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Question: Simplify the following: \(\left( {1 + \cos \dfrac{\pi }{8}} \right)\left( {1 + \cos \dfrac{{3\pi ...

Simplify the following:
(1+cos⁡π8)(1+cos⁡3π8)(1+cos⁡5π8)(1+cos⁡7π8)=\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) =
A. 12\dfrac{1}{2}
B. 14\dfrac{1}{4}
C. 18\dfrac{1}{8}
D. 116\dfrac{1}{{16}}

Explanation

Solution

We can see that in the above question we have trigonometric ratios.So we will first try to convert the values in the similar terms, so that we can create the trigonometric identity as: sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 . We will use this identity to solve the above question.

Complete step by step answer:
Here we have,
(1+cos⁡π8)(1+cos⁡3π8)(1+cos⁡5π8)(1+cos⁡7π8)=\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) = ?
Let us take the third term, in this we can write it as
5π8\dfrac{{5\pi }}{8} as the difference of two numbers.
So we have:
cos⁡5π8=cos⁡(π−3π8)\cos \dfrac{{5\pi }}{8} = \cos \left( {\pi - \dfrac{{3\pi }}{8}} \right)
We know the identity that:
cos⁡(π−θ)=−cos⁡θ\cos (\pi - \theta ) = - \cos \theta
From comparing we have
θ=3π8\theta = \dfrac{{3\pi }}{8}
So we can write that:
cos⁡(π−3π8)=−cos⁡3π8\cos \left( {\pi - \dfrac{{3\pi }}{8}} \right) = - \cos \dfrac{{3\pi }}{8}

Similarly we can write
cos⁡(7π8)=cos⁡(π−π8)\cos \left( {\dfrac{{7\pi }}{8}} \right) = \cos \left( {\pi - \dfrac{\pi }{8}} \right)
By applying the same identity as above we have :
cos⁡(π−π8)=−cos⁡π8\cos \left( {\pi - \dfrac{\pi }{8}} \right) = - \cos \dfrac{\pi }{8}
By putting all the terms back together we have:
(1+cos⁡π8)(1+cos⁡3π8)(1−cos⁡3π8)(1−cos⁡π8)\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)
We will group the similar terms together and we have:
(1+cos⁡π8)(1−cos⁡π8)(1+cos⁡3π8)(1−cos⁡3π8)\left( {1 + \cos \dfrac{\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)
Now we can multiply and write it as
(1−cos⁡2π8)(1−cos⁡23π8)\left( {1 - {{\cos }^2}\dfrac{\pi }{8}} \right)\left( {1 - {{\cos }^2}\dfrac{{3\pi }}{8}} \right)

We know the identity that:
1−cos⁡2θ=sin⁡2θ1 - {\cos ^2}\theta = {\sin ^2}\theta
So by applying this we have
θ=π8\theta = \dfrac{\pi }{8}
So we can write
(1−cos⁡2π8)=sin⁡2π8\left( {1 - {{\cos }^2}\dfrac{\pi }{8}} \right) = {\sin ^2}\dfrac{\pi }{8}
Again we have
(1−cos⁡23π8)\left( {1 - {{\cos }^2}\dfrac{{3\pi }}{8}} \right)
By comparing with the identity we can write this also as
(1−cos⁡23π8)=sin⁡23π8\left( {1 - {{\cos }^2}\dfrac{{3\pi }}{8}} \right) = {\sin ^2}\dfrac{{3\pi }}{8}
By putting the terms back together we have:
sin⁡2π8⋅sin⁡23π8{\sin ^2}\dfrac{\pi }{8} \cdot {\sin ^2}\dfrac{{3\pi }}{8}

