Question

If A + B + C = $\pi$, then prove that $\cos \left( 2A \right)+\cos \left( 2B \right)+\cos \left( 2C \right)=-1-4\cos \left( A \right)\cos \left( B \right)\cos \left( C \right)$.

Answer 1

Hint: We will use some trigonometric formulas given by$\cos \left( B \right)+\cos \left( C \right)=2\cos \left( \dfrac{B+C}{2} \right)\cos \left( \dfrac{B-C}{2} \right)$ and $\cos \left( 2A \right)=2{{\cos }^{2}}\left( A \right)-1$ for solving the expression. Also, we will apply $\cos \left( B+C \right)=\cos B\cos C-\sin B\sin C,\cos \left( B-C \right)=\cos B\cos C+\sin A\sin B$ to solve the question further. The property A + B + C = $\pi$, given to us is going to be very useful here for making substitutions in terms of $\pi$.

Complete step-by-step answer:

We come to know that the angles satisfy the property of a triangle in which the sum of interior angles of a triangle is equal to $\pi$. Here, we are given this property as A + B + C = $\pi$. Now, we will consider the equation $\cos \left( 2A \right)+\cos \left( 2B \right)+\cos \left( 2C \right)=-1-4\cos \left( A \right)\cos \left( B \right)\cos \left( C \right)$...(i). Now we will consider the left side of the equation (i). That is we now have $\cos \left( 2A \right)+\cos \left( 2B \right)+\cos \left( 2C \right)$.
Now, we will use the formula given by $\cos \left( B \right)+\cos \left( C \right)=2\cos \left( \dfrac{B+C}{2} \right)\cos \left( \dfrac{B-C}{2} \right)$ in the expression $\cos \left( 2A \right)+\cos \left( 2B \right)+\cos \left( 2C \right)$. Therefore, we get
$\begin{aligned} & \cos \left( 2A \right)+\cos \left( 2B \right)+\cos \left( 2C \right)=\cos \left( 2A \right)+2\cos \left( \dfrac{2B+2C}{2} \right)\cos \left( \dfrac{2B-2C}{2} \right) \\\ & \Rightarrow \cos \left( 2A \right)+\cos \left( 2B \right)+\cos \left( 2C \right)=\cos \left( 2A \right)+2\cos \left( B+C \right)\cos \left( B-C \right) \\\ \end{aligned}$.
Now we will apply the formula given by $\cos \left( 2A \right)=2{{\cos }^{2}}\left( A \right)-1$ in the above trigonometric equation. Therefore, we get $\cos \left( 2A \right)+\cos \left( 2B \right)+\cos \left( 2C \right)=2{{\cos }^{2}}\left( A \right)-1+2\cos \left( B+C \right)\cos \left( B-C \right)$...(ii).
Since, we are given that A + B + C = $\pi$. Now, we will take the term A to the right side of the equation. Thus, we get B + C = $\pi$ - A. After this step we will substitute this value in the equation (ii). Therefore, we now have $\cos \left( 2A \right)+\cos \left( 2B \right)+\cos \left( 2C \right)=2{{\cos }^{2}}\left( A \right)-1+2\cos \left( \pi -A \right)\cos \left( B-C \right)$...(iii). As we know that the angle $\pi -A$ lies in the second quadrant also we know that cosine is negative in the second quadrant thus the value of $\cos \left( \pi -A \right)$ becomes $-\cos \left( A \right)$. Now we are going to put this value in equation (iii). This results into $\begin{aligned} & \cos \left( 2A \right)+\cos \left( 2B \right)+\cos \left( 2C \right)=2{{\cos }^{2}}\left( A \right)-1+2\left( -\cos \left( A \right) \right)\cos \left( B-C \right) \\\ & \Rightarrow \cos \left( 2A \right)+\cos \left( 2B \right)+\cos \left( 2C \right)=2{{\cos }^{2}}\left( A \right)-1-2\cos \left( A \right)\cos \left( B-C \right) \\\ \end{aligned}$
