Question

Find the value of \[{\cos ^3}\left( {\dfrac{\pi }{8}} \right)\cos \left( {\dfrac{{3\pi }}{8}} \right) + {\sin ^3}\left( {\dfrac{\pi }{8}} \right)\sin \left( {\dfrac{{3\pi }}{8}} \right)\] .
A.\[\dfrac{1}{4}\]
B.\[\dfrac{1}{{2\sqrt 2 }}\]
C.\[\dfrac{1}{2}\]
D.\[\dfrac{1}{{\sqrt 2 }}\]

Answer 1

Hint: First write the formula of sin3x and cos3x then obtain the value of \[{\cos ^3}x,{\sin ^3}x\] and substitute in the given equation and calculate to obtain the required value.

Formula used:
\[\begin{array}{l}\cos 3x = 4{\cos ^3}x - 3\cos x\\{\cos ^3}x = \dfrac{1}{4}\left( {\cos 3x + 3\cos x} \right)\end{array}\]
And
\[\begin{array}{l}\sin 3x = 3\sin x - 4{\sin ^3}x\\{\sin ^3}x = \dfrac{1}{4}(3\sin x - \sin 3x)\end{array}\]
\[\cos 2A = {\cos ^2}A - {\sin ^2}A\]
\[\cos (A - B) = \cos A\cos B + \sin A\sin B\]

Complete step by step solution:
Given trigonometry expression is \[{\cos ^3}\left( {\dfrac{\pi }{8}} \right)\cos \left( {\dfrac{{3\pi }}{8}} \right) + {\sin ^3}\left( {\dfrac{\pi }{8}} \right)\sin \left( {\dfrac{{3\pi }}{8}} \right)\]
Apply the formula \[{\cos ^3}x = \dfrac{1}{4}\left( {\cos 3x + 3\cos x} \right)\] and \[{\sin ^3}x = \dfrac{1}{4}(3\sin x - \sin 3x)\]
\[ = \dfrac{1}{4}\left( {\cos \dfrac{{3\pi }}{8} + 3\cos \dfrac{\pi }{8}} \right)\cos \left( {\dfrac{{3\pi }}{8}} \right) + \dfrac{1}{4}\left( {3\sin \dfrac{\pi }{8} - \sin \dfrac{{3\pi }}{8}} \right)\sin \left( {\dfrac{{3\pi }}{8}} \right)\]
Apply distributive property:
\[ = \dfrac{1}{4}{\cos ^2}\dfrac{{3\pi }}{8} + \dfrac{3}{4}\cos \dfrac{\pi }{8}\cos \left( {\dfrac{{3\pi }}{8}} \right) + \dfrac{3}{4}\sin \dfrac{\pi }{8}\sin \left( {\dfrac{{3\pi }}{8}} \right) - \dfrac{1}{4}{\sin ^2}\left( {\dfrac{{3\pi }}{8}} \right)\]
\[ = \dfrac{1}{4}\left( {{{\cos }^2}\dfrac{{3\pi }}{8} - {{\sin }^2}\dfrac{{3\pi }}{8}} \right) + \dfrac{3}{4}\left( {\cos \dfrac{\pi }{8}\cos \dfrac{{3\pi }}{8} + \sin \dfrac{\pi }{8}\sin \dfrac{{3\pi }}{8}} \right)\]
\[ = \dfrac{1}{4}\cos \dfrac{{6\pi }}{8} + \dfrac{3}{4}\cos \left( {\dfrac{{3\pi }}{8} - \dfrac{\pi }{8}} \right)\]
\[ = \dfrac{1}{4}\cos \dfrac{{3\pi }}{4} + \dfrac{3}{4}\cos \dfrac{\pi }{4}\]
Rewrite \[\cos \dfrac{{3\pi }}{4}\]:
\[ = \dfrac{1}{4}\cos \left( {\pi - \dfrac{\pi }{4}} \right) + \dfrac{3}{4}\cos \dfrac{\pi }{4}\]
Apply trigonometry supplementary angles
\[ = - \dfrac{1}{4}\cos \dfrac{\pi }{4} + \dfrac{3}{4}\cos \dfrac{\pi }{4}\]
Substitute \[\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}\]in the above equation:
\[ = - \dfrac{1}{{4\sqrt 2 }} + \dfrac{3}{{4\sqrt 2 }}\]
\[ = \dfrac{2}{{4\sqrt 2 }}\]
\[ = \dfrac{1}{{2\sqrt 2 }}\]
The correct option is B.

Note: Sometime students did not understand how to find the values of \[{\cos ^3}x,{\sin ^3}x\]. So for this use the formula of cos3x and sin3x then from the formula obtain the required value and substitute in the given expression for further calculation.