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Quiz 8
Math 1151, Precalculus 2
Name:
Summer 2012, Section 002
Problem 1.(15 points) Solve the following trigonometric equation on the interval
[0, 2π).
sin(2θ) sin(θ) = cos(θ)
First we use a trigonometric identity to rewrite the equation as 2 sin2 θ cos θ = cos θ.
Now if cos θ = 0 the same θ value is a solution to the original equation. So
π
2
and
3π
2
are solutions. Now if cos θ 6= 0 we can cancel them from both sides of the equation and
√
then solve for sine obtaining the equation sin θ = ±
are π4 , 3π
, 5π
and
4
4
2
.
2
The solutions to this equation
7π
.
4
Problem 2.(15 points) A 20 foot ladder leaning against a building makes a 65◦ angle
with the ground. How far up the building does the ladder touch? Draw and label a
picture of the situation.
The situation creates a right triangle with a hypotenuse of 20 and a 65 degree angle.
Thus the height to which the ladder reaches is 20 sin(65◦ ) = 18.13 ft.
Extra Credit(8 points) Find the vertical asymptotes of
x2 +x−12
.
x2 −x−6
Is there a horizontal
asymptote or an oblique asymptote? Explain, but do not find the asymptote.
To find the vertical asymptotes we look for when the denominator is zero. We factored
the denominator obtaining (x − 3)(x + 2). We also factored the numerator to get
(x − 3)(x + 4). So there is one vertical asymptote at x = −2.
The degree of the numerator is the same as the degree of the denominator and so the
function will have a horizontal asymptote.