Solution : The general solution of \(cot \theta\) = 0 is given by \(\theta\) = \((2n + 1){\pi\over 2}\), n \(\in\) Z. Proof : We have, \(cot \theta\) = \(OM\over PM\) \(\therefore\) \(cot \theta\) = 0 \(\implies\) \(OM\over PM\) = 0 \(\implies\) OM = 0 \(\implies\) OP coincides with OY or OY’ \(\implies\) \(\theta\) = …
What is the General Solution of \(Cot \theta\) = 0 ? Read More »
Solution : The general solution of \(cos \theta\) = 0 is given by \(\theta\) = \((2n + 1){\pi\over 2}\), n \(\in\) Z. Proof : We have, \(cos \theta\) = \(PM\over OP\) \(\therefore\) \(cos \theta\) = 0 \(\implies\) \(OM\over OP\) = 0 \(\implies\) OM = 0 \(\implies\) OP coincides with OY or OY’ \(\implies\) \(\theta\) = …
What is the General Solution of \(Cos \theta\) = 0 ? Read More »
Solution : The general solution of \(tan \theta\) = 0 is given by \(\theta\) = \(n\pi\), n \(\in\) Z. Proof : We have, \(tan \theta\) = \(PM\over OM\) \(\therefore\) \(tan \theta\) = 0 \(\implies\) \(PM\over OM\) = 0 \(\implies\) PM = 0 \(\implies\) OP coincides with OX or OX’ \(\implies\) \(\theta\) = 0, \(\pi\), \(2\pi\), …
What is the General Solution of \(Tan \theta\) = 0 ? Read More »
Solution : The general solution of \(sin \theta\) = 0 is given by \(\theta\) = \(n\pi\), n \(\in\) Z. Proof : We have, \(sin \theta\) = \(PM\over OP\) \(\therefore\) \(sin \theta\) = 0 \(\implies\) \(PM\over OP\) = 0 \(\implies\) PM = 0 \(\implies\) OP coincides with OX or OX’ \(\implies\) \(\theta\) = 0, \(\pi\), \(2\pi\), …
What is the General Solution of \(Sin \theta\) = 0 ? Read More »
Solution : The value of cos 54 degrees is \(\sqrt{10 – 2\sqrt{5}}\over 4\). Proof : We know that value of sin 36 degrees is \(\sqrt{10 – 2\sqrt{5}}\over 4\). Since 54 degree is the complement of 36 degree. \(\therefore\) cos 54 = sin(90 – 36) = sin 36 = \(\sqrt{10 – 2\sqrt{5}}\over 4\) Hence, cos 54 …
Solution : The value of sin 54 degrees is \(\sqrt{5} + 1\over 4\). Proof : We know that value of cos 36 degrees is \(\sqrt{5} + 1\over 4\). Since 54 degree is the complement of 36 degree. \(\therefore\) sin 54 = sin(90 – 36) = cos 36 = \(\sqrt{5} + 1\over 4\) Hence, sin 54 …
Solution : The value of sin 36 degrees is \(\sqrt{10 – 2\sqrt{5}}\over 4\). Proof : We know that the value of cos 36 degrees is \(\sqrt{5} + 1\over 4\). Putting \(\theta\) = 36 in \(cos \theta\) = \(\sqrt{1 – sin^2 \theta}\), we get cos 36 = \(\sqrt{1 – sin^2 36}\) = \(\sqrt{1 – ({\sqrt{5} + …
Solution : The value of cos 36 degrees is \(\sqrt{5} + 1\over 4\). Proof : We know that the value sin 18 degrees is \(\sqrt{5} – 1\over 4\). We have, \(cos 2\theta\) = \(1 – 2 sin^2 \theta\) \(\therefore\) cos 36 = 1 – \(2 sin^2 18\) \(\implies\) cos 36 = 1 – 2\(({\sqrt{5} – …
Solution : The value of cos 72 degrees is \(\sqrt{5} – 1\over 4\). Proof : We know that value of sin 18 degrees is \(\sqrt{5} – 1\over 4\). Since 72 degree is the complement of 18 degree. \(\therefore\) cos 72 = cos(90 – 18) = sin 18 = \(\sqrt{5} – 1\over 4\) Hence, cos 72 …
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 – 18) = cos 18 = \(\sqrt{10 + 2\sqrt{5}}\over 4\) Hence, sin 72 …
