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Question: Prove that in any triangle \({{\cos }^{2}}A+{{\cos }^{2}}B-{{\cos }^{2}}C=1-2\sin A\sin B\sin C\)...

Prove that in any triangle cos⁡2A+cos⁡2B−cos⁡2C=1−2sin⁡Asin⁡Bsin⁡C{{\cos }^{2}}A+{{\cos }^{2}}B-{{\cos }^{2}}C=1-2\sin A\sin B\sin C

Explanation

Solution

We will first use the trigonometric identity that sin⁡2B+cos⁡2B=1{{\sin }^{2}}B+{{\cos }^{2}}B=1 to simplify the left hand side of the equation to be proved then we will use the identity that cos⁡2A−sin⁡2B=cos⁡(A+B)cos⁡(A−B){{\cos }^{2}}A-{{\sin }^{2}}B=\cos \left( A+B \right)\cos \left( A-B \right) to combine the cosine terms and take cos⁡C\cos C common in the left hand side then we will use the trigonometric identity that cos⁡C+cos⁡D=2cos⁡(C+D2)sin⁡(D−C2)\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{D-C}{2} \right) to further simplify the left hand side and made it equal to the right hand side.

Complete step-by-step answer:
We have been given a triangle ABC and we have to prove that cos⁡2A+cos⁡2B−cos⁡2C=1−2sin⁡Asin⁡Bsin⁡C{{\cos }^{2}}A+{{\cos }^{2}}B-{{\cos }^{2}}C=1-2\sin A\sin B\sin C
We will first draw a ΔABC\Delta ABC and label its sides as a, b, c

Now, we will take the LHS of the question and prove it to be equal to RHS. In LHS we have cos⁡2A+cos⁡2B−cos⁡2C{{\cos }^{2}}A+{{\cos }^{2}}B-{{\cos }^{2}}C
Now we know that sin⁡2B+cos⁡2B=1{{\sin }^{2}}B+{{\cos }^{2}}B=1
Therefore cos⁡2B=1−sin⁡2B{{\cos }^{2}}B=1-{{\sin }^{2}}Busing this we have,
cos⁡2A+cos⁡2B−cos⁡2C=cos⁡2A+1−sin⁡2B−cos⁡2C=1+(cos⁡2A−sin⁡2B)−cos2C{{\cos }^{2}}A+{{\cos }^{2}}B-{{\cos }^{2}}C={{\cos }^{2}}A+1-{{\sin }^{2}}B-{{\cos }^{2}}C=1+\left( {{\cos }^{2}}A-{{\sin }^{2}}B \right)-co{{s}^{2}}C
Now we know the trigonometric identity that,
cos⁡2A−sin⁡2B=cos⁡(A+B)cos⁡(A−B){{\cos }^{2}}A-{{\sin }^{2}}B=\cos \left( A+B \right)\cos \left( A-B \right)
So, we have
cos⁡2A+cos⁡2B−cos⁡2C=1+(cos⁡(A+B)cos⁡(A−B))−cos⁡2C{{\cos }^{2}}A+{{\cos }^{2}}B-{{\cos }^{2}}C=1+\left( \cos \left( A+B \right)\cos \left( A-B \right) \right)-{{\cos }^{2}}C
Also, now we know that according to angle sum property of triangle we have
A+B+C=π A+B=π−C \begin{aligned} & A+B+C=\pi \\\ & A+B=\pi -C \\\ \end{aligned}
So, using this we have,
cos⁡2A+cos⁡2B−cos⁡2C=1+cos⁡(π−C)cos⁡(A−B)−cos⁡2C{{\cos }^{2}}A+{{\cos }^{2}}B-{{\cos }^{2}}C=1+\cos \left( \pi -C \right)\cos \left( A-B \right)-{{\cos }^{2}}C
Also, we know that
cos⁡(π−C)=−cos⁡C\cos \left( \pi -C \right)=-\cos C
So we have
cos⁡2A+cos⁡2B−cos⁡2C=1−cos⁡C(cos⁡(A−B)+cos⁡C){{\cos }^{2}}A+{{\cos }^{2}}B-{{\cos }^{2}}C=1-\cos C\left( \cos \left( A-B \right)+\cos C \right)
Also, we know that
cos⁡C=cos⁡(π−(A+B))\cos C=\cos \left( \pi -\left( A+B \right) \right)
cos⁡2A+cos⁡2B−cos⁡2C=1−cos⁡C(cos⁡(A−B)+cos⁡(π−(A+B))){{\cos }^{2}}A+{{\cos }^{2}}B-{{\cos }^{2}}C=1-\cos C\left( \cos \left( A-B \right)+\cos \left( \pi -\left( A+B \right) \right) \right)
=1−cos⁡C(cos⁡(A−B)+cos⁡(A+B))=1-\cos C\left( \cos \left( A-B \right)+\cos \left( A+B \right) \right)
Now using the trigonometric identity that
cosC⁡+cosD=2sin(C+D2)sin⁡(D−C2)\operatorname{cosC}+cosD=2sin\left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{D-C}{2} \right)
We have,
cos⁡2A+cos⁡2B−cos⁡2C=1−cos⁡C(2sin⁡(A−B+A+B2)sin⁡(A+B−A+B2)){{\cos }^{2}}A+{{\cos }^{2}}B-{{\cos }^{2}}C=1-\cos C\left( 2\sin \left( \dfrac{A-B+A+B}{2} \right)\sin \left( \dfrac{A+B-A+B}{2} \right) \right)
=1−cos⁡C(2sin⁡(2A2)sin⁡(2B2))=1-\cos C\left( 2\sin \left( \dfrac{2A}{2} \right)\sin \left( \dfrac{2B}{2} \right) \right)
=1−cos⁡C(2sin⁡Asin⁡B)=1-\cos C\left( 2\sin A\sin B \right)
=1−2sin⁡Asin⁡Bcos⁡C=1-2\sin A\sin B\cos C
Since, LHS = RHS
Hence proved

Note: In these types of questions it is very important to remember trigonometric identities like
cos⁡A+cosB⁡=2sin⁡(A+B2)sin⁡(B−A2) sin⁡2A+cos⁡2A=1 A+B+C=π \begin{aligned} & \cos A+\operatorname{cosB}=2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{B-A}{2} \right) \\\ & {{\sin }^{2}}A+{{\cos }^{2}}A=1 \\\ & A+B+C=\pi \\\ \end{aligned}