Re: Various Trig Problems

Have you read your textbook or tried any of these 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).

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.

Re: Various Trig Problems

Hello, whittwhitt!

Quote:

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

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

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

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

Quote:

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

$\displaystyle \text{Evaluate: }\:\csc\theta$

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

$\displaystyle \theta$ is in a right triangle with: $\displaystyle opp = 1,\;adj=3.$

Pythagorus tells us that: $\displaystyle hyp = \sqrt{10}.$

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