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Solution to the reviewing problems.
You are supposed to be able to do the following problems. If not, just refer
to the solution, if you are still confused with the solution, do come to my office
and I am glad to answer your questions.
- What is the absolute value of (1 + i
3)^20.
Solution:
The absolute value of 1 + i
3 is r =
3)^2 = 2. So the polar form of 1 + i
is (2, θ). Note that in polar form, the 20-th power of (2, θ) is (2^20 , 20 θ), it is clear that the
absolute value of (1 + i
3)^20 is 2^20.
- (2, 3) is on the terminal of angle θ, find the exact value of sin θ and sec θ.
Solution:
sin θ =
y
r
y √ x^2 + y^2
22 + 3^2
cos θ =
x
r
x √ x^2 + y^2
22 + 3^2
sec θ =
cos θ
- If sin x =
and
π
2
< x < π, find the exact value of sin
x
2
Solution:
As sin x =
, by Pythagorean theorem, we know that cos x = ±
1 − sin^2 x = ±
. As x is in quadrant II, cos x is negative, that is cos x = −
By half-angle identity, sin
x
2
1 − cos x
2
1 − cos x
2
Note that as
π
2
< x < π, we have
π
4
x
2
π
2
, thus
x
2
is in quadrant I, so sin
x
2
thus sin x =
- Find all the solutions to z^6 = i and sketch all the solutions in the coordinate plane. Solution:
Assume (r, θ) is the polar form of z, then the polar form of z^6 is just (r^6 , 6 θ). Note
that the polar form of i is (1,
π
2
), and z^6 = i, we have (in polar form)
(r
6 , 6 θ) = (1,
π
2
Thus r^6 = 1 and 6θ =
π
2
- 2πk, which means r = 1 and θ =
π
12
π
3
k.
As r = 1, all the solutions must be on the unit circle (with radius 1). Just enumerate
k = 0, 1 , 2 , 3 , ... for the argument θ =
π
12
π
3
k, we can sketch all the solutions on the unit
circle, just have a try.
- f (x) = − sin x is even, odd or neither?
Solution: First of all, apply the negative angle formula for sine, which says that sin(−x) = − sin x,
thus sin x is odd. (or just sketch the graph of sin x, you will find that it an odd function).
f (x) = − sin x is also odd because f (−x) = − sin(−x) = −(− sin x) = −f (x).
- If you know cos
π
4
, then cos
5 π
4
=? cos
7 π
4
Solution: Apply the plus π identity for cosine, that is cos(x + π) = − cos x, just plug in x =
π
4
we get cos
5 π
4
= − cos
π
4
To calculate cos
7 π
4
, just note that
7 π
4
= 2π −
π
4
. Thus cos
7 π
4
= cos(2π −
π
4
cos(−
π
4
) = cos
π
4
- In a triangle ABC (with three sides a, b and c), if A = 60◦, b = 3 and c = 2, a =?
Solution:
Nothing tricky, just apply the cosine law, we have
a
2 = b
2
2 − 2 bc cos A
Plug in the value of b, c and A, we have
2 = 2
2
2 − 2 · 2 ·
3 cos A
thus cos A =
, so A = cos−^1
= 30◦^ (Note: you don’t need to worry about the
domain/range mess of trig inverse function while you are solving triangles, just use trig
inverse for angles!)
Similarly, we have
b^2 = a^2 + c^2 − 2 · a · c cos B
Plug in the values of a, b and c, we have cos B = 0, thus B = cos−^1 0 = 90◦.
Now we know A = 30◦^ and B = 90◦, note that A + B + C = 180◦, it is clear that
C = 60◦, done!
