# multiply through with sin and cos

• Nov 6th 2010, 05:09 AM
skoker
multiply through with sin and cos
I am fallowing the solution for a verify identity problem in my textbook, it looks like they multiply through with sin then cos in order to create tan functions under the radical. is this normal method to use? if they are using another method or identity can you explain.

thanks

this is exactly how it is in the book 4 I dont get or step 2.

$
\begin{document}

\sqrt{\frac{2}{1 + \cos \theta} ^{}}

\sqrt{\frac{\tmtextit{2 \sin} \theta}{\tmtextit{\sin} \theta \left( 1 +
\tmtextit{\cos} \theta \right) }}

\sqrt{\frac{\tmtextit{2 \sin} \theta}{\tmtextit{\sin} \theta + \tmtextit{\sin}
\theta \tmtextit{\cos} \theta}}

\sqrt{\frac{\frac{2 \sin \theta}{\cos \theta}}{ \frac{\tmtextit{\sin}
\theta}{\tmtextit{\cos} \theta} + \frac{\tmtextit{\sin} \theta \tmtextit{\cos}
\theta}{ \tmtextit{\cos} \theta}}^{}}

$
• Nov 6th 2010, 06:46 AM
Unknown008
I'm not sure what you don't understand, but what you posted is good.

Let's consider only the part within the square root.

$\begin{array}{ccl}
\dfrac{2}{1 + \cos\theta} &=& \dfrac{2}{1 + \cos\theta} \times 1 \\
&=& \dfrac{2}{1 + \cos\theta} \times \dfrac{\sin\theta}{\sin\theta} \\
&=& \dfrac{2\sin\theta}{\sin\theta(1 + \cos\theta)} \\
&=& \dfrac{2\sin\theta}{\sin\theta + \sin\theta\cos\theta}\\
&=& \dfrac{2\sin\theta}{\sin\theta + \sin\theta\cos\theta}\times1 \\
&=& \dfrac{2\sin\theta}{\sin\theta + \sin\theta\cos\theta} \times \dfrac{\frac{1}{\cos\theta}}{\frac{1}{\cos\theta}} \\
&=& \dfrac{\frac{2\sin\theta}{\cos\theta}}{\frac{\sin\ theta}{\cos\theta} + \frac{\sin\theta\cos\theta}{\cos\theta}} \end{array}$
• Nov 6th 2010, 07:09 AM
skoker
I was not sure if multiplying through with sin or cos in order to advance to the next step was 'legal'. I was looking for a formula or a rule that would introduce sin or cos. but I guess you can just multiply through with trig functions as a general rule, correct?
• Nov 6th 2010, 07:14 AM
Unknown008
As I showed you, I only multiplied by 1. This is possible, right?

Well, $\dfrac{\cos\theta}{\cos\theta} = 1$

Hence, there is no rule violated. The same thing applies to sin and other such things.
• Nov 6th 2010, 08:54 AM
skoker
I never thought about that formality in this context, 1=x/x. that makes sense thanks.
• Nov 6th 2010, 08:57 AM
Unknown008
Yes, those sorts of methods are commonly used in rationalisation, be it concerning trigonometry, or complex numbers, or surds where they are most commonly used.