt{3}}{2}sinx+cosx]$

$=co{s}^{2}x+2(\frac{co{s}^{2}x}{4}+\frac{3si{n}^{2}x}{4})[\because (a+b{)}^2+(a-b{)}^2=2(a^2+b^2\left)\right]$

$=\frac{4co{s}^{2}x+2co{s}^{2}x+6si{n}^{2}x}{4}$

$=\frac{6(si{n}^{2}x+c{o}^{2}x)}{4}=\frac{3}{2}=RHS[\because si{n}^2\theta +co{s}^2\theta =1]$