Question

The‌ ‌value‌ ‌of‌ ‌$\left(‌ ‌{1‌ ‌+‌ ‌\cos‌ ‌\dfrac{\pi‌ ‌}{8}}‌ ‌\right)\left(‌ ‌{1‌ ‌+‌ ‌\cos‌ ‌\dfrac{{3\pi‌ ‌}}{8}}‌ ‌\right)\left(‌ ‌{1‌ ‌+‌ ‌ \cos‌ ‌\dfrac{{5\pi‌ ‌}}{8}}‌ ‌\right)\left(‌ ‌{1‌ ‌+‌ ‌\cos‌ ‌\dfrac{{7\pi‌ ‌}}{8}}‌ ‌\right)$‌ ‌is‌ ‌
$\left(‌ ‌1‌ ‌\right)$‌ ‌$\dfrac{1}{2}$‌ ‌
$\left(‌ ‌2‌ ‌\right)$‌ ‌$\dfrac{1}{4}$‌ ‌
$\left(‌ ‌3‌ ‌\right)$‌ ‌$\dfrac{1}{8}$‌ ‌
$\left(‌ ‌4‌ ‌\right)$‌ ‌$\dfrac{{\sqrt‌ ‌{2‌ ‌+‌ ‌1}‌ ‌}}{{2\sqrt‌ ‌2‌ ‌}}$‌

Answer 1

We‌ ‌have‌ ‌to‌ ‌find‌ ‌the‌ ‌value‌ ‌of‌ ‌the‌ ‌given‌ ‌trigonometric‌ ‌function.‌ ‌We‌ ‌solve‌ ‌this‌ ‌question‌ ‌using‌ ‌the‌ ‌concept‌ ‌of‌ ‌the‌ ‌multiplication‌ ‌of‌ ‌terms,‌ ‌the‌ ‌concept‌ ‌of‌ ‌the‌ ‌formulas‌ ‌of‌ ‌the‌ ‌trigonometric‌ ‌function‌ ‌and‌ ‌the‌ ‌various‌ ‌values‌ ‌of‌ ‌the‌ ‌trigonometric‌ ‌function.‌ ‌And‌ ‌the‌ ‌conversion‌ ‌of‌ ‌the‌ ‌angle‌ ‌of‌ ‌the‌ ‌trigonometric‌ ‌function.‌ ‌First,‌ ‌we‌ ‌will‌ ‌take‌ ‌the‌ ‌terms‌ ‌and‌ ‌convert‌ ‌the‌ ‌angle‌ ‌of‌ ‌the‌ ‌terms‌ ‌such‌ ‌that‌ ‌we‌ ‌get‌ ‌the‌ ‌value‌ ‌similar‌ ‌to‌ ‌that‌ ‌of‌ ‌the‌ ‌others.‌ ‌Then‌ ‌we‌ ‌will‌ ‌take‌ ‌two‌ ‌terms‌ ‌of‌ ‌the‌ ‌function‌ ‌then‌ ‌multiply‌ ‌them‌ ‌using‌ ‌the‌ ‌formula‌ ‌of‌ ‌the‌ ‌difference‌ ‌of‌ ‌squares‌ ‌of‌ ‌two‌ ‌numbers‌ ‌and‌ ‌then‌ ‌simplifying‌ ‌the‌ ‌expression‌ ‌by‌ ‌using‌ ‌the‌ ‌formula‌ ‌of‌ ‌relation‌ ‌between‌ ‌sine‌ ‌and‌ ‌cosine‌ ‌functions.‌ ‌Further‌ ‌we‌ ‌will‌ ‌use‌ ‌the‌ ‌formula‌ ‌of‌ ‌the‌ ‌sum‌ ‌of‌ ‌two‌ ‌sine‌ ‌functions‌ ‌and‌ ‌then‌ ‌substituting‌ ‌the‌ ‌values‌ ‌in‌ ‌the‌ ‌expression,‌ ‌we‌ ‌will‌ ‌obtain‌ ‌the‌ ‌answer.‌ ‌

