Write the expression as the sine, cosine, tangent of an angle (in radians.)
(tan(pi/5)-tan(pi/3))/(1+tan(pi/5)tan(pi/3))
recall the tangent addition/subtraction formulas. You have
tan(pi/5 - pi/3) = tan(-2pi/15) = -tan(2pi/15)
To express the given expression in terms of sine, cosine, and tangent of an angle, we can use the following trigonometric identities:
1. $\tan(x) = \frac{\sin(x)}{\cos(x)}$
2. $\tan(a - b) = \frac{\tan(a) - \tan(b)}{1 + \tan(a)\tan(b)}$
Let's start by rewriting the expression using the identity in step 2:
$\frac{\tan(\frac{\pi}{5}) - \tan(\frac{\pi}{3})}{1 + \tan(\frac{\pi}{5})\tan(\frac{\pi}{3})}$
Now, let's substitute the values of $\tan(\frac{\pi}{5})$ and $\tan(\frac{\pi}{3})$ using the identity in step 1:
$\frac{\frac{\sin(\frac{\pi}{5})}{\cos(\frac{\pi}{5})} - \frac{\sin(\frac{\pi}{3})}{\cos(\frac{\pi}{3})}}{1 + \frac{\sin(\frac{\pi}{5})}{\cos(\frac{\pi}{5})} \cdot \frac{\sin(\frac{\pi}{3})}{\cos(\frac{\pi}{3})}}$
Now, let's simplify the expression:
$\frac{\sin(\frac{\pi}{5})\cos(\frac{\pi}{3}) - \sin(\frac{\pi}{3})\cos(\frac{\pi}{5})}{\cos(\frac{\pi}{5})\cos(\frac{\pi}{3}) + \sin(\frac{\pi}{5})\sin(\frac{\pi}{3})}$
Now, we can use the following trigonometric identities:
1. $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$
2. $\cos(\frac{\pi}{3}) = \frac{1}{2}$
3. $\sin(\frac{\pi}{5}) = \frac{1}{2}\sqrt{\frac{5}{2}-\frac{1}{2}\sqrt{5}}$
4. $\cos(\frac{\pi}{5}) = \frac{1}{2}\sqrt{\frac{5}{2}+\frac{1}{2}\sqrt{5}}$
Substituting these values into the expression:
$\frac{\frac{1}{2}\sqrt{\frac{5}{2}-\frac{1}{2}\sqrt{5}} \cdot \frac{1}{2} - \frac{\sqrt{3}}{2} \cdot \frac{1}{2}\sqrt{\frac{5}{2}+\frac{1}{2}\sqrt{5}}}{\frac{1}{2}\sqrt{\frac{5}{2}+\frac{1}{2}\sqrt{5}} \cdot \frac{1}{2} + \frac{1}{2}\sqrt{\frac{5}{2}-\frac{1}{2}\sqrt{5}} \cdot \frac{\sqrt{3}}{2}}$
Now, simplifying the expression further will give us the final answer.