1. ## Various Trig Problems

1. Find the x and y-intercepts of X² / x²-x-6
2. Find the vertical asymptote of (x-8)(x+4) / x²-1
3. Find the slant asymptote of x²+2x-3 / x-5
4. Assume that θ is an acute angle in a right triangle. Evaluate the indicated trigonometric function

tan θ =1/3; csc θ

5. Phase shift of -2+2sin(3x+pi/2)
6. Exact value of arcsin(-(sqrt3)/2)

2. ## Re: Various Trig Problems

To find the x-intercept, set x=0 and solve for y.
To find the y-intercept, set y=0 and solve for x (you may get multiple solutions).

3. ## Re: Various Trig Problems

Q 5. Compare The expression with the standard from i.e., f(x) = A sin ( Bx + C ) + D the phase shift is given by C. In this case the phase shift is pi/2
Q 6. arcsin(-(sqrt3)/2) that means we want to find that angle for which sin has the value -(-(sqrt3)/2).
We have learnt that Sinθ = √3/2 for θ= π/3 and also trigonometric ratio sine is negative in 3rd and 4th Quadrant.
Therefore Sinθ = - √3/2 for θ = π+ π/3= 4π/3 and 2π- π/3= 5π/3.

4. ## Re: Various Trig Problems

Hello, whittwhitt!

3. Find the slant asymptote of: . $y \:=\:\frac{x^2+2x-3}{x-5}$

Long division: . $\frac{x^2+2x-3}{x-5} \:=\:x+7 + \frac{32}{x-5}$

As $x\to\infty,\;\frac{32}{x-5} \to 0.$

Therefore, the slant asymptote is: . $y \:=\:x+7$

$\text{4. Assume that }\theta\text{ is an acute angle in a right triangle and }\tan\theta = \tfrac{1}{3}$
$\text{Evaluate: }\:\csc\theta$

We have: . $\tan\theta \:=\:\frac{1}{3} \:=\:\frac{opp}{adj}$

$\theta$ is in a right triangle with: $opp = 1,\;adj=3.$
Pythagorus tells us that: $hyp = \sqrt{10}.$

Therefore: . $\csc\theta \:=\:\frac{hyp}{opp} \:=\:\frac{\sqrt{10}}{1} \:=\:\sqrt{10}$