JEE MAIN - Mathematics (2019 - 9th April Morning Slot - No. 12)
The value of cos210° – cos10°cos50° + cos250° is
$${3 \over 2} + \cos {20^o}$$
$${3 \over 4}$$
$${3 \over 2}(1 + \cos {20^o})$$
$${3 \over 2}$$
Explanation
cos210° – cos10°cos50° + cos250°
= $${1 \over 2}$$[ 2cos210° – 2cos10°cos50° + 2cos250°]
= $${1 \over 2}$$[ 1 + cos20° - cos60° - cos40° + 1 + cos100°]
= $${1 \over 2}$$[ 2 - $${1 \over 2}$$ + cos20° + cos100° - cos40°]
= $${1 \over 2}$$[ $${3 \over 2}$$ + 2cos60°cos40° - cos40°]
= $${1 \over 2}$$[ $${3 \over 2}$$ + cos40° - cos40°]
= $${3 \over 4}$$
= $${1 \over 2}$$[ 2cos210° – 2cos10°cos50° + 2cos250°]
= $${1 \over 2}$$[ 1 + cos20° - cos60° - cos40° + 1 + cos100°]
= $${1 \over 2}$$[ 2 - $${1 \over 2}$$ + cos20° + cos100° - cos40°]
= $${1 \over 2}$$[ $${3 \over 2}$$ + 2cos60°cos40° - cos40°]
= $${1 \over 2}$$[ $${3 \over 2}$$ + cos40° - cos40°]
= $${3 \over 4}$$
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