# Trig identity Help

• Jan 8th 2008, 01:13 PM
nugiboy
Trig identity Help
Hi im stuck on part two of this question:

http://i137.photobucket.com/albums/q220/e3tiger/Sin.jpg

I know that:

$\displaystyle tanx = sinx/cosx$

and

$\displaystyle sinx^2 + cosx^2 = 1$

Any ideas?
• Jan 8th 2008, 01:41 PM
Peritus
4*sin(x) = 3*cos(x)

<=> tan(x) = 3/4

we have divided by cos(x), so we have to verify that it's well defined, the values of x that can cause problems are x = pi/2 + pi*k, (k is an integer)

the equation obviously doesn't hold in this set of points so we're ok.

now as you have already seen in part 1 the tan function is periodic with a period of pi, thus the equation also true for:

tan(x + pi*k) = 3/4
<=> x+pi*k = atan(3/4)
<=> x = atan(3/4) + pi*k

now choose the values in the given range namely (0 <= x <= 2*pi)...
• Jan 8th 2008, 02:16 PM
wingless
We'll sketch it from 0 to $\displaystyle 2\pi$. If you are allowed to use a calculator or a trigonometric table, find some points and place them. Find where the graph intersects the x-axis. Then, if you know how to do, find the vertical asymptotes.

As $\displaystyle Tan(x) = \frac{Sin(x)}{Cos(x)}$, the vertical asymptotes will be at $\displaystyle Cos(x) = 0$, which makes the tangent $\displaystyle \mp \infty$.

So,
$\displaystyle Cos(x) = 0$
$\displaystyle x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2} ...$
We'll just need $\displaystyle x = \frac{\pi}{2}, \frac{3\pi}{2}$, as $\displaystyle 0\leq x \leq 2\pi$.

So, the vertical asymptotes are at $\displaystyle x = \frac{\pi}{2}$ and $\displaystyle \frac{3\pi}{2}$. You can use one sided limits to find how do these asymptotes look like.

You can find the critical points and find whether the graph increases or decreases using first and second derivatives. Or if you have enough points to guess how the graph looks like, you can finish it now.