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Question: If \(A+B+C=\pi \), then \({{\cos }^{2}}A+{{\cos }^{2}}B+{{\cos }^{2}}C=1-2\cos A\cos B\cos C\), A....

If A+B+C=πA+B+C=\pi , then cos⁡2A+cos⁡2B+cos⁡2C=1−2cos⁡Acos⁡Bcos⁡C{{\cos }^{2}}A+{{\cos }^{2}}B+{{\cos }^{2}}C=1-2\cos A\cos B\cos C,
A. True
B. False

Explanation

Solution

We start solving the problem by replacing by recalling the identity ′′cos⁡2θ=1−2cos⁡2θ′′''\cos 2\theta =1-2{{\cos }^{2}}\theta ''. We then substitute cos⁡2A{{\cos }^{2}}A, cos⁡2B{{\cos }^{2}}B and cos⁡2C{{\cos }^{2}}C with (1+cos⁡2A)2\dfrac{\left( 1+\cos 2A \right)}{2}, (1+cos⁡2B)2 \dfrac{\left( 1+\cos 2B \right)}{2}\ and (1+cos⁡2C)2\dfrac{\left( 1+\cos 2C \right)}{2} in the left hand side. We then use cos⁡C+cos⁡D=2cos⁡(C+D2)cos⁡(C−D2)\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right) to proceed further through the problem. We then use the given condition A+B+C=πA+B+C=\pi and make subsequent calculations to get the required result.

Complete step by step answer:
According to the problem, we have given the condition A+B+C=πA+B+C=\pi , then we need to prove cos⁡2A+cos⁡2B+cos⁡2C=1−2cos⁡Acos⁡Bcos⁡C{{\cos }^{2}}A+{{\cos }^{2}}B+{{\cos }^{2}}C=1-2\cos A\cos B\cos C.
Proof:
⇒LHS=cos⁡2A+cos⁡2B+cos⁡2C\Rightarrow LHS={{\cos }^{2}}A+{{\cos }^{2}}B+{{\cos }^{2}}C ---(1).
We know cos⁡2θ=2cos⁡2θ−1\cos 2\theta =2{{\cos }^{2}}\theta -1.
⇒2cos⁡2θ=1+cos⁡2θ\Rightarrow 2{{\cos }^{2}}\theta =1+\cos 2\theta .
Dividing both sides by 2, we will get,
⇒1+cos⁡2θ2=cos⁡2θ\Rightarrow \dfrac{1+\cos 2\theta }{2}={{\cos }^{2}}\theta .
We use this result to make replacements for cos⁡2A{{\cos }^{2}}A, cos⁡2B{{\cos }^{2}}Band cos⁡2C{{\cos }^{2}}C in equation (1).
⇒LHS=(1+cos⁡2A2)+(1+cos⁡2B2)+(1+cos⁡2C2)\Rightarrow LHS=\left( \dfrac{1+\cos 2A}{2} \right)+\left( \dfrac{1+\cos 2B}{2} \right)+\left( \dfrac{1+\cos 2C}{2} \right).
Taking all the constants terms together and all the terms containing cosine together, we will get,
⇒LHS=(12+12+12)+(cos⁡2A2)+(cos⁡2B2)+(cos⁡2C2)\Rightarrow LHS=\left( \dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2} \right)+\left( \dfrac{\cos 2A}{2} \right)+\left( \dfrac{\cos 2B}{2} \right)+\left( \dfrac{\cos 2C}{2} \right).
⇒LHS=(32)+(cos⁡2A2)+(cos⁡2B2)+(cos⁡2C2)\Rightarrow LHS=\left( \dfrac{3}{2} \right)+\left( \dfrac{\cos 2A}{2} \right)+\left( \dfrac{\cos 2B}{2} \right)+\left( \dfrac{\cos 2C}{2} \right).

