Question

If $A+B+C={{270}^{\circ }}$, then $\cos 2A+\cos 2B+\cos 2C+4\sin A\sin B\sin C$ is equal to
(a)0
(b)1
(c)2
(d)3

Answer 1

Hint: Use the following trigonometric identities to simplify the given equation.
$\cos (A+B)=\cos A\cos B-\sin A\sin B$
$\cos (A-B)=\cos A\cos B+\sin A\sin B$
$\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$
$\cos A-\cos B=2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)$
$\cos 2A=1-2{{\sin }^{2}}A$

Complete step by step answer:
We have, if $A+B+C={{270}^{\circ }}$, then $\cos 2A+\cos 2B+\cos 2C+4\sin A\sin B\sin C=?$
Let $\cos 2A+\cos 2B+\cos 2C+4\sin A\sin B\sin C=k$.
Now,
$k=\cos 2A+\cos 2B+\cos 2C+4\sin A\sin B\sin C...................(1)$
We know the trigonometric identity, $\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$
Using this trigonometric identity, equation (1) becomes,
$k=\cos 2A+(\cos 2B+\cos 2C)+4\sin A\sin B\sin C$
\[=\cos 2A+\left( 2\cos \left( \dfrac{2B+2C}{2} \right)\cos \left( \dfrac{2B-2C}{2} \right) \right)+4\sin A\sin B\sin C\]
\[=\cos 2A+\left( 2\cos \left( B+C \right)\cos \left( B-C \right) \right)+4\sin A\sin B\sin C\]
We are given that, $A+B+C={{270}^{\circ }}$.
On rearranging, the above can be rewritten as $B+C={{270}^{\circ }}-A$.
Substituting $B+C={{270}^{\circ }}-A$ in the equation, we obtain,
\[k=\cos 2A+2\cos \left( {{270}^{\circ }}-A \right)\cos \left( B-C \right)+4\sin A\sin B\sin C\]
We know that, $\cos ({{270}^{\circ }}-\theta )=-\sin \theta $(a proof for the same is discussed later on in this article)
Hence,
\[k=\cos 2A-2\sin A\cos \left( B-C \right)+4\sin A\sin B\sin C\]
We also know that, $\cos 2A=1-2{{\sin }^{2}}A$
Substituting $\cos 2A=1-2{{\sin }^{2}}A$ in the equation, we obtain,
\[k=1-2{{\sin }^{2}}A-2\sin A\cos \left( B-C \right)+4\sin A\sin B\sin C\]
We take $2\sin A$ common for the second and third term. The equation then becomes,
\[k=1-2\sin A\left( \sin A+\cos \left( B-C \right) \right)+4\sin A\sin B\sin C\]
We are given that, $A+B+C={{270}^{\circ }}$.
On rearranging, the above can be rewritten as $A={{270}^{\circ }}-(B+C)$.
Substituting $A={{270}^{\circ }}-(B+C)$ in the equation, we get,
\[k=1-2\sin A\left( \sin \left( {{270}^{\circ }}-(B+C) \right)+\cos \left( B-C \right) \right)+4\sin A\sin B\sin C\]
We know that, $\sin ({{270}^{\circ }}-\theta )=-\cos \theta $
Hence, the equation can be rewritten as,
\[k=1-2\sin A\left( -\cos \left( B+C \right)+\cos \left( B-C \right) \right)+4\sin A\sin B\sin C\]
\[k=1-2\sin A\left( \cos \left( B-C \right)-\cos \left( B+C \right) \right)+4\sin A\sin B\sin C\]
We know the trigonometric identity, $\cos A-\cos B=2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)$
Using the above trigonometric identity, the equation can be rewritten as,
\[k=1-2\sin A\left( 2\sin \left( \dfrac{B+C+B-C}{2} \right)\sin \left( \dfrac{B+C-B+C}{2} \right) \right)+4\sin A\sin B\sin C\]
\[k=1-2\sin A\left( 2\sin \left( \dfrac{2B}{2} \right)\sin \left( \dfrac{2C}{2} \right) \right)+4\sin A\sin B\sin C\]
\[k=1-2\sin A.2\sin B\sin C+4\sin A\sin B\sin C\]
\[k=1-4\sin A\sin B\sin C+4\sin A\sin B\sin C\]
\[\therefore k=1\]
But, from equation (1),
$k=\cos 2A+\cos 2B+\cos 2C+4\sin A\sin B\sin C$
$\therefore \cos 2A+\cos 2B+\cos 2C+4\sin A\sin B\sin C=1$
Hence, the value of $\cos 2A+\cos 2B+\cos 2C+4\sin A\sin B\sin C$at $A+B+C={{270}^{\circ }}$ is equal to 1.

So, the correct answer is “Option B”.

Note: Remembering a lot of trigonometric equations may seem tricky and confusing. But, we can easily derive any of the trigonometric equations from the very basic trigonometric equations.
For example,
$\cos \left( {{270}^{\circ }}-\theta \right)=-\sin \theta $. One can easily go wrong with signs of such equations. If we look at the LHS of this equation we can see that it is in the form of the trigonometric identity$\cos (A-B)$. Using the this trigonometric equation $\cos ({{270}^{\circ }}-\theta )$ can be written as $\cos {{270}^{\circ }}\cos \theta +\sin {{270}^{\circ }}\sin \theta $.
Value of $\cos {{270}^{\circ }}=0$ and value of $\sin {{270}^{\circ }}=-1$
Hence, $\cos ({{270}^{\circ }}-\theta )=(-1)\sin \theta =-\sin \theta $