Solution : The value of cos 36 degrees is \(\sqrt{5} + 1\over 4\). Proof : We know that the value sin 18 degrees is \(\sqrt{5} โ 1\over 4\). We have, \(cos 2\theta\) = \(1 โ 2 sin^2 \theta\) \(\therefore\)ย ย cos 36 = 1 โ \(2 sin^2 18\) \(\implies\)ย cos 36 = 1 โ 2\(({\sqrt{5} โ โฆ
Solution : The value of cos 72 degrees is \(\sqrt{5} โ 1\over 4\). Proof : We know that value of sin 18 degrees is \(\sqrt{5} โ 1\over 4\). Since 72 degree is the complement of 18 degree. \(\therefore\) cos 72 = cos(90 โ 18) = sin 18 = \(\sqrt{5} โ 1\over 4\) Hence, cos 72 โฆ
Solution : The value of sin 72 degrees is \(\sqrt{10 + 2\sqrt{5}}\over 4\). Proof : We know that value of cos 18 degrees is \(\sqrt{10 + 2\sqrt{5}}\over 4\). Since 72 degree is the complement of 18 degree. \(\therefore\) sin 72 = sin(90 โ 18) = cos 18 = \(\sqrt{10 + 2\sqrt{5}}\over 4\) Hence, sin 72 โฆ
Solution : The value of cos 18 degrees is \(\sqrt{10 + 2\sqrt{5}}\over 4\). Proof : We know that the value of sin 18 degrees is \(\sqrt{5} โ 1\over 4\). Putting \(\theta\) = 18 inย \(cos \theta\) = \(\sqrt{1 โ sin^2 \theta}\), we get cos 18 = \(\sqrt{1 โ sin^2 18}\) = \(\sqrt{1 โ ({\sqrt{5} โ โฆ
Solution : The value of sin 18 degrees is \(\sqrt{5} โ 1\over 4\). Proof :ย Let \(\theta\) = 18 degrees. Then, \(5\theta\) = 90 degrees \(\implies\)ย ย \(2\theta\) + \(3\theta\) = 90 \(\implies\)ย ย \(2\theta\) = 90 โ \(3\theta\) \(\implies\)ย ย \(sin 2\theta\) = \(sin (90 โ 3\theta)\) \(\implies\)ย \(sin 2\theta\) = \(cos 3\theta\) By using the formula โฆ
Solution : We have, L.H.S = cot A + cot (60 + A) โย cot (60 โ A) \(\implies\) L.H.S = \(1\over tan A\) + \(1\over tan (60 + A)\) โ \(1\over tan (60 โ A)\) \(\implies\) L.H.S = \(1\over tan A\)+ \(1 โ \sqrt{3} tan A\over \sqrt{3} + tan A\) โ \(1 + \sqrt{3} โฆ
Prove that cot A + cot (60 + A) โย cot (60 โ A) = 3 cot 3A. Read More ยป
Solution : We have, L.H.S = tan A + tan (60 + A) โย tan (60 โ A) \(\implies\) L.H.S = tan A + \(\sqrt{3} + tan A\over 1 โ \sqrt{3} tan A\) โ \(\sqrt{3} โ tan A\over 1 + \sqrt{3} tan A\) [ By using this formula, tan (A + B) = \(tan A โฆ
Prove that tan A + tan (60 + A) โย tan (60 โ A) = 3 tan 3A. Read More ยป
Solution : We have, L.H.S = cos A cos (60 โ A) cos (60 + A) \(\implies\) L.H.S = cos A (\(cos^2 60 โ sin^2 A\)) [ By using this formula, cos (A + B) cos (A โ B) = \(cos^2 A\) โ \(sin^2 B\)ย above ] \(\implies\) L.H.S = cos A (\(1\over 4\) โ โฆ
Prove that cos A cos (60 โ A) cos (60 + A) = \(1\over 4\) cos 3A. Read More ยป
Solution : We have, L.H.S = sin A sin (60 โ A) sin (60 + A) \(\implies\) L.H.S = sin A (\(sin^2 60 โ sin^2 A\)) [ By using this formula, sin (A + B) sin (A โ B) = \(sin^2 A\) โ \(sin^2 B\)ย above ] \(\implies\) L.H.S = sin A (\(3\over 4\) โ โฆ
Prove that sin A sin (60 โ A) sin (60 + A) = \(1\over 4\) sin 3A. Read More ยป
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 = \(2 tan 60\over 1 โ tan^2 60\) = \(2 \times \sqrt{3}\over 1 โ 3\) \(\implies\)ย tan 120 = \(-\sqrt{3}\)
