Hi

Can someone tell me how to do this question:

1) Convert 4sin2x+5cos2x into a single cosine wave form Acos(2x+C)

P.S

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- Jun 4th 2010, 05:52 AMPaymemoneyQuick Trig Question
Hi

Can someone tell me how to do this question:

1) Convert 4sin2x+5cos2x into a single cosine wave form Acos(2x+C)

P.S - Jun 4th 2010, 05:54 AMmathaddict
in general ,

$\displaystyle a\sin \theta+b\cos \theta=R\sin(\theta+\alpha)$

where $\displaystyle R=\sqrt{a^2+b^2}$

and $\displaystyle \tan \alpha=\frac{b}{a}$

It works the same for $\displaystyle a\cos \theta+b\sin \theta=R\cos (\theta-\alpha)$

but you flip the sign - Jun 4th 2010, 07:31 AMGrandad
Hello PaymemoneyJust to expand a little on

*mathaddict*'s reply, use the identity$\displaystyle \cos(P+Q) = \cos P \cos Q - \sin P \sin Q$as follows:$\displaystyle A\cos(2x+C) = 4\sin 2x +5\cos 2x$Square (1) and (2) and add:

$\displaystyle \Rightarrow A\cos2x\cos C -A\sin2x\sin C = 4\sin2x+5\cos2x$

$\displaystyle \Rightarrow \left\{\begin{array}{l l}A\cos C = 5 & \quad\text{(1)}\\ A\sin C = -4&\quad\text{(2)}\end{array}\right.$

$\displaystyle A^2 = 5^2+(-4)^2$Divide (2) by (1):

$\displaystyle \Rightarrow A = \sqrt{41}$

$\displaystyle \tan C = -\tfrac45$Grandad