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Question: If \(\cos A = \dfrac{3}{4}\), then \(32\sin (\dfrac{A}{2})\sin (\dfrac{{5A}}{2}) = \)? \(A)7\) \...

If cos⁡A=34\cos A = \dfrac{3}{4}, then 32sin⁡(A2)sin⁡(5A2)=32\sin (\dfrac{A}{2})\sin (\dfrac{{5A}}{2}) = ?
A)7A)7
B)8B)8
C)11C)11
D)D)None of these

Explanation

Solution

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=34\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.
2sin⁡xsin⁡y=cos⁡(x−y)cos⁡(x+y)2\sin x\sin y = \cos (x - y)\cos (x + y)
cos⁡2A=2cos⁡2A−1\cos 2A = 2{\cos ^2}A - 1
cos⁡3A=4cos⁡3A−3cos⁡A\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=34\cos A = \dfrac{3}{4}and we have to find the generalization of the trigonometric form of 32sin⁡(A2)sin⁡(5A2)32\sin (\dfrac{A}{2})\sin (\dfrac{{5A}}{2})
Starting with taking common of 1616in the trigonometric form, then we get 32sin⁡(A2)sin⁡(5A2)⇒16[2sin⁡(A2)sin⁡(5A2)]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 2sin⁡xsin⁡y=cos⁡(x−y)−cos(x+y)2\sin x\sin y = \cos (x - y)\\-cos (x + y)where x=A2,y=5A2x = \dfrac{A}{2},y = \dfrac{{5A}}{2}
Thus, we get 16[2sin⁡(A2)sin⁡(5A2)]⇒16[cos⁡(A2−5A2)−cos⁡(A2+5A2)]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)]16[\cos (2A) - \cos (3A)]
Now applying the second and third formula of trigonometric we get; 16[cos⁡(2A)−cos⁡(3A)]⇒16[2cos⁡2A−1−(4cos⁡3A−3cos⁡A)]16[\cos (2A) - \cos (3A)] \Rightarrow 16[2{\cos ^2}A - 1 - (4{\cos ^3}A - 3\cos A)]
Simplifying we get, 16[2cos⁡2A−1−4cos⁡3A+3cos⁡A]16[2{\cos ^2}A - 1 - 4{\cos ^3}A + 3\cos A]
Since the value of cos⁡A=34\cos A = \dfrac{3}{4}and square of the cos is cos⁡2A=(34)2⇒916{\cos ^2}A = {(\dfrac{3}{4})^2} \Rightarrow \dfrac{9}{{16}}and cube of cos is cos⁡3A=(34)3⇒2764{\cos ^3}A = {(\dfrac{3}{4})^3} \Rightarrow \dfrac{{27}}{{64}}
Applying these values in the 16[2cos⁡2A−1−4cos⁡3A+3cos⁡A]16[2{\cos ^2}A - 1 - 4{\cos ^3}A + 3\cos A]then we get, 16[2(916)−1−4(2764)+3(34)]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(916)−1−4(2764)+3(34)]=16[98−1−2716+94]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[98−1−2716+94]=18−16−27+36⇒1116[\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⁡2A{\cos ^2}Ais 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{(\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⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B
b)sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B