# Trig

• October 22nd 2013, 07:44 AM
srirahulan
Trig
$5cos^2{\theta}+18cos{\theta}sin{\theta}+29sin^2{\t heta}$ Simplify this into $a+bcos(2{\theta}+{\alpha})$ a,b are constant and $\alpha$ is completely different from $\theta$
• October 22nd 2013, 09:42 AM
SlipEternal
Re: Trig
Show what work you have so far.
• October 22nd 2013, 05:05 PM
srirahulan
Re: Trig
I write that like this $2cos^2{\theta}-1+1+3cos^2{\theta}+9sin2{\theta}+29sin^2{\theta}$ But then i could not find out the $\alpha$ It is very difficult to arrange them

A correction--- that $29sin^2{\theta}$
• October 24th 2013, 04:35 PM
srirahulan
Re: Trig
• October 24th 2013, 09:30 PM
SlipEternal
Re: Trig
Try using the sum of angles formula for $\cos$ and see what you get: $a+b\cos(2\theta+\alpha) = a+b[\cos(2\theta)\cos(\alpha)-\sin(2\theta)\sin(\alpha)]$... Keep trying manipulations until you get something close to what you started with.
• October 24th 2013, 09:52 PM
ibdutt
Re: Trig
• October 25th 2013, 05:59 AM
Soroban
Re: Trig
Hello, srirahulan!

Quote:

$5\cos^2\theta+18\cos\theta\sin\theta+29\sin^2\thet a$

$\text{Simplify to }\,a+b\cos(2\theta+\alpha),\:\text{ where }a,b\text{ are constants and }\alpha \ne \theta$

$5\cos^2\theta + 18\cos\theta\sin\theta + 29\sin^2\theta$

. . $=\;5\left(\frac{1+\cos2\theta}{2}\right) + 9\sin2\theta + 29\left(\frac{1-\cos2\theta}{2}\right)$

. . $=\;\tfrac{1}{2}\big(5 + 5\cos2\theta + 18\sin2\theta + 29 - 29\cos2\theta\big)$

. . $=\;\tfrac{1}{2}\big[34 - 24\cos2\theta + 18\sin2\theta\big]$

. . $=\;\tfrac{1}{2}\bigg[34 - 6\big(4\cos2\theta - 3\sin2\theta\big)\bigg]$

. . $=\;\tfrac{1}{2}\bigg[34 - 30\left(\tfrac{4}{5}\cos2\theta - \tfrac{3}{5}\sin2\theta\right)\bigg]$

$\text{Let }\cos\alpha = \tfrac{4}{5},\;\sin\alpja = \tfrac{3}{5}$

$\text{We have: }\:\tfrac{1}{2}\bigg[34 - 30\left(\cos\alpha\cos2\theta - \sin\alpha\sin2\theta\right)\bigg]$

. . . . . $=\;\tfrac{1}{2}\bigg[34 - 30\cps(2\theta + \alpha)\bigg]$

. . . . . $=\;17 - 15\cos(2\theta + \alpha)$