Here you will learn what is an empty set in math, its symbol and definition with examples. Let’s begin – What is the Empty Set in Math ? Definition : A set is said to be empty set if it has no element. It is also called null or void set. It is denoted by […]
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Here you will learn what is set builder form and how to represent sets in set builder form with examples. Let’s begin – Set Builder Form Definition : In this form, a set is described by a characterizing property P(x) of its elements x. In such a case the set is described by {x :
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Here you will learn what is roster form in sets and how to represent sets in roaster form with examples. Let’s begin – Roster Form in Sets Definition : In this method a set is described by listing elements, separated by comma and enclose then by curly brackets { }. Example 1 : The set
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Solution : The general solution of \(tan \theta\) = \(tan \alpha\) is given by \(\theta\) = \(n\pi + \alpha\), n \(\in\) Z. Proof : We have, \(tan \theta\) = \(tan \alpha\) \(\implies\) \(sin \theta\over cos \theta\) = \(sin \alpha\over cos \alpha\) \(\implies\) \(sin \theta cos \alpha\) – \(cos \theta sin \alpha\) = 0 \(\implies\) \(sin
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Solution : The general solution of \(cos \theta\) = \(cos \alpha\) is given by \(\theta\) = \(2n\pi \pm \alpha\), n \(\in\) Z. Proof : We have, \(cos \theta\) = \(cos \alpha\) \(\implies\) \(cos \theta\) – \(cos \alpha\) = 0 \(\implies\) -\(2 sin ({\theta + \alpha\over 2}) sin({\theta – \alpha\over 2})\) = 0 \(\implies\) \(sin ({\theta
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Solution : The general solution of \(sin \theta\) = \(sin \alpha\) is given by \(\theta\) = \(n\pi + (-1)^n \alpha\), n \(\in\) Z. Proof : We have, \(sin \theta\) = \(sin \alpha\) \(\implies\) \(sin \theta\) – \(sin \alpha\) = 0 \(\implies\) \(2 sin ({\theta – \alpha\over 2}) cos({\theta + \alpha\over 2})\) = 0 \(\implies\) \(sin
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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, \(\sqrt{3}\) =
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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
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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, \(cos^2 \theta\) =\(cos^2 \alpha\) \(\implies\) \(2 cos^2 \theta\) =\(2 cos^2 \alpha\) \(\implies\) \(1 + cos 2\theta\) = \(1 + cos 2\alpha\) \(\implies\) \(cos 2\theta\) = \(cos 2\alpha\) \(\implies\) \(2\theta\)
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