MATH 100 V1A
October 10th – Practice problems
Hints and Solutions
1. (a) Sketch the graph of sin x1 . Is this function continuous? (You may assume that
trigonometric functions are continuous on their domain; see Spiderwire for a proof.)
Solution: The graph of sin
1
x
looks something like:
y
1
x
−1.5 −1 −0.5
0.5
1
1.5
−1
First note that the domain of sin x1 is {x ∈ R | x 6= 0}, since the domain of x1 is
{x ∈ R | x 6= 0} and the domain of sin(x) is R, so sin x1 will be defined provided
1
is defined.
x
Now, since rational functions are continuous on their domain, and trigonometric
functions are continuous on their domain, the composite of a trigonometric function
with a rational function will also be continuous on its domain. Hence, sin x1 is
continuous on {x ∈ R | x 6= 0}.
(b) Is the function
(
sin
f (x) =
0
1
x
if x 6= 0
if x = 0
continuous?
Hint: Consider the graph of the function. Can you find a δ > 0 so that every x ∈ (−δ, δ)
is such that f (x) is within, say, 21 of 0? No! Justify this by showing that no matter
how small δ > 0 is, there is always an x within δ of 0 (i.e. x ∈ (−δ, δ)) for which
f (x) = ±1.
2. In class, we claimed
that the function g(x) = ax3 + bx2 + 1, where a =
b = π1 π3 − 2 , makes the function


if x < 0
tan(x)
f (x) = g(x)
if 0 ≤ x ≤ π


cos(sin(x)) if x > π
1
π2
1−
2
π
and
differentiable for all x > − π4 . Prove that this is true.
Hint: To show that f is differentiable for all x > − π4 , we need to show that
f (x + h) − f (x)
exists,
h→0
h
lim
for all x > − π4 . First explain why this limit must exist when x > − π4 but x 6= 0, π.
To show that the limit exists when x = 0, π, show that the left and right-hand limits
are equal. Note: When considering the left (or right) hand limit, you should be able
to recognize that limit as a special limit whose value you know how to find (without
evaluating the limit).
3. Sketch and find the equation of the line tangent to y = sin x at x = π.
Hint: The derivative of sin x is cos x, so the slope of the tangent line at x = π is just
cos(π) = −1.
4. Find the equation of the line tangent to y = cos2 x at x = π3 .
Hint: The derivative of cos2 x is 2 cos x · (− sin x) (Chain
Rule), so the slope of the
√ √
3
π
π
1
tangent line is just 2 cos 3 · − sin 3 = 2 · 2 · − 2 = − 23 .
5. Explain why sin
π
6
= sin − 11π
.
6
Hint: If we rotate the positive x-axis counter-clockwise by an angle θ, then we can
think of sin θ as the y-coordinate
of the point that lies on the rotated half-line and the
π
unit circle. So sin 6 is the y-coordinate of the point on the unit circle and the line
obtained by rotating the half-line counter-clockwise by an angle π6 , and sin − 11π
is
6
the y-coordinate of the point on the unit circle and the line obtained by rotating the
half-line clockwise by an angle 11π
. Why are these the same?
6
6. Sketch the graphs of csc x =
1
,
sin x
sec x =
Solution: The function csc x looks like
2
1
cos x
and cot x =
1
.
tan x
y
4
2
x
−6
−4
−2
2
4
6
−2
−4
The function sec x looks like
y
4
2
x
−6
−4
−2
2
4
6
−2
−4
The function cot x looks like
y
4
2
x
−6
−4
−2
2
−2
−4
3
4
6