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Question

$I n space capital delta A B C sum a cubed c o s invisible function application left parenthesis B minus C right parenthesis equals$

0
abc
3abc
1

hint Hint:

We will rewrite the expression using these formulas $a cos open parentheses B minus C close parentheses space equals space b cos open parentheses B close parentheses plus space c cos open parentheses C close parentheses b cos open parentheses C minus A close parentheses space equals space c cos open parentheses C close parentheses plus space a cos open parentheses A close parentheses c cos open parentheses A minus B close parentheses space equals space a cos open parentheses A close parentheses plus space b cos open parentheses B close parentheses$
and then we will rearrange similar terms and then again use these formulas to get the final expression.

The correct answer is: 3abc

Step1: Rewriting the expression
The given expression can be rewritten as $a cubed cos open parentheses B minus C close parentheses space plus space b cubed cos open parentheses C minus A close parentheses space plus space c cubed cos open parentheses A minus B close parentheses$
Step2: Using the trigonometric formula
$a cos open parentheses B minus C close parentheses space equals space b cos open parentheses B close parentheses plus space c cos open parentheses C close parentheses b cos open parentheses C minus A close parentheses space equals space c cos open parentheses C close parentheses plus space a cos open parentheses A close parentheses c cos open parentheses A minus B close parentheses space equals space a cos open parentheses A close parentheses plus space b cos open parentheses B close parentheses$
By putting these values in the given expression we get.
$a squared left parenthesis b cos open parentheses B close parentheses plus c cos open parentheses C close parentheses right parenthesis plus space b squared left parenthesis a cos open parentheses A close parentheses plus c cos open parentheses C close parentheses right parenthesis plus space c squared left parenthesis a cos open parentheses A close parentheses plus b cos open parentheses B close parentheses right parenthesis$
=> $a squared b cos open parentheses B close parentheses plus a squared c cos open parentheses C close parentheses space plus space b squared a cos open parentheses A close parentheses plus b squared c cos open parentheses C close parentheses plus space c squared a cos open parentheses A close parentheses plus c squared b cos open parentheses B close parentheses$
Rearranging the above expression we get,
=> $a squared b cos open parentheses B close parentheses plus space b squared a cos open parentheses A close parentheses plus a squared c cos open parentheses C close parentheses plus c squared a cos open parentheses A close parentheses plus c squared b cos open parentheses B close parentheses plus b squared c cos open parentheses C close parentheses$
Step3: Again using the trigonometric formula.
=> $a b left parenthesis a cos open parentheses B close parentheses plus space b cos open parentheses A close parentheses right parenthesis plus a c left parenthesis a cos open parentheses C close parentheses plus c cos open parentheses A close parentheses right parenthesis plus c b left parenthesis c cos open parentheses B close parentheses plus b cos open parentheses C close parentheses right parenthesis$
=> $a b c plus a c b plus c b a$
=> $3 a b c$

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