Wave function
Watch this video to learn about wave function.
acosx + bsinx
Given any expression of the form \(a\cos x + b\sin x\), it can be rewritten into any one of the following forms:
- \(k\cos (x - \alpha )\)
- \(k\cos (x + \alpha )\)
- \(k\sin (x - \alpha )\)
- \(k\sin (x + \alpha )\)
The form you should use may be given to you in a question, but if not, any one will do. If in doubt, \(k\cos (x - \alpha )\) usually works.
These worked examples show the processes you'll need to go through to rewrite an expression in this form.
Question
Write \(2\sin x^\circ + 5\cos x^\circ\) in the form \(k\sin (x + \alpha )^\circ\) where \(k\textgreater0\) and \(0^\circ \le \alpha ^\circ \textless 360^\circ\).
You've been given the form to write the equation in, so now equate both expressions. This just means, make them equal each other.
\(2\sin x^\circ + 5\cos x^\circ = k\sin (x + \alpha )^\circ\)
Expand the brackets on the right side of the equation.
\(= k\sin x^\circ \cos \alpha ^\circ + k\cos x^\circ \sin \alpha ^\circ\)
Equate the coefficients:
\(k\cos \alpha ^\circ = 2\)
\(k\sin \alpha ^\circ = 5\)
Calculate \(k\) using the values for the coefficients:
\(k = \sqrt {{a^2} + {b^2}}\)
\(= \sqrt {4 + 25}\)
\(= \sqrt {29}\)
Calculate \(\alpha\)
\(\tan \alpha ^\circ =\frac{{k\sin \alpha ^\circ }}{{k\cos \alpha ^\circ }} = \frac{5}{2}\)
\(\alpha = {\tan ^{ - 1}}\left( {\frac{5}{2}} \right)\)
We know that \(\alpha\) is in the first quadrant as \(k\cos \alpha ^\circ \textgreater 0\) and \(k\sin \alpha ^\circ \textgreater 0\)
\(= 68.2^\circ\)
Finally write the equation in the form it was asked for:
\(2\sin x^\circ + 5\cos x^\circ = \sqrt {29} \sin (x + 68.2)^\circ\)
Question
Write \(\cos 2x - \sqrt 3 \sin 2x\) in the form \(k\cos (2x + \alpha )\) where \(k\textgreater0\) and \(0 \le \alpha \le 2\pi\)
\(\cos 2x - \sqrt 3 \sin 2x = k\cos (2x + \alpha )\)
\(= k\cos 2x\cos \alpha - k\sin 2x\sin \alpha\)
\(= k\cos \alpha \cos 2x - k\sin \alpha \sin 2x\)
\(k\cos \alpha = 1\)
\(k\cos \alpha\) is the co-efficient of the \(\cos 2x\) term
\(k\sin \alpha = \sqrt 3\)
\(k\sin \alpha\) is the co-efficient of the \(\sin 2x\) term
\(k = \sqrt {{1^2} + {{\left( {\sqrt 3 } \right)}^2}}\)
\(= \sqrt {1 + 3}\)
\(= \sqrt 4\)
\(= 2\)
We know that \(\alpha\) is in the first quadrant because \(k\cos \alpha \textgreater0\) and \(k\sin \alpha \textgreater0\).
\(a = {\tan ^{ - 1}}\sqrt 3\)
\(= \frac{\pi }{3}\)
Therefore:
\(\cos 2x - \sqrt 3 \sin 2x = 2\cos \left( {2x + \frac{\pi }{3}} \right)\)
