JEE MAIN - Mathematics (2020 - 9th January Morning Slot - No. 16)
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 :
$${\cos ^3}\left( {{\pi \over 8}} \right)$$$${\cos}\left( {{3\pi \over 8}} \right)$$+$${\sin ^3}\left( {{\pi \over 8}} \right)$$$${\sin}\left( {{3\pi \over 8}} \right)$$
is :
$${1 \over {\sqrt 2 }}$$
$${1 \over 2}$$
$${1 \over 4}$$
$${1 \over 2{\sqrt 2 }}$$
Explanation
$${\cos ^3}\left( {{\pi \over 8}} \right)$$$${\cos}\left( {{3\pi \over 8}} \right)$$+$${\sin ^3}\left( {{\pi \over 8}} \right)$$$${\sin}\left( {{3\pi \over 8}} \right)$$
= $${\cos ^3}\left( {{\pi \over 8}} \right)\sin \left( {{\pi \over 8}} \right) + {\sin ^3}\left( {{\pi \over 8}} \right)\cos \left( {{\pi \over 8}} \right)$$
= $$\sin \left( {{\pi \over 8}} \right)\cos \left( {{\pi \over 8}} \right)\left[ {{{\cos }^2}\left( {{\pi \over 8}} \right) + {{\sin }^2}\left( {{\pi \over 8}} \right)} \right]$$
= $$\sin \left( {{\pi \over 8}} \right)\cos \left( {{\pi \over 8}} \right)$$ $$ \times $$ 1
= $${1 \over 2} \times 2\sin \left( {{\pi \over 8}} \right)\cos \left( {{\pi \over 8}} \right)$$
= $${1 \over 2}\sin \left( {{\pi \over 4}} \right)$$
= $${1 \over {2\sqrt 2 }}$$
= $${\cos ^3}\left( {{\pi \over 8}} \right)\sin \left( {{\pi \over 8}} \right) + {\sin ^3}\left( {{\pi \over 8}} \right)\cos \left( {{\pi \over 8}} \right)$$
= $$\sin \left( {{\pi \over 8}} \right)\cos \left( {{\pi \over 8}} \right)\left[ {{{\cos }^2}\left( {{\pi \over 8}} \right) + {{\sin }^2}\left( {{\pi \over 8}} \right)} \right]$$
= $$\sin \left( {{\pi \over 8}} \right)\cos \left( {{\pi \over 8}} \right)$$ $$ \times $$ 1
= $${1 \over 2} \times 2\sin \left( {{\pi \over 8}} \right)\cos \left( {{\pi \over 8}} \right)$$
= $${1 \over 2}\sin \left( {{\pi \over 4}} \right)$$
= $${1 \over {2\sqrt 2 }}$$
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