Question

If $\cos A = \dfrac{3}{4}$, then $32\sin (\dfrac{A}{2})\sin (\dfrac{{5A}}{2}) = $?
$A)7$
$B)8$
$C)11$
$D)$None of these

Answer 1

Hint: Since from the given that we need to find the values of given trigonometric functions.
So, we need to analyze the given information so that we are able to solve the problem which is given as $\cos A = \dfrac{3}{4}$. The trigonometric Identities are useful whenever trigonometric functions are involved in an expression or an equation and these identities are useful whenever expressions involving trigonometric functions need to be simplified.
Formula to be used:
The trigonometric identities that are used to solve the given problem are as follows.
$2\sin x\sin y = \cos (x - y)\cos (x + y)$
$\cos 2A = 2{\cos ^2}A - 1$
$\cos 3A = 4{\cos ^3}A - 3\cos A$

Complete step by step answer:
Since from the given that we have the value of $\cos A = \dfrac{3}{4}$and we have to find the generalization of the trigonometric form of $32\sin (\dfrac{A}{2})\sin (\dfrac{{5A}}{2})$
Starting with taking common of $16$in the trigonometric form, then we get $32\sin (\dfrac{A}{2})\sin (\dfrac{{5A}}{2}) \Rightarrow 16[2\sin (\dfrac{A}{2})\sin (\dfrac{{5A}}{2})]$
Now, applying the first formula which is given as $2\sin x\sin y = \cos (x - y)\-cos (x + y)$where $x = \dfrac{A}{2},y = \dfrac{{5A}}{2}$
Thus, we get $16[2\sin (\dfrac{A}{2})\sin (\dfrac{{5A}}{2})] \Rightarrow 16[\cos (\dfrac{A}{2} - \dfrac{{5A}}{2}) - \cos (\dfrac{A}{2} + \dfrac{{5A}}{2})]$
Simplifying the equation, we get $16[\cos (2A) - \cos (3A)]$
Now applying the second and third formula of trigonometric we get; $16[\cos (2A) - \cos (3A)] \Rightarrow 16[2{\cos ^2}A - 1 - (4{\cos ^3}A - 3\cos A)]$
Simplifying we get, $16[2{\cos ^2}A - 1 - 4{\cos ^3}A + 3\cos A]$
Since the value of $\cos A = \dfrac{3}{4}$and square of the cos is \[{\cos ^2}A = {(\dfrac{3}{4})^2} \Rightarrow \dfrac{9}{{16}}\]and cube of cos is \[{\cos ^3}A = {(\dfrac{3}{4})^3} \Rightarrow \dfrac{{27}}{{64}}\]
Applying these values in the $16[2{\cos ^2}A - 1 - 4{\cos ^3}A + 3\cos A]$then we get, $16[2(\dfrac{9}{{16}}) - 1 - 4(\dfrac{{27}}{{64}}) + 3(\dfrac{3}{4})]$
Thus, solving this with the help of division, multiplication, and subtraction operation we get; $16[2(\dfrac{9}{{16}}) - 1 - 4(\dfrac{{27}}{{64}}) + 3(\dfrac{3}{4})] = 16[\dfrac{9}{8} - 1 - \dfrac{{27}}{{16}} + \dfrac{9}{4}]$
$16[\dfrac{9}{8} - 1 - \dfrac{{27}}{{16}} + \dfrac{9}{4}] = 18 - 16 - 27 + 36 \Rightarrow 11$

So, the correct answer is “Option C”.

Note: Don’t forget that in trigonometry the square of the values \[{\cos ^2}A\]is the cos square because the square will only apply to the trigonometric values, not to the given constants.
If we apply the square as ${(\cos A)^2}$then it is wrong and the answers will be incorrect too.
Some other trigonometric formulas with the sine and cos are
a)$\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$
b)$\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$