Now we will take the term – 1 to the right most side of the right hand side only. Thus, we now have $\cos \left( 2A \right)+\cos \left( 2B \right)+\cos \left( 2C \right)=2{{\cos }^{2}}\left( A \right)-2\cos \left( A \right)\cos \left( B-C \right)-1$. At this step we will take $2\cos \left( A \right)$ term as a common term. So, we get $\cos \left( 2A \right)+\cos \left( 2B \right)+\cos \left( 2C \right)=2\cos \left( A \right)\left[ \cos \left( A \right)-\cos \left( B-C \right) \right]-1$...(iv).
We can clearly see that we are halfway proving $\cos \left( 2A \right)+\cos \left( 2B \right)+\cos \left( 2C \right)=-1-4\cos \left( A \right)\cos \left( B \right)\cos \left( C \right)$. We only need to make the rest of the expression, except 1, into $-4\cos \left( A \right)\cos \left( B \right)\cos \left( C \right)$. For this we will use A + B + C = $\pi$ information again by keeping B and C to the right side of the equal sign. So, we will now substitute A = - (B + C). This results into
$\cos \left( 2A \right)+\cos \left( 2B \right)+\cos \left( 2C \right)=2\cos \left( A \right)\left[ \cos \left( \pi -\left( B+C \right) \right)-\cos \left( B-C \right) \right]-1$and by $\cos \left( \pi -\theta \right)=-\cos \left( \theta \right)$ this gives $\cos \left( 2A \right)+\cos \left( 2B \right)+\cos \left( 2C \right)=2\cos \left( A \right)\left[ -\cos \left( B+C \right)-\cos \left( B-C \right) \right]-1$
By the formulas $\cos \left( B+C \right)=\cos B\cos C-\sin B\sin C,\cos \left( B-C \right)=\cos B\cos C+\sin A\sin B$ into the equation we will have
$\begin{aligned} & \cos \left( 2A \right)+\cos \left( 2B \right)+\cos \left( 2C \right)=2\cos \left( A \right)\left[ -\left( \cos B\cos C-\sin B\sin C \right)-\left( \cos B\cos C+\sin B\sin C \right) \right]-1 \\\ & \Rightarrow \cos \left( 2A \right)+\cos \left( 2B \right)+\cos \left( 2C \right)=2\cos \left( A \right)\left[ -\cos B\cos C+\sin B\sin C-\cos B\cos C-\sin B\sin C \right]-1 \\\ & \Rightarrow \cos \left( 2A \right)+\cos \left( 2B \right)+\cos \left( 2C \right)=2\cos \left( A \right)\left[ -\cos B\cos C-\cos B\cos C \right]-1 \\\ & \Rightarrow \cos \left( 2A \right)+\cos \left( 2B \right)+\cos \left( 2C \right)=2\cos \left( A \right)\left[ -2\cos B\cos C \right]-1 \\\ & \Rightarrow \cos \left( 2A \right)+\cos \left( 2B \right)+\cos \left( 2C \right)=-4\cos \left( A \right)\cos B\cos C-1 \\\ \end{aligned}$
Clearly we can see that the expression is equal to the right side of the equation $\cos \left( 2A \right)+\cos \left( 2B \right)+\cos \left( 2C \right)=-1-4\cos \left( A \right)\cos \left( B \right)\cos \left( C \right)$.
Hence, we have proved the required expression.

Note: We can also use the formula by $\cos \left( B \right)+\cos \left( C \right)=2\cos \left( \dfrac{B+C}{2} \right)\cos \left( \dfrac{B-C}{2} \right)$ as in the form of by $\cos \left( A \right)+\cos \left( B \right)=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$ and after that we can also put by $\cos \left( 2C \right)=2{{\cos }^{2}}\left( C \right)-1$ instead of by $\cos \left( 2A \right)=2{{\cos }^{2}}\left( A \right)-1$. Also we can use the formula $\cos \left( A \right)+\cos \left( B \right)=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$ and $\cos \left( A \right)-\cos \left( B \right)=-2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)$ formulas instead of $\cos \left( B+C \right)=\cos B\cos C-\sin B\sin C,\cos \left( B-C \right)=\cos B\cos C+\sin A\sin B$while solving $\cos \left( 2A \right)+\cos \left( 2B \right)+\cos \left( 2C \right)=2\cos \left( A \right)\left[ -\cos \left( B+C \right)-\cos \left( B-C \right) \right]-1$. Solving the question like this also gives the required answer.