Skip to main content
Back to Trigonometry
JEE Main 2019
Trigonometry
Trigonometric Ratio and Identites
Medium

Question

The value of {\cos ^3}\left( {{\pi \over 8}} \right)$$$${\cos}\left( {{3\pi \over 8}} \right)+{\sin ^3}\left( {{\pi \over 8}} \right)$$$${\sin}\left( {{3\pi \over 8}} \right) is :

Options

Solution

{\cos ^3}\left( {{\pi \over 8}} \right)$$$${\cos}\left( {{3\pi \over 8}} \right)+{\sin ^3}\left( {{\pi \over 8}} \right)$$$${\sin}\left( {{3\pi \over 8}} \right) = cos3(π8)sin(π8)+sin3(π8)cos(π8){\cos ^3}\left( {{\pi \over 8}} \right)\sin \left( {{\pi \over 8}} \right) + {\sin ^3}\left( {{\pi \over 8}} \right)\cos \left( {{\pi \over 8}} \right) = sin(π8)cos(π8)[cos2(π8)+sin2(π8)]\sin \left( {{\pi \over 8}} \right)\cos \left( {{\pi \over 8}} \right)\left[ {{{\cos }^2}\left( {{\pi \over 8}} \right) + {{\sin }^2}\left( {{\pi \over 8}} \right)} \right] = sin(π8)cos(π8)\sin \left( {{\pi \over 8}} \right)\cos \left( {{\pi \over 8}} \right) ×\times 1 = 12×2sin(π8)cos(π8){1 \over 2} \times 2\sin \left( {{\pi \over 8}} \right)\cos \left( {{\pi \over 8}} \right) = 12sin(π4){1 \over 2}\sin \left( {{\pi \over 4}} \right) = 122{1 \over {2\sqrt 2 }}

Practice More Trigonometry Questions

View All Questions