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Math 1500 Fall 2010
Optional Trig Problems
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1. Graph the functions below by hand, not by plotting points, but by starting with the
graph of one of the standard functions and then applying the appropriate transformation.
(a) y = 1 + 2 cos x
(b) y = 4 sin 3x
(c) y = 14 tan x − π4
The green curves above are the original curves and the red curves are the shifted curves.
2. Find the functions (a) f ◦ g, (b) g ◦ f , (c) f ◦ f , (d) g ◦ g and their domains.
(i) f (x) = 1 − 3x, g(x) = cos x
(a) 1 − 3 · cos x
(b) cos(1 − 3x)
The domain of all four are all real numbers.
(ii) f (x) =
(a)
sin 2x
1+sin 2x
x
1+x ,
(c) 1 − 3(1 − 3x)
(d) cos(cos x)
g(x) = sin 2x
x
(b) sin 2 1+x
(c)
x
1+x
x
1+ 1+x
(d) sin(2(sin 2x))
The domain of (a) is all real numbers except when sin 2x = −1. If θ = 2x this happens when sin θ = −1
(4n+3)π
−π
7π
−5π
or when θ = 3π
where n is an
2 or 2 or 2 or 2 and so on. One way to write these is that θ =
2
integer. Now θ = 2x so to solve for x we need to divide by 2. So the domain is all real numbers except
x = (4n+3)π
where n is an integer.
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The domain of (b) is all real numbers except where f is undefined, when x = −1.
x
The domain of (c) is all real numbers except where 1 + x = 0 and where 1 + 1+x
= 0. This is when x = −1
1
and when 1 + 2x = 0 (so when x = −1 and x = − 2 ).
The domain of (d) is all real numbers since the domain of the sin function is all real numbers.
√
3. Express f (x) = sin( x) in the form h ◦ g.
h(x) = sin(x) and g(x) =
√
x.
1
4. Find the domain of g(t) = sin(e−t ).
The domain is all real numbers since we can plug any value into the exponential function and the sin
function.
5. Find the exact value of each expression without using a calculator.
√
(a) sin−1 ( 3/2)
(b) arctan 1
(c) tan(sec−1 4)
(d) sec−1 2
(a)
π
3
(b)
π
4
(c) Let θ = (sec−1 4), Then cos θ = 14 . Since cosine is the ratio of the adjacent angle and the hypotenuse
we have the following triangle:
Using Pythagorean Theorem we find the third side is
−1
(d) If sec
2 = θ for some angle θ then cos θ =
1
2
√
√
15 and so tan θ =
which is when θ =
15
1
=
√
15.
π
3.
6. Simplify the expressions below.
(a) tan(sin−1 x)
Let θ = sin−1 x so sin θ = x. Filling in the proper triangle we get:
Using Pythagorean Theorem we get that the adjacent side has length
√ x
.
1−x2
√
1 − x2 and so tan θ = tan(sin−1 x) =
(b) cos(2 tan−1 x)
Let tan−1 x = θ so tan θ = x. Filling in the proper triangle we get:
√
Using Pythagorean Theorem we get that the hypotenuse has length 1 + x2 . Now cos(2 tan−1 x) = cos(2θ)
and by the Double Angle Formula we can write this as cos 2θ = cos2 θ − sin2 θ and using the triangle above
1
x2
1−x2
we get 1+x
2 − 1+x2 = 1+x2 .
7. Determine the infinite limits.
(a) lim x csc x
x→2π −
As we approach 2π from the left x is positive. Approaching 2π from the left means we are in the fourth
quadrant. In this quadrant, sin x is negative and so csc x is negative as well. So the limit approaches −∞
(b) lim x cot x
x→π −
As we approach π from the left x is positive. Approaching π from the left means we are in the second
quadrant. In this quadrant, tan x is negative and so cot x is negative as well. So the limit approaches −∞
2