mathemerize

Solution : We have, L.H.S = cos A cos (60 – A) cos (60 + A) \(\implies\) L.H.S = cos A (\(cos^2 60 – sin^2 A\)) [ By using this formula, cos (A + B) cos (A – B) = \(cos^2 A\) – \(sin^2 B\)  above ] \(\implies\) L.H.S = cos A (\(1\over 4\) – …

Prove that cos A cos (60 – A) cos (60 + A) = \(1\over 4\) cos 3A. Read More »

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) = \(sin^2 A\) – \(sin^2 B\)  above ] \(\implies\) L.H.S = sin A (\(3\over 4\) – …

Prove that sin A sin (60 – A) sin (60 + A) = \(1\over 4\) sin 3A. Read More »

Solution : The value of tan 120 degrees is \(-\sqrt{3}\). Proof : We Know that tan 60 = \(\sqrt{3}\). By using formula of tan 2A, tan 120 = \(2 tan 60\over 1 – tan^2 60\) = \(2 \times \sqrt{3}\over 1 – 3\) \(\implies\)  tan 120 = \(-\sqrt{3}\)

Solution : The value of cos 120 degrees is \(-1\over 2\). Proof : We Know that sin 60 = \(\sqrt{3}\over 2\) and cos 60 = \(1\over 2\) By using cos 2A formula, cos 120 = \(cos^2 60\) – \(sin^2 60\) = \(1\over 4\) – \(3\over 4\) \(\implies\)  cos 120 = \(-1\over 2\)

Solution : The value of sin 120 degrees is \(\sqrt{3}\over 2\). Proof : We Know that sin 60 = \(\sqrt{3}\over 2\) and cos 60 = \(1\over 2\) By using the formula of sin 2A, sin 120 = 2 sin 60 cos 60 = 2 \(\times\) \(\sqrt{3}\over 2\) \(\times\) \(1\over 2\) \(\implies\)  sin 120 = \(\sqrt{3}\over …

What is the Value of Sin 120 Degrees ? Read More »