weird trig question?

**mathswannabe** · February 23rd 2011, 07:00 PM

Determine 'a' if

$\cos(\theta+{\frac{\pi}{4}}) = a(\cos\theta + \sin\theta)$

Using the identity:

$\cos(\theta+\phi)$ $=$ $\cos\theta\cdot\cos\phi-\sin\theta\cdot\sin\phi$

I have got
$\begin{eqnarray} {\dfrac{1}{\sqrt{2}}}\left(\cos\theta-\sin\theta\right) &=&a\left(\cos\theta + \sin\theta\right)\nonumber \end{eqnarray} $

But don't know where to go from here??

**Prove It** · February 23rd 2011, 07:13 PM

Are you sure it wasn't asking you to find $\displaystyle a$ if

$\displaystyle \cos{\left(\theta - \frac{\pi}{4}\right)} = a(\cos{\theta} + \sin{\theta})$ ?

**harish21** · February 23rd 2011, 07:16 PM

Originally Posted by mathswannabe

Determine 'a' if

$\cos(\theta+{\frac{\pi}{4}}) = a(\cos\theta + \sin\theta)$

Using the identity:

$\cos(\theta+\phi)$ $=$ $\cos\theta\cdot\cos\phi-\sin\theta\cdot\sin\phi$

I have got
$\begin{eqnarray} {\dfrac{1}{\sqrt{2}}}\left(\cos\theta-\sin\theta\right) &=&a\left(\cos\theta + \sin\theta\right)\nonumber \end{eqnarray} $

But don't know where to go from here??

$\dfrac{1}{\sqrt{2}}(\cos(\theta)-\sin(\theta))=a(\cos(\theta)+\sin(\theta))$

$a = \dfrac{1}{\sqrt{2}} \cdot \dfrac{(\cos(\theta)-\sin(\theta))}{(\cos(\theta)+\sin(\theta))}$

you can simplify further as:

$a = \dfrac{1}{\sqrt{2}} \cdot \dfrac{(\cos(\theta)-\sin(\theta))}{(\cos(\theta)+\sin(\theta))} \times \dfrac{(\cos(\theta)+\sin(\theta))}{(\cos(\theta)+ \sin(\theta))}=...$

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