Download y= x sin( ) y= x cos( )

2.2 The graphs of sin(x) and cos(x).
Now I am going to define the two basic trig functions:
sin(x)
and
cos(x).
P
1
x
O
sin(x) the distance from P to the horizontal axis
cos(x) the distance from P to the vertical axis
sin(x)
cos(x)
Study the diagram at the right. The circle has radius 1. The arm OP
starts at the positive horizontal axis and rotates in a counter-clockwise
direction. When it has turned through an angle of x radians, we define:
Now is a good moment to stop and get the class to try to draw what the
graphs of sin(x) and cos(x) ought to look like. These graphs are
displayed below, for x ≥ 0, along with their “generator.” Study these
diagrams carefully. Note the periodicity––it’s clear that when the arm
has turned through a complete revolution (2π radians) and keeps on
turning, the values of sin and cos repeat. Note also the relationship
between the two graphs. Can you formulate that algebraically?
1
y=sin(x)
0.5
π/6
0
sin( π )
6
π π
6 2
π
3π
2
2π
5π
2
3π
2
2π
5π
2
3π
7π
2
4π
9π
2
5π
-0.5
-1
π/6
1
0.5
0
y=cos(x)
cos( π )
6
π π
6 2
π
3π
7π
2
4π
9π
2
5π
-0.5
-1
Note that in the upper graph, the circle is exactly the one from the original definition at the top. But the
circle in the lower graph is turned from that through 90º (or π/2). Why has that happened? Well, function
graphs are conventionally plotted with f(x) on the vertical axis. The original definition of sin(x) gives it as
the length of a vertical segment, so it projects directly onto the graph. But cos(x) is defined as the length of
a horizontal segment, and to generate the cos graph I had to rotate the circle so that the x-axis was vertical.
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sin&cos
8/16/2007
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In summary, here are more complete versions of the graphs, plotted together, emphasizing the relationship
between them, that the cos graph is obtained by translating the sin graph an amount π/2 to the left.
1
y=sinx
y=cosx
−π
−π
2
0
π
2
π
3π
2
2π
5π
2
3π
4π
7π
2
-1
There are lots of simple arithmetic relationships involving sin and cos, and I will record some of these here.
But I warn you not to overburden your memory with these. When you need to use them, look at the circle
definition or look at the graph (or look at them both!)
Observe that sin is an odd function (graph symmetric through the
origin) and cos is an even function (graph symmetric about the y-axis).
Algebraically, this tells us that:
sin(–x) = –sin(x)
cos(–x) = cos(x)
π−x
sin(x) = sin(π–x) = –sin(π+x) = –sin(–x).
cos(x) = cos(–x) = –cos(π+x) = –cos(π –x).
This can be seen both from the symmetry of the graphs and from the
circle diagram at the right.
x
cos(x)
sin(x)
The internal symmetries of each of the graphs give us:
O
π+x
−x
There are two important ways of getting a relationship between sin and
cos. The first is through translation. The fact that if we move the singraph π/2 units left we get the cos-graph can be written:
cos(x) = sin( π +x)
2
and since cos(x) = cos(–x), this can be also written:
The second relationship between sin and cos comes from the
Pythagorean Theorem applied to the bold triangle at the right:
O
sin2(x) + cos2(x) = 1.
sin&cos
1
8/16/2007
x
sin(x)
cos(x) = sin( π –x).
2
cos(x)
2
Some time ago a student submitted an
What should sin(x+y) be? Should it be the sum of sin(x) and
sin(y)? The answer is emphatically no! If it were, we would
have a great proof that 2 = 0:
sin(π) = sin( π ) + sin( π )
2
2
0 = 1 + 1 !!!
assignment with the assertion sin(
π
4
½. When questioned, he replied that
)=
π
since sin( π ) = 1, should not sin( ) be
2
4
half that?
In fact, here are the formulas for the sin and cos of the sum
and difference of angles.
sin(x+y) = sin(x)cos(y) + cos(x)sin(y)
sin(x–y) = sin(x)cos(y) – cos(x)sin(y)
In fact, he was implicitly assuming that
sin(x) is a linear function of the form kx.
But it isn’t.
That same error leads “popular” misstatements such as:
cos(x+y) = cos(x)cos(y) – sin(x)sin(y)
cos(x–y) = cos(x)cos(y) + sin(x)sin(y)
sin(2x) = 2 sin(x)
sin(x+y) = sin(x) + sin(y)
There are a number of ways to establish these formulas, but
the most elegant I’ve seen I pulled off the web at
neither of which are true. See below.
http://www.ies.co.jp/math/java/trig/kahote/kahote.html
It uses the simple fact that if you take the basic sin-cos triangle
and scale it by a factor of k, you will have ksinx and kcosx
appearing as the lengths of the legs of the triangle.
Let’s try to establish the sin(x+y) formula. For this we would
want to organize things so that sin(x) and cos(x) appeared in
the role of k in a couple of angle y triangles. Well here’s the
diagram.
sinx
⎫
⎪sinx cosy⎫
1
⎪sin(x+y)
⎬
y⎪
⎬
⎭
cosx ⎫⎬ cosx siny ⎪
xy
⎭
⎭
sin(x+y) = sinx cosy + cosx siny
This is a lovely “proof without words.”
However, for those who need a kick
start, I’ll add a few words.
There are three right-angled triangles.
The big one, with angle x, has
hypotenuse 1 and sides cosx and sinx.
