Hi,
Does the answer key section here for number 2 makes sense? It seems like they went from cos^2 - sin^2 to cos^2 + sin^2 for no reason.
http://i.imgur.com/bdyQPrJ.png
Thanks.
Hi,
Does the answer key section here for number 2 makes sense? It seems like they went from cos^2 - sin^2 to cos^2 + sin^2 for no reason.
http://i.imgur.com/bdyQPrJ.png
Thanks.
Hello, KevinShaughnessy!
What they did was correct.
$\displaystyle \text{Prove: }\:\frac{\sin2t}{\sin t} - \frac{\cos2t}{\cos t} \:=\:\sec t$
$\displaystyle \frac{\sin2t}{\sin t} - \frac{\cos2t}{\cos t} \;=\;\frac{2\sin t\cos t}{\sin t} - \frac{(\cos^2\!t-\sin^2\!t)}{\cos t}$
n . . . . . . . . . $\displaystyle =\;2\cos t - \left(\frac{\cos^2\!t-\sin^2\!t}{\cos t}\right)$
n . . . . . . . . . $\displaystyle =\;\frac{2\cos^2\!t - (\cos^2\!t - \sin^2\!t)}{\cos t}$
n . . . . . . . . . $\displaystyle =\;\frac{2\cos^2\!t - \cos^2\!t + \sin^2t}{\cos t}$
n . . . . . . . . . $\displaystyle =\;\frac{\cos^2\!t + \sin^2\!t}{\cos t}$
n . . . . . . . . . $\displaystyle =\;\frac{1}{\cos t}$
n . . . . . . . . . $\displaystyle =\;\sec t$
Yes, the answer key is perfectly correct. The problem is to prove the trig identity $\displaystyle \frac{sin(2t)}{sin(t)}- \frac{cos(2t)}{cos(t)}= sec(t)$. They do that by using the "double angle" identities: $\displaystyle sin(2t)= 2 sin(t)cos(t)$ and $\displaystyle cos(2t)= cos^2(t)- sin^2(t)$ to write the left side as $\displaystyle \frac{2sin(t)cos(t)}{sin(t)}- \frac{cos^2(t)- sin^2(t)}{cos(t)}$.
Of course we can cancel the "sin(t)" terms in the first fraction to get
$\displaystyle 2 cos(t)- \frac{cos^2(t)- sin^2(t)}{cos(t)}$
and turn the first "2cos(t)" into a fraction by multiplying both numerator and denominator by cos(t):
$\displaystyle \frac{2cos^2(t)}{cos(t)}- \frac{cos^2(t)- sin^2(t)}{cos(t)}$.
Now add the fractions, remembering to distribute that "-" in front of the second fraction:
$\displaystyle \frac{2cos^2(t)- cos^2(t)+ sin^2(t)}{cos(t)}= \frac{cos^2(t)+ sin^2(t)}{cos(t)}= \frac{1}{cos(t)}= sec(t)$.