Complete step-by-step solution:
Given:
The value of $\left( {1 + \cos \dfrac{\pi }{8}} \right)\left( {1 + \cos \dfrac{{3\pi }}{8}} \right)\left( {1 + \cos \dfrac{{5\pi }}{8}} \right)\left( {1 + \cos \dfrac{{7\pi }}{8}} \right)$.
We know that the we can convert the angle of the functions as:
$\cos \dfrac{{5\pi }}{8} = \cos \left( {\pi - \dfrac{{3\pi }}{8}} \right)$
$\cos \dfrac{{7\pi }}{8} = \cos \left( {\pi - \dfrac{\pi }{8}} \right)$
As, we know that the value of the cosine function is negative in the second quadrant, so we can write it as:
$\cos \dfrac{{5\pi }}{8} = - \cos \left( {\dfrac{{3\pi }}{8}} \right)$
$\cos \dfrac{{7\pi }}{8} = - \cos \left( {\dfrac{\pi }{8}} \right)$
Now, substituting the values in the expression we get
$\left( {1 + \cos \dfrac{\pi }{8}} \right)\left( {1 + \cos \dfrac{{3\pi }}{8}} \right)\left( {1 + \cos \dfrac{{5\pi }}{8}} \right)\left( {1 + \cos \dfrac{{7\pi }}{8}} \right) = \left( {1 + \cos \dfrac{\pi }{8}} \right)\left( {1 + \cos \dfrac{{3\pi }}{8}} \right)\left( {1 - \cos \dfrac{{3\pi }}{8}} \right)\left( {1 - \cos \dfrac{\pi }{8}} \right)$
Also, we know that the formula for difference of square of two terms is given as:
${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$
Using the formula, we can write the expression as:
$\left( {1 + \cos \dfrac{\pi }{8}} \right)\left( {1 + \cos \dfrac{{3\pi }}{8}} \right)\left( {1 + \cos \dfrac{{5\pi }}{8}} \right)\left( {1 + \cos \dfrac{{7\pi }}{8}} \right) = \left( {1 - {{\cos }^2}\dfrac{\pi }{8}} \right)\left( {1 - {{\cos }^2}\dfrac{{3\pi }}{8}} \right)$
Now we know that
${\sin ^2}x + {\cos ^2}x = 1$
${\sin ^2}x = 1 - {\cos ^2}x$
Using the relation of sine and cosine function, we get the expression as:
$\left( {1 + \cos \dfrac{\pi }{8}} \right)\left( {1 + \cos \dfrac{{3\pi }}{8}} \right)\left( {1 + \cos \dfrac{{5\pi }}{8}} \right)\left( {1 + \cos \dfrac{{7\pi }}{8}} \right) = {\sin ^2}\dfrac{\pi }{8} \times {\sin ^2}\dfrac{{3\pi }}{8}$
Multiplying both numerator and denominator by $4$ we get the expression as:
$\left( {1 + \cos \dfrac{\pi }{8}} \right)\left( {1 + \cos \dfrac{{3\pi }}{8}} \right)\left( {1 + \cos \dfrac{{5\pi }}{8}} \right)\left( {1 + \cos \dfrac{{7\pi }}{8}} \right) = \dfrac{1}{4}\left( {4{{\sin }^2}\dfrac{\pi }{8} \times {{\sin }^2}\dfrac{{3\pi }}{8}} \right)$
$\left( {1 + \cos \dfrac{\pi }{8}} \right)\left( {1 + \cos \dfrac{{3\pi }}{8}} \right)\left( {1 + \cos \dfrac{{5\pi }}{8}} \right)\left( {1 + \cos \dfrac{{7\pi }}{8}} \right) = \dfrac{1}{4}{\left( {2\sin \dfrac{\pi }{8} \times \sin \dfrac{{3\pi }}{8}} \right)^2}$
We know that the formula for difference of two cosine terms is given as:
$- 2\sin A\sin B = \cos \left( {A + B} \right) - \cos \left( {A - B} \right)$
Using the formula, we get the expression as:
$\left( {1 + \cos \dfrac{\pi }{8}} \right)\left( {1 + \cos \dfrac{{3\pi }}{8}} \right)\left( {1 + \cos \dfrac{{5\pi }}{8}} \right)\left( {1 + \cos \dfrac{{7\pi }}{8}} \right) = \dfrac{1}{4}{\left[ { - \left( {\cos \left( {\dfrac{{3\pi }}{8} + \dfrac{\pi }{8}} \right) - \cos \left( {\dfrac{{3\pi }}{8} - \dfrac{\pi }{8}} \right)} \right)} \right]^2}$
$\left( {1 + \cos \dfrac{\pi }{8}} \right)\left( {1 + \cos \dfrac{{3\pi }}{8}} \right)\left( {1 + \cos \dfrac{{5\pi }}{8}} \right)\left( {1 + \cos \dfrac{{7\pi }}{8}} \right) = \dfrac{1}{4}{\left[ { - \left( {\cos \dfrac{\pi }{2} - \cos \dfrac{\pi }{4}} \right)} \right]^2}$
Also, we can write it as:
$\left( {1 + \cos \dfrac{\pi }{8}} \right)\left( {1 + \cos \dfrac{{3\pi }}{8}} \right)\left( {1 + \cos \dfrac{{5\pi }}{8}} \right)\left( {1 + \cos \dfrac{{7\pi }}{8}} \right) = \dfrac{1}{4}{\left[ { - \cos \dfrac{\pi }{2} + \cos \dfrac{\pi }{4}} \right]^2}$
As we know that the values of trigonometric function are as:
$\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$ and $\cos \dfrac{\pi }{2} = 0$
Substituting the values, we get the expression as:
$\left( {1 + \cos \dfrac{\pi }{8}} \right)\left( {1 + \cos \dfrac{{3\pi }}{8}} \right)\left( {1 + \cos \dfrac{{5\pi }}{8}} \right)\left( {1 + \cos \dfrac{{7\pi }}{8}} \right) = \dfrac{1}{4}{\left[ {\dfrac{1}{{\sqrt 2 }}} \right]^2}$
$\left( {1 + \cos \dfrac{\pi }{8}} \right)\left( {1 + \cos \dfrac{{3\pi }}{8}} \right)\left( {1 + \cos \dfrac{{5\pi }}{8}} \right)\left( {1 + \cos \dfrac{{7\pi }}{8}} \right) = \dfrac{1}{4} \times \dfrac{1}{2}$
Thus, we get the value of the function as:
$\left( {1 + \cos \dfrac{\pi }{8}} \right)\left( {1 + \cos \dfrac{{3\pi }}{8}} \right)\left( {1 + \cos \dfrac{{5\pi }}{8}} \right)\left( {1 + \cos \dfrac{{7\pi }}{8}} \right) = \dfrac{1}{8}$
Hence the value of the given trigonometric function is $\dfrac{1}{8}$.
Thus, the correct option is $\left( 3 \right)$.

Note: We can also find the value of the given trigonometric function by taking different values for the difference of two angles , like we could also have taken the two angles as 60° and 45° .
The various trigonometric formulas are given as :
$\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$
$\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$
$\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$
$\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$
All the trigonometric functions are positive in first quadrant , the sin function are positive in second quadrant and rest are negative , the tan function are positive in third quadrant and rest are negative , the cos function are positive in fourth quadrant and rest are negative .