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Question: If \({\cos ^2}A + co{s^2}B + {\cos ^2}C = 1,\) then \(\Delta ABC\) is A) Equilateral B) Isosce...

If cos⁡2A+cos2B+cos⁡2C=1,{\cos ^2}A + co{s^2}B + {\cos ^2}C = 1, then ΔABC\Delta ABC is
A) Equilateral
B) Isosceles
C) Scalene
D) Right angled

Explanation

Solution

Given cos⁡2A+cos2B+cos⁡2C=1,{\cos ^2}A + co{s^2}B + {\cos ^2}C = 1, we know that cos⁡2θ=2cos⁡2θ−1\cos 2\theta = 2{\cos ^2}\theta - 1 then convert cos⁡2A{\cos ^2}A and cos2Bco{s^2}B into cos⁡2A\cos 2A and cos⁡2B\cos 2B. Then equation will be cos⁡2A+cos⁡2B+2cos⁡2C=0\cos 2A + \cos 2B + 2{\cos ^2}C = 0. we will use cos⁡A+cos⁡B=2cos⁡(A+B2)cos⁡(A−B2)\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right) for simplification. After simplification we will get cos⁡Ccos⁡Bcos⁡A=0\cos C\cos B\cos A = 0 this is possible only when any one angle is the right angle triangle.

Complete step by step solution:
Given cos⁡2A+cos2B+cos⁡2C=1,{\cos ^2}A + co{s^2}B + {\cos ^2}C = 1,
Multiplying 2 to both side
2cos⁡2A+2cos2B+2cos⁡2C=22{\cos ^2}A + 2co{s^2}B + 2{\cos ^2}C = 2
Convert cos⁡2A{\cos ^2}A and cos2Bco{s^2}B into cos⁡2A\cos 2A and cos⁡2B\cos 2B by using cos⁡2θ=2cos⁡2θ−1\cos 2\theta = 2{\cos ^2}\theta - 1
cos⁡2A+1+cos⁡2B+1+2cos⁡2C=2\cos 2A + 1 + \cos 2B + 1 + 2{\cos ^2}C = 2
cos⁡2A+cos⁡2B+2cos⁡2C=0\cos 2A + \cos 2B + 2{\cos ^2}C = 0
It is known that cos⁡A+cos⁡B=2cos⁡(A+B2)cos⁡(A−B2)\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right) then equation will be
2cos⁡(2A+2B2)cos⁡(2A−2B2)+2cos⁡2C=02\cos \left( {\dfrac{{2A + 2B}}{2}} \right)\cos \left( {\dfrac{{2A - 2B}}{2}} \right) + 2{\cos ^2}C = 0
2cos⁡(A+B)cos⁡(A−B)+2cos⁡2C=02\cos \left( {A + B} \right)\cos \left( {A - B} \right) + 2{\cos ^2}C = 0
ABC is a triangle sum of all the angle is 180∘{180^ \circ } so A+B=π−CA + B = \pi - C and cos(π−θ)=−cos⁡θcos\left( {\pi - \theta } \right) = - \cos \theta
−2cos⁡Ccos⁡(A−B)+2cos⁡2C=0- 2\cos C\cos \left( {A - B} \right) + 2{\cos ^2}C = 0
Taking common 2cos⁡C2\cos C
−2cos⁡C(cos⁡(A−B)−cos⁡C)=0- 2\cos C\left( {\cos \left( {A - B} \right) - \cos C} \right) = 0
Again using cos⁡A−cos⁡B=2sin⁡(A+B2)sin⁡(B−A2)\cos A - \cos B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{B - A}}{2}} \right) we get
cos⁡Csin⁡(A−B+C2)sin⁡(A−B−C2)=0\cos C\sin \left( {\dfrac{{A - B + C}}{2}} \right)\sin \left( {\dfrac{{A - B - C}}{2}} \right) = 0
A+C=π−BA + C = \pi - B and B+C=π−AB + C = \pi - A
cos⁡Csin⁡(π2−B)sin⁡(A−π2)=0\cos C\sin \left( {\dfrac{\pi }{2} - B} \right)\sin \left( {A - \dfrac{\pi }{2}} \right) = 0
We know that sin⁡(π2−θ)=cos⁡θ\sin \left( {\dfrac{\pi }{2} - \theta } \right) = \cos \theta then
cos⁡Ccos⁡Bcos⁡A=0\cos C\cos B\cos A = 0
This can be 0 only if and only if one angle of the triangle is the right angle.

Hence proved ABC is a right angled triangle.

Note:
Formula used in this question are
cos⁡A+cos⁡B=2cos⁡(A+B2)cos⁡(A−B2)\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)
cos⁡A−cos⁡B=2sin⁡(A+B2)sin⁡(B−A2)\cos A - \cos B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{B - A}}{2}} \right)
sin⁡(π2−θ)=cos⁡θ\sin \left( {\dfrac{\pi }{2} - \theta } \right) = \cos \theta