Now we will solve this expression. We know the identity that states:
sin⁡(π2−x)=cos⁡x\sin \left( {\dfrac{\pi }{2} - x} \right) = \cos x
So it also true for:
sin⁡2(π2−x)=cos⁡2x{\sin ^2}\left( {\dfrac{\pi }{2} - x} \right) = {\cos ^2}x
So from the above we can write
sin⁡23π8=sin⁡2(π2−π8){\sin ^2}\dfrac{{3\pi }}{8} = {\sin ^2}\left( {\dfrac{\pi }{2} - \dfrac{\pi }{8}} \right)
By applying the identity it gives us:
sin⁡2(π2−3π8)=cos⁡2π8{\sin ^2}\left( {\dfrac{\pi }{2} - \dfrac{{3\pi }}{8}} \right) = {\cos ^2}\dfrac{\pi }{8}
Now we have terms
sin⁡2π8⋅cos⁡2π8{\sin ^2}\dfrac{\pi }{8} \cdot {\cos ^2}\dfrac{\pi }{8}
We can multiply and divide the term by same number i.e. 44, it can be written as:
44×sin⁡2π8cos⁡2π8\dfrac{4}{4} \times {\sin ^2}\dfrac{\pi }{8}{\cos ^2}\dfrac{\pi }{8}

We can take 14\dfrac{1}{4} out of the bracket and write it as:
14(4×sin⁡2π8⋅cos⁡2π8)\dfrac{1}{4}\left( {4 \times {{\sin }^2}\dfrac{\pi }{8} \cdot {{\cos }^2}\dfrac{\pi }{8}} \right)
The above expression can also be written as:
14(2×sin⁡π8⋅cos⁡π8)2\dfrac{1}{4}{\left( {2 \times \sin \dfrac{\pi }{8} \cdot \cos \dfrac{\pi }{8}} \right)^2}
We can see that we get an identity inside the bracket i.e.
2sin⁡θcos⁡θ=2sin⁡θ2\sin \theta \cos \theta = 2\sin \theta , here we have θ=π8\theta = \dfrac{\pi }{8}
So we can write this as:
14(sin⁡2×π8)2\dfrac{1}{4}{\left( {\sin 2 \times \dfrac{\pi }{8}} \right)^2}

On simplifying we have,
14(sin⁡π4)2\dfrac{1}{4}{\left( {\sin \dfrac{\pi }{4}} \right)^2}
We know the value of the function:
sin⁡π4=12\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}
So by putting this value in the expression we have:
14(12)2=14×12\dfrac{1}{4}{\left( {\dfrac{1}{{\sqrt 2 }}} \right)^2} = \dfrac{1}{4} \times \dfrac{1}{2}
It gives us the value: 18\dfrac{1}{8}

Hence the correct option is C.

Note: We should note that in the above solution we have use the multiplication with the algebraic formula i.e.
(a+b)(a−b)=a2−b2(a + b)(a - b) = {a^2} - {b^2}
In the above solution we have:
(1+cos⁡π8)(1−cos⁡π8)\left( {1 + \cos \dfrac{\pi }{8}} \right)\left( {1 - \cos \dfrac{\pi }{8}} \right)
On comparing here we have
a=1,b=cos⁡π8a = 1,b = \cos \dfrac{\pi }{8}
So we can write the above as :
(1+cos⁡π8)(1−cos⁡π8)=(1)2−(cos⁡π8)2\left( {1 + \cos \dfrac{\pi }{8}} \right)\left( {1 - \cos \dfrac{\pi }{8}} \right) = {(1)^2} - {\left( {\cos \dfrac{\pi }{8}} \right)^2}
It gives us the value
1−cos⁡2π81 - {\cos ^2}\dfrac{\pi }{8}.
Similarly when we multiply the second term we have,
(1+cos⁡3π8)(1−cos⁡3π8)=(1)2−(cos⁡3π8)2\left( {1 + \cos \dfrac{{3\pi }}{8}} \right)\left( {1 - \cos \dfrac{{3\pi }}{8}} \right) = {(1)^2} - {\left( {\cos \dfrac{{3\pi }}{8}} \right)^2}
On simplifying it gives us the value
1−cos⁡23π81 - {\cos ^2}\dfrac{{3\pi }}{8}