Question -
Answer -
Let us consider LHS:
4 (cos3┬а10o┬а+ sin3┬а20o)
We know that, sin 60o┬а=┬атИЪ3/2 = cos30o
Sin 30o┬а= cos 60o┬а=1/2
So,
Sin (3├Ч20o) = cos (3├Ч10o)
3sin 20┬░тАУ 4sin320┬░ = 4cos310┬░ тАУ3cos 10┬░
(we know, sin 3╬╕ = 3sin ╬╕ тАУ 4sin3┬а╬╕and cos 3╬╕ = 4cos3╬╕ тАУ 3cos╬╕)
So,
4(cos310┬░+sin320┬░) = 3(sin20┬░+cos 10┬░)
= RHS
Hence proved.