Taking 12\dfrac{1}{2} common from last three terms, we will get,
⇒LHS=32+12[cos⁡2A+cos⁡2B+cos⁡2C]\Rightarrow LHS=\dfrac{3}{2}+\dfrac{1}{2}\left[ \cos 2A+\cos 2B+\cos 2C \right] ---(2).
We know that cos⁡C+cos⁡D=2cos⁡(C+D2)cos⁡(C−D2)\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right) and we use this result to calculate cos⁡2A+cos⁡2B\cos 2A+\cos 2B.
So, cos⁡2A+cos⁡2B=2cos⁡(2A+2B2)cos⁡(2A−2B2)\cos 2A+\cos 2B=2\cos \left( \dfrac{2A+2B}{2} \right)\cos \left( \dfrac{2A-2B}{2} \right)
⇒cos⁡2A+cos⁡2B=2cos⁡(A+B)cos⁡(A−B)\Rightarrow \cos 2A+\cos 2B=2\cos \left( A+B \right)\cos \left( A-B \right)
Replacing cos⁡2A+cos⁡2B\cos 2A+\cos 2B with 2cos⁡(A+B)cos⁡(A−B)2\cos \left( A+B \right)\cos \left( A-B \right) in equation (2) we get,
⇒LHS=32+12[2cos⁡(A+B)cos⁡(A−B)+cos⁡2C]\Rightarrow LHS=\dfrac{3}{2}+\dfrac{1}{2}\left[ 2\cos \left( A+B \right)\cos \left( A-B \right)+\cos 2C \right].
We know, cos⁡2C=2cos⁡2C−1\cos 2C=2{{\cos }^{2}}C-1
⇒LHS=32+12[2cos⁡(A+B)cos⁡(A−B)+2cos⁡2C−1]\Rightarrow LHS=\dfrac{3}{2}+\dfrac{1}{2}\left[ 2\cos \left( A+B \right)\cos \left( A-B \right)+2{{\cos }^{2}}C-1 \right].
⇒LHS=32+12×2cos⁡(A+B)cos⁡(A−B)+12×2cos⁡2C−12\Rightarrow LHS=\dfrac{3}{2}+\dfrac{1}{{ {2}}}\times {2}\cos \left( A+B \right)\cos \left( A-B \right)+\dfrac{1}{{ {2}}}\times {2}{{\cos }^{2}}C-\dfrac{1}{2}.
Taking all the constant terms together, we will get,
⇒LHS=(32−12)+cos⁡(A+B)cos⁡(A−B)+cos⁡2C\Rightarrow LHS=\left( \dfrac{3}{2}-\dfrac{1}{2} \right)+\cos \left( A+B \right)\cos \left( A-B \right)+{{\cos }^{2}}C.
⇒LHS=1+cos⁡(A+B)cos⁡(A−B)+cos⁡2C\Rightarrow LHS=1+\cos \left( A+B \right)\cos \left( A-B \right)+{{\cos }^{2}}C.
According to the problem, we have A+B+C=πA+B+C=\pi .
⇒A+B=π−C\Rightarrow A+B=\pi -C.
Taking cos both sides,
⇒cos⁡(A+B)=cos⁡(π−C)\Rightarrow \cos \left( A+B \right)=\cos \left( \pi -C \right).
⇒cos⁡(A+B)=−cos⁡C\Rightarrow \cos \left( A+B \right)=-\cos C.
So, LHS will change to,
⇒LHS=1+[(−cos⁡C)cos⁡(A−B)+cos⁡2C]\Rightarrow LHS=1+\left[ \left( -\cos C \right)\cos \left( A-B \right)+{{\cos }^{2}}C \right].
Taking (−cos⁡C)\left( -\cos C \right) common from the last two terms, we will get,
⇒LHS=1−cos⁡C[cos⁡(A−B)+cos⁡C]\Rightarrow LHS=1-\cos C\left[ \cos \left( A-B \right)+\cos C \right].
Now, replacing (−cos⁡C)\left( -\cos C \right) with cos⁡(A+B)\cos \left( A+B \right), we will get, [As derived above cos⁡(A+B)=−cos⁡C]\left[ As\ derived\ above\ \cos \left( A+B \right)=-\cos C \right].
⇒LHS=1−cos⁡C[cos⁡(A−B)+cos⁡(A+B)]\Rightarrow LHS=1-\cos C\left[ \cos \left( A-B \right)+\cos \left( A+B \right) \right].
Now, we use the u=identity cos⁡C+cos⁡D=2cos⁡(C+D2)cos⁡(C−D2)\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right).
⇒cos⁡(A−B)+cos⁡(A+B)=2cos⁡[A−B+A+B2]cos⁡[(A−B)−(A+B)2]\Rightarrow \cos \left( A-B \right)+\cos \left( A+B \right)=2\cos \left[ \dfrac{A-B+A+B}{2} \right]\cos \left[ \dfrac{\left( A-B \right)-\left( A+B \right)}{2} \right].
⇒cos⁡(A−B)+cos⁡(A+B)=2cos⁡Acos⁡B\Rightarrow \cos \left( A-B \right)+\cos \left( A+B \right)=2\cos A\cos B.
Putting this in LHS, we will get,
⇒LHS=1−cos⁡C[2cos⁡Acos⁡B]\Rightarrow LHS=1-\cos C\left[ 2\cos A\cos B \right].
⇒LHS=1−2cos⁡Acos⁡Bcos⁡C=RHS\Rightarrow LHS=1-2\cos A\cos B\cos C=RHS.
Thus, the given statement is true.

Note:
We can solve this proof by assigning the values for the angles A, B and C which on adding give us π\pi . We should make mistakes in signs and make multiplication operations which give us wrong results. We can also expect problems that contain sine, tangent functions in place of the cosine function given in the present problem.