Question -
Answer -
Let us consider LHS:
cos 78o┬аcos 42o┬аcos 36o
Let us multiply and divide by 2 we get,
cos 78o┬аcos 42o┬аcos 36o┬а=1/2 (2 cos 78o┬аcos 42o┬аcos 36o)
We know, 2 cos A cos B = cos (A + B) + cos (A тАУ B)
Then the above equation becomes,
= 1/2 (cos (78o┬а+ 42o) +cos (78o┬атАУ 42o)) ├Ч cos 36o
= 1/2 (cos 120o┬а+ cos 36o)├Ч cos 36o
= 1/2 (cos (180o┬атАУ 60o) +cos 36o) ├Ч cos 36o
= 1/2 (-cos (60o) + cos 36o) ├Чcos 36o┬а[since, cos(180┬░ тАУ A) = тАУ A]
= 1/2 (-1/2 + (тИЪ5 + 1)/4) ((тИЪ5 + 1)/4) [since, cos 36o┬а=(тИЪ5 + 1)/4]
= 1/2 (тИЪ5 + 1 тАУ 2)/4 ((тИЪ5 + 1)/4)
= 1/2 (тИЪ5 тАУ 1)/4) ((тИЪ5 + 1)/4)
= 1/2 ((тИЪ5)2┬атАУ 12)/16
= 1/2 (5-1)/16
= 1/2 (4/16)
= 1/8
= RHS
Hence proved.