We have,<\/p>\n
L.H.S = sin A sin (60 – A) sin (60 + A)<\/p>\n
\\(\\implies\\) L.H.S = sin A (\\(sin^2 60 – sin^2 A\\))<\/p>\n
[ By using this formula, sin (A + B) sin (A – B) = \\(sin^2 A\\) – \\(sin^2 B\\)\u00a0 above ]<\/p>\n
\\(\\implies\\) L.H.S = sin A (\\(3\\over 4\\) – \\(sin^2 A\\)) = \\(1\\over 4\\) sin A \\((3 – 4 sin^2 A)\\)<\/p>\n
\\(\\implies\\) L.H.S = \\(1\\over 4\\)(3 sin A – \\(4 sin^3 A\\))<\/p>\n
Since 3 sin A – \\(4 sin^3 A\\) = sin 3A,<\/p>\n
\\(\\implies\\) L.H.S = \\(1\\over 4\\) sin 3A = R.H.S<\/p>\n\n\n
<\/p>\n","protected":false},"excerpt":{"rendered":"
Solution : We have, L.H.S = sin A sin (60 – A) sin (60 + A) \\(\\implies\\) L.H.S = sin A (\\(sin^2 60 – sin^2 A\\)) [ By using this formula, sin (A + B) sin (A – B) = \\(sin^2 A\\) – \\(sin^2 B\\)\u00a0 above ] \\(\\implies\\) L.H.S = sin A (\\(3\\over 4\\) – …<\/p>\n