Solution : We have, \(\sqrt{3} cos \theta\) + \(sin \theta\) = \(\sqrt{2}\) ………….(i) This is of the form \(a cos\theta\) + \(b sin \theta\) = c, where a = \(\sqrt{3}\), b = 1 and c = \(\sqrt{2}\). Let a = \(r cos\alpha\) and b = \(r sin\alpha\). Then, \(\sqrt{3}\) = …
Solve : \(\sqrt{3} cos \theta\) + \(sin \theta\) = \(\sqrt{2}\) Read More »
General Solution of the equation \(a cos\theta\) + \(b sin \theta\) = c, where a, b, c \(\in\) R such that | c | \(\le\) \(\sqrt{a^2 + b^2}\). To solve this type of equations, we first reduce them in the form \(cos \theta\) = \(cos \alpha\), or \(sin \theta\) = \(sin \alpha\). The following algorithm …
General Solution of the Equation of form \(a cos\theta\) + \(b sin \theta\) = c Read More »
Solution : The general solution of \(sin^2 \theta\) = \(sin^2 \alpha\) is given by \(\theta\) = \(n\pi \pm \alpha\), n \(\in\) Z. Proof : We have, \(sin^2 \theta\) =\(sin^2 \alpha\) \(\implies\) \(2 sin^2 \theta\) =\(2 sin^2 \alpha\) \(\implies\) \(1 – cos 2\theta\) = \(1 – cos 2\alpha\) \(\implies\) \(cos 2\theta\) = \(cos 2\alpha\) \(\implies\) \(2\theta\) …
What is the General Solution of \(sin^2 \theta\) =\(sin^2 \alpha\) ? Read More »
Solution : The general solution of \(cos^2 \theta\) = \(cos^2 \alpha\) is given by \(\theta\) = \(n\pi \pm \alpha\), n \(\in\) Z. Proof : We have, \(cos^2 \theta\) =\(cos^2 \alpha\) \(\implies\) \(2 cos^2 \theta\) =\(2 cos^2 \alpha\) \(\implies\) \(1 + cos 2\theta\) = \(1 + cos 2\alpha\) \(\implies\) \(cos 2\theta\) = \(cos 2\alpha\) \(\implies\) \(2\theta\) …
What is the General Solution of \(cos^2 \theta\) =\(cos^2 \alpha\) ? Read More »
Solution : The general solution of \(tan^2 \theta\) = \(tan^2 \alpha\) is given by \(\theta\) = \(n\pi \pm \alpha\), n \(\in\) Z. Proof : We have, \(tan^2 \theta\) =\(tan^2 \alpha\) \(\implies\) \(1 – tan^2\theta\over 1 + tan^2 \theta\) =\(1 – tan^2\alpha\over 1 + tan^2 \alpha\) \(\implies\) \(cos 2\theta\) = \(cos 2\alpha\) \(\implies\) \(2\theta\) = \(2n\pi …
What is the General Solution of \(tan^2 \theta\) =\(tan^2 \alpha\) ? Read More »
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 …
