Question

If $A+B+C=270°$, then $\cos 2A+\cos 2B+\cos 2C+4\sin \left(A\right)\sin \left(B\right)\sin \left(C\right)=?$

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is B

$1$

Explanation for correct option:

Step1. Finding value for $\mathit{c}\mathit{o}\mathit{s}\mathbf{2}\mathit{A}\mathbf{+}\mathit{c}\mathit{o}\mathit{s}\mathbf{2}\mathit{B}\mathbf{+}\mathit{c}\mathit{o}\mathit{s}\mathbf{2}\mathit{C}\mathbf{+}\mathbf{4}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left(}\mathit{A}\mathbf{\right)}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left(}\mathit{B}\mathbf{\right)}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left(}\mathit{C}\mathbf{\right)}$:

Given that,

$A+B+C=270°$

Now

$\begin{array}{rll}A+B+C& =& 270°......\left(i\right)\\ & & \cos 2A+\cos 2B+\cos 2C+4\sin \left(A\right)\sin \left(B\right)\sin \left(C\right)\\ & =& 2\cos (A+B)\cos (A–B)+\cos 2C+4\sin \left(A\right)\sin \left(B\right)s\mathrm{in}\left(C\right)\\ & =& 2cos(270°–C)cos(A–B)+cos2C+4sin\left(A\right)sin\left(B\right)sin\left(C\right)\end{array}$

Step2. Using the formula $\mathit{c}\mathit{o}\mathit{s}\mathbf{2}\mathit{x}\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{2}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{x}$:

$\begin{array}{rcl}& & –2\sin C\cos (A–B)+1–2{\sin }^{2}C+4\sin A\sin B\sin C\\ & =& 1–2\sin C\left[\cos \right(A–B)+\sin C]+4\sin A\sin B\sin C\\ & =& 1–2\sin C\left[\cos \right(A–B)+\sin (270°–(A+B)\left)\right]+4\sin A\sin B\sin C\\ & =& 1–2\sin C\left[\cos \right(A–B)–\cos (A+B\left)\right]+4\sin A\sin B\sin C\\ & =& 1–2\sin C[2\sin A\sin B]+4\sin A\sin B\sin C\\ & =& 1–4\sin A\sin B\sin C+4\sin A\sin B\sin C\\ & =& 1\end{array}$

Hence, correct option is $\mathbf{\left(}\mathit{B}\mathbf{\right)}$.