Question

If $\sin \left(\theta +\alpha \right)=a$ and $\sin \left(\theta +\beta \right)=b$, then $\cos 2\left(\alpha -\beta \right)-\left(4ab\right)\cos \left(\alpha -\beta \right)$ is equal to

• A. $1–a^2–b^2$
• B. $1–2a^2–2b^2$
• C. $2+a^2+b^2$
• D. $2-a^2-b^2$

Answer 1

The correct option is B

$1–2a^2–2b^2$

Explanation for the correct option:

Step 1:Simplify the function using trigonometric identities

As we know that $\sin \left(\theta +\alpha \right)=a$ and $\sin \left(\theta +\beta \right)=b$.

Then, by the trigonometric identity ${\mathbf{sin}}^{\mathbf{2}}\left(A\right)\mathbf{+}{\mathbf{cos}}^{\mathbf{2}}\left(A\right)\mathbf{=}\mathbf{1}$, we have

$\begin{array}{rcl}\therefore \cos \left(\theta +\alpha \right)& =& \sqrt{1-{\sin }^2\left(\theta +\alpha \right)}\\ & =& \sqrt{1-a^2}\dots \left(1\right)\end{array}$

and

$\begin{array}{rcl}\therefore \cos \left(\theta +\beta \right)& =& \sqrt{1-{\sin }^2(\theta +\beta )}\\ & =& \sqrt{1-b^2}\dots \left(2\right)\end{array}$

The angle that is required is $\alpha -\beta$ which can be given as

$\alpha -\beta =\left(\theta +\alpha \right)-\left(\theta +\beta \right)$

Here, we will apply the trigonometric function $\cos$ on both sides. The equation will become

$\cos \left(\alpha -\beta \right)=\cos \left(\left(\theta +\alpha \right)-\left(\theta +\beta \right)\right)$

Step 2: Apply the trigonometric identity $\mathbf{cos}\left(A-B\right)\mathbf{=}\mathbf{cos}\left(A\right)\mathbf{cos}\left(B\right)\mathbf{+}\mathbf{sin}\left(A\right)\mathbf{sin}\left(B\right)$ on the right hand side

$\cos \left(\alpha -\beta \right)=\cos \left(\theta +\alpha \right)\cos \left(\theta +\beta \right)+\sin \left(\theta +\alpha \right)\sin \left(\theta +\beta \right)$

From the given information and equations $\left(1\right)$ and $\left(2\right)$, we have

$\begin{array}{rcl}\therefore \cos \left(\alpha -\beta \right)& =& \left(\sqrt{1-a^2}\right)\left(\sqrt{1-b^2}\right)+ab\\ & =& \sqrt{\left(1-a^2\right)\left(1-b^2\right)}+ab\\ & =& \sqrt{1-a^2-b^2+a^2{b}^2}+ab\dots \left(1\right)\end{array}$

Step 3: Derivation of the expression

The given expression is $\cos 2\left(\alpha -\beta \right)-\left(4ab\right)\cos \left(\alpha -\beta \right)$.

Since, $\mathbf{cos}\mathbf{2}\left(\theta \right)\mathbf{=}\mathbf{2}{\mathbf{cos}}^{\mathbf{2}}\left(\theta \right)\mathbf{-}\mathbf{1}$, then the above expression will be

$\begin{array}{rcl}\therefore \cos 2\left(\alpha -\beta \right)-\left(4ab\right)\cos \left(\alpha -\beta \right)& =& 2{\cos }^2\left(\alpha -\beta \right)-1-\left(4ab\right)\cos \left(\alpha -\beta \right)\\ & =& 2\cos \left(\alpha -\beta \right)\left[\cos \left(\alpha -\beta \right)-2ab\right]-1\end{array}$

From equation $\left(3\right)$, we have

$\begin{array}{rcl}\therefore \cos 2\left(\alpha -\beta \right)-\left(4ab\right)\cos \left(\alpha -\beta \right)& =& 2\left(\sqrt{1-a^2-b^2+a^2{b}^2}+ab\right)\left[\sqrt{1-a^2-b^2+a^2{b}^2}+ab-2ab\right]-1\\ & =& 2\left(\sqrt{1-a^2-b^2+a^2{b}^2}+ab\right)\left[\sqrt{1-a^2-b^2+a^2{b}^2}-ab\right]-1\end{array}$

Here, we can apply the algebraic identity, $\left(a^2-b^2\right)\mathbf{=}\left(a+b\right)\left(a-b\right)$,

$\begin{array}{rcl}\therefore \cos 2\left(\alpha -\beta \right)-\left(4ab\right)\cos \left(\alpha -\beta \right)& =& 2\left[{\left(\sqrt{1-a^2-b^2+a^2{b}^2}\right)}^2-{\left(ab\right)}^2\right]-1\\ & =& 2\left[1-a^2-b^2+a^2{b}^2-a^2{b}^2\right]-1\\ & =& 2-2a^2-2b^2-1\\ & =& 1-2a^2-2b^2\end{array}$

Hence, the correct option is (B).