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Maths-

General

Easy

Question

In any triangle $A B C comma sin to the power of 2 end exponent invisible function application A minus sin to the power of 2 end exponent invisible function application B plus sin to the power of 2 end exponent invisible function application C$ is always equal to

$2 \sin A \sin B \cos C$
$2 \sin A \cos B \sin C$
$2 \sin A \cos B \cos C$
$2 \sin A \sin B \sin C$

The correct answer is: $2 sin invisible function application A cos invisible function application B sin invisible function application C$

$sin to the power of 2 end exponent invisible function application A minus sin to the power of 2 end exponent invisible function application B plus sin to the power of 2 end exponent invisible function application C equals sin invisible function application open parentheses A plus B close parentheses sin invisible function application open parentheses A minus B close parentheses plus sin to the power of 2 end exponent invisible function application C equals sin invisible function application C open parentheses sin invisible function application open parentheses A minus B close parentheses plus sin invisible function application C close parentheses$
$equals sin invisible function application C open parentheses sin invisible function application open parentheses A minus B close parentheses plus sin invisible function application open parentheses A plus B close parentheses close parentheses equals 2 sin invisible function application A cos invisible function application B sin invisible function application C$

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