Question

Simplify \[2{\sin ^2}\beta + 4\cos \left( {\alpha + \beta } \right)\sin \alpha \sin \beta + \cos 2\left( {\alpha + \beta } \right)\].
A. \[\sin 2\alpha \]
B. \[\cos 2\beta \]
C. \[\cos 2\alpha \]
D. \[\sin 2\beta \]

Answer 1

Hint: First we will apply the formula \[\cos \left( {x + y} \right) = \cos x\cos y - \sin x\sin y\] to simply the terms \[\cos \left( {\alpha + \beta } \right)\] and \[\cos 2\left( {\alpha + \beta } \right)\]. Then apply the distributive property \[a \cdot \left( {b + c} \right) = a \cdot b + a \cdot c\]. Apply the formulas \[2\sin x\cos x = \sin 2x\] and \[2{\sin ^2}x = 1 - \cos 2x\] to convert the given expression in \[\sin 2\alpha \], \[\sin 2\beta \], \[\cos 2\beta \], and \[\sin 2\beta \].

Formula Used:
\[\cos \left( {x + y} \right) = \cos x\cos y - \sin x\sin y\]
Distributive property: \[a \cdot \left( {b + c} \right) = a \cdot b + a \cdot c\]
\[2\sin x\cos x = \sin 2x\]
\[2{\sin ^2}x = 1 - \cos 2x\]

Complete step by step solution:
Given expression is \[2{\sin ^2}\beta + 4\cos \left( {\alpha + \beta } \right)\sin \alpha \sin \beta + \cos 2\left( {\alpha + \beta } \right)\]
\[ = 2{\sin ^2}\beta + 4\cos \left( {\alpha + \beta } \right)\sin \alpha \sin \beta + \cos \left( {2\alpha + 2\beta } \right)\]
Apply the formula \[\cos \left( {x + y} \right) = \cos x\cos y - \sin x\sin y\] in \[\cos \left( {\alpha + \beta } \right)\] and \[\cos \left( {2\alpha + 2\beta } \right)\].
\[ = 2{\sin ^2}\beta + 4\left( {\cos \alpha \cos \beta - \sin \alpha \sin \beta } \right)\sin \alpha \sin \beta + \left( {\cos 2\alpha \cos 2\beta - \sin 2\alpha \sin 2\beta } \right)\]
Apply the distributive property \[a \cdot \left( {b + c} \right) = a \cdot b + a \cdot c\]
\[ = 2{\sin ^2}\beta + 4\cos \alpha \cos \beta \sin \alpha \sin \beta - 4{\sin ^2}\alpha {\sin ^2}\beta + \left( {\cos 2\alpha \cos 2\beta - \sin 2\alpha \sin 2\beta } \right)\]
\[ = 2{\sin ^2}\beta + \left( {2\cos \alpha \sin \alpha } \right)\left( {2\cos \beta \sin \beta } \right) - 4{\sin ^2}\alpha {\sin ^2}\beta + \left( {\cos 2\alpha \cos 2\beta - \sin 2\alpha \sin 2\beta } \right)\]
Apply the formula \[2\sin x\cos x= \sin 2x\]
\[ = 2{\sin ^2}\beta + \sin 2\alpha \sin 2\beta - 4{\sin ^2}\alpha {\sin ^2}\beta + \cos 2\alpha \cos 2\beta - \sin 2\alpha \sin 2\beta \]
Cancel out \[\sin 2\alpha \sin 2\beta \]
\[ = 2{\sin ^2}\beta - 4{\sin ^2}\alpha {\sin ^2}\beta + \cos 2\alpha \cos 2\beta \]
\[ = 2{\sin ^2}\beta - \left( {2{{\sin }^2}\alpha } \right)\left( {2{{\sin }^2}\beta } \right) + \cos 2\alpha \cos 2\beta \]
Now applying \[2{\sin ^2}x = 1 - \cos 2x\]
\[ = \left( {1 - \cos 2\beta } \right) - \left( {1 - \cos 2\alpha } \right)\left( {1 - \cos 2\beta } \right) + \cos 2\alpha \cos 2\beta \]
Apply distributive property \[a \cdot \left( {b + c} \right) = a \cdot b + a \cdot c\]
\[ = 1 - \cos 2\beta - 1 + \cos 2\alpha + \cos 2\beta - \cos 2\beta \cos 2\alpha + \cos 2\alpha \cos 2\beta \]
Now cancel out 1, \[\cos 2\beta \], and \[\cos 2\alpha \cos 2\beta \]
\[ = \cos 2\alpha \]

Hence option C is the correct option.

Note: Students often confused with the formulas \[\cos \left( {x + y} \right) = \cos x\cos y - \sin x\sin y\] and \[\cos \left( {x + y} \right) = \cos x\cos y + \sin x\sin y\]. The correct formulas are \[\cos \left( {x + y} \right) = \cos x\cos y - \sin x\sin y\] and \[\cos \left( {x - y} \right) = \cos x\cos y + \sin x\sin y\].