Question

Simplify the trigonometric expression: ${\left( {\cos \alpha + \cos \beta } \right)^2} + {\left( {\sin \alpha + \sin \beta } \right)^2}$.
(A) $4{\cos ^2}\left( {\dfrac{{\alpha - \beta }}{2}} \right)$
(B) $4{\sin ^2}\left( {\dfrac{{\alpha - \beta }}{2}} \right)$
(C) $4{\cos ^2}\left( {\dfrac{{\alpha + \beta }}{2}} \right)$
(D) $4{\sin ^2}\left( {\dfrac{{\alpha + \beta }}{2}} \right)$

Answer 1

The given question deals with basic simplification of trigonometric functions by using some of the simple trigonometric formulae and identity such as ${\sin ^2}x + {\cos ^2}x = 1$. Basic algebraic rules and trigonometric identities are to be kept in mind while doing simplification in the given problem. We evaluate the whole squares of the terms. We must know the double angle formulae for trigonometric functions in order to get the required result and match the options given in the problem.

Complete answer: In the given problem, we have to simplify the trigonometric expression: ${\left( {\cos \alpha + \cos \beta } \right)^2} + {\left( {\sin \alpha + \sin \beta } \right)^2}$ .
So, we have, ${\left( {\cos \alpha + \cos \beta } \right)^2} + {\left( {\sin \alpha + \sin \beta } \right)^2}$
Using ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ to evaluate the squares of the terms, we get,
$=$ $\left( {{{\cos }^2}\alpha + 2\cos \alpha \cos \beta + {{\cos }^2}\beta } \right) + \left( {{{\sin }^2}\alpha + 2\sin \alpha \sin \beta + {{\sin }^2}\beta } \right)$
On opening bracket and grouping the terms of same angle together, we get,
$=$ $\left( {{{\cos }^2}\alpha + {{\sin }^2}\alpha } \right) + 2\cos \alpha \cos \beta + 2\sin \alpha \sin \beta + \left( {{{\cos }^2}\beta + {{\sin }^2}\beta } \right)$
Now, we know the trigonometric identity ${\sin ^2}x + {\cos ^2}x = 1$. So, we get,
$=$ $1 + 2\cos \alpha \cos \beta + 2\sin \alpha \sin \beta + 1$
Now, adding up the like terms, we get,
$=$ $2 + 2\cos \alpha \cos \beta + 2\sin \alpha \sin \beta$
Now, taking $2$ common from the last two terms, we get,
$=$ $2 + 2\left( {\cos \alpha \cos \beta + \sin \alpha \sin \beta } \right)$
Now, we know the trigonometric formula for cosine of difference of angles as $\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$. So, we get,
$=$ $2 + 2\cos \left( {\alpha - \beta } \right)$
Now, taking $2$ common from the two terms, we get,
$=$ $2\left( {1 + \cos \left( {\alpha - \beta } \right)} \right)$
Now, we know the trigonometric identity $\cos 2x = 2{\cos ^2}x - 1$. So, we get,
$=$ $2\left( {1 + 2{{\cos }^2}\left( {\dfrac{{\alpha - \beta }}{2}} \right) - 1} \right)$
Subtracting the like terms with opposite signs and simplifying the expression, we get,
$=$ $2\left( {2{{\cos }^2}\left( {\dfrac{{\alpha - \beta }}{2}} \right)} \right)$
$=$ $4{\cos ^2}\left( {\dfrac{{\alpha - \beta }}{2}} \right)$
So, the simplified expression of ${\left( {\cos \alpha + \cos \beta } \right)^2} + {\left( {\sin \alpha + \sin \beta } \right)^2}$ is $4{\cos ^2}\left( {\dfrac{{\alpha - \beta }}{2}} \right)$ by the use of simple trigonometric identities and formulae.
Hence, option (A) is the correct answer.

Additional information: Trigonometric functions are also called Circular functions. Trigonometric functions are the functions that relate an angle of a right angled triangle to the ratio of two side lengths. There are $6$ trigonometric functions, namely: $\sin (x)$,$\cos (x)$,$\tan (x)$,$\cos ec(x)$,$\sec (x)$ and $\cot \left( x \right)$ . Also, $\cos ec(x)$ ,$\sec (x)$ and $\cot \left( x \right)$ are the reciprocals of $\sin (x)$,$\cos (x)$ and $\tan (x)$ respectively.

Note:
The problem deals with Trigonometric functions. For solving such problems, trigonometric formulae should be remembered by heart such as: ${\sin ^2}x + {\cos ^2}x = 1$ and $\cos 2x = 2{\cos ^2}x - 1$ . Besides these simple trigonometric formulae, trigonometric identities are also of significant use in such types of questions where we have to simplify trigonometric expressions with help of basic knowledge of algebraic rules and operations. However, questions involving this type of simplification of trigonometric ratios may also have multiple interconvertible answers.