Question

$\cos 2(\theta +\Phi )+4\cos (\theta +\Phi )\sin \theta \sin \Phi +2{\sin }^2\Phi =?$

• A. $\cos 2\theta$
• B. $\cos 3\theta$
• C. $\sin 2\theta$
• D. $\sin 3\theta$

Answer 1

The correct option is A

$\cos 2\theta$

Explanation for the correct option:

Step 1. solve the given equation:

$\cos 2(\theta +\varphi )+4\cos (\theta +\varphi )\sin \theta \sin \varphi +2{\sin }^2\varphi$

$=2{\cos }^2(\theta +\varphi )-1+4\cos (\theta +\varphi )\sin \theta \sin \varphi +2{\sin }^2\varphi$ ; $\left[\mathbf{\because }\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{2}\mathit{\theta }\mathbf{=}\mathbf{2}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{\theta }\mathbf{-}\mathbf{1}\right]$

$=2cos(\theta +\varphi )\left[\cos (\theta +\varphi )+2\sin \theta \sin \varphi \right]-1+2{\sin }^2\varphi$

Step 2. Use formula $\mathbf{2}\mathbf{sinA}\mathbf{}\mathbf{sinB}\mathbf{}\mathbf{=}\mathbf{}\mathbf{cos}\mathbf{}\mathbf{(}\mathbf{A}\mathbf{-}\mathbf{B}\mathbf{)}\mathbf{-}\mathbf{cos}\mathbf{}\mathbf{(}\mathbf{A}\mathbf{+}\mathbf{B}\mathbf{)}\mathbf{,}\mathbf{}\mathbf{1}\mathbf{-}\mathbf{2}{\mathbf{sin}}^{\mathbf{2}}\mathbf{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathbf{cos}\mathbf{}\mathbf{2}\mathbf{\theta }$

$=2cos(\theta +\varphi )\left[\cos (\theta +\varphi )+\cos (\theta -\varphi )-\cos (\theta +\varphi )\right]-cos2\varphi$

$=2\cos (\theta +\varphi )\cos (\theta -\varphi )-\cos 2\varphi$

Step 3. Use formula $\mathbf{2}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathit{A}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathit{B}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{(}\mathit{A}\mathbf{+}\mathit{B}\mathbf{)}\mathbf{}\mathbf{+}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{(}\mathit{A}\mathbf{-}\mathit{B}\mathbf{)}$

$=\cos \left(\theta +\varphi +\theta -\varphi \right)+\cos \left(\theta +\varphi -\theta +\varphi \right)-\cos 2\varphi$

$=\cos 2\theta +\cos 2\varphi -\cos 2\varphi$

$=\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{2}\mathit{\theta }$

Hence, Option ‘A’ is Correct.