The two smaller ones have angle y and
would have side lengths cosy and siny,
except they are scaled down, one by the
factor cosx and the other by the factor
sinx. Thus, instead of side length cosy
and siny, these triangles have side length
cosx siny and sinx cosy.
If we now step back and imagine the
triangle with angle x+y (which is not
drawn) the length of the vertical side is
the sum of these. And that gives us the
sin(x+y) formula.
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Double and half angle.
Setting x=y in the sum formulae, we get:
sin(2x) = 2sin(x)cos(x)
The cos(2x) formula has three
forms, the 2nd and 3rd obtained
from the 1st with a substitution
from sin2 + cos2 = 1.
cos(2x) = cos2x – sin2x.
= 2cos2x – 1
= 1 – 2sin2x
The last two formulae for cos(2x) give us the half-angle
formulas (replace x by x/2).
cos
x
1 + cos x
=
2
2
sin
x
1 − cos x
=
2
2
The last two forms are useful
because they can be solved for
cosx and sinx in terms of
cos(2x), and that gives us the
half-angle formulae.
Special angles
Recall that the special 30-60-90 and 45-45-90 triangles give us
simple exact expressions for the sin and cos of the angles. If
we write the angles in radians, we get the same formulae with
a different look.
π/6 π/6
2
We summarize the results in the following table.
sin 0 = 0
sin π/6 = 1/2
cos 0 = 0
3 /2
cos π/6 = 3 /2
cos π/3 = 1/2
sin π/4 = 1/ 2
sin π/2 = 1
cos π/4 = 1/ 2
cos π/2 = 1
sin π/3 =
2
3
π/3
π/3
1
1
π/4
These angles are unusual in that we get exact expressions for
sin and cos. Are there other angles for which this is also true?
Of course there are the corresponding angles in other
quadrants such as 2π/3 and –π/4, but what about angles other
than these?
Well the sum and difference formulae for sin and cos will give
us expressions for any angle that can be built out of special
ones by adding or subtracting. And they in turn give rise to
the double- and half-angle formulae. These all give us ways
of generating expressions for the sin and cos of new angles.
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2
1
π/4
1
4
Example 1. Here are two different ways to write 15° in terms
of the special angles
15º = 60º–45º.
15° = 30°/2
In radians:
π
12
=
π
12
π
3
−
π
4
1π
2 6
=
Use both the sin difference formula and the half-angle formula
to get an exact expression for sin(π/12). Compare your
answers.
Solution: The sin difference formula gives us:
sin
π
12
= sin(
π
3
−
π
4
) = sin
=
π
3
cos
π
4
− cos
π
3
sin
π
4
3 1 1 1
−
2 2 2 2
=
3 −1
2 2
The half-angle formula gives us:
1 − cos(π / 6)
1 − ( 3 / 2)
sin
=
=
12
2
2
π
=
2− 3
4
=
2− 3
2
Now this is interesting. The
two standard ways we might do
this calculation give us two
different answers and it’s not at
all clear they are the same.
There’s an interesting new
technique comes out of this:
how to take the square root of
something like a + b 3 .
That’s an answer of sorts. But it’s a bit more complicated than
it ought to be, and besides it’s not at all clear that the two
answers are the same! We certainly have to do something
about that. There are a number of things we might do at this
point.
1) Show directly that the two answers are the same. The
simplest way to do that is to square both answers and
compare.
2) Suppose we didn’t have the first answer and ask how we
might simplify the second answer. That seems like an
interesting (and possibly useful) question. How do you take
the square root of something of the form: a + b 3 ? See
problem .
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Problems.
1. Calculate the following. Give exact answers. Work with a picture, either the cos and sin graphs, or the
circle diagram (or both!)
(b) cos(3π/4)
(a) sin(2π/3)
(c) sin(7π/4)
(d) sin(–5π/4)
(e) cos(4π/3)
(f) cos(–5π/6)
(g) sin(11π/6)
(h) cos(14π/3)
(i) sin(–19π/6)
(j) sin(21π/4)
2. Find all values of x in the interval [–2π, 4π] which satisfy the equation sinx = sin
π
9
. Display all your
solutions on a copy of the graph y = sinx.
3. Use the internal symmetry of the sin and cos graphs to simplify the following:
(a) sin(x + π)
(b) cos(x – π/2)
(c) sin(x + 3π/2)
(d) sin(5π/2 – x)
(e) cos(4π/3 – x)
4. Use the addition laws for sin and cos to calculate the following. Simplify as much as possible.
(a) sin(x + π/4)
(b) cos(x – π/3)
(c) sin(x – 5π/4)
(d) sin(5π/6 – x)
(e) cos(3π/4 – x)
5.(a) Use the sin sum formula to calculate sin(5π/12).
(b) Use the half-angle formula to calculate sin(π/8). Simplify.
(c) Noting that 1/6 – 1/8 = 1/24, calculate cos(π/24).
6. Investigate the solutions of the equation
sin(x) = cos(x/2) (0 ≤ x ≤ 4π)
using two approaches:
(a) On the same set of axes, draw rough graphs of sin(x) and cos(x/2) and approximately locate the
intersections.
(b) Use the half-angle formula for cos(x/2) and work algebraically with the resulting equation to get more
exact decimal approximations for the solutions. Use the cos–1 button on your calculator.
7. How many solutions are there to the equation:
2x = 9π sinx ?
Illustrate your answer on a graph of sinx, and find exact values for as many solutions as you can.
8. Use the formula sin2(x) + cos2(x) = 1 to find all solutions of the equation
in the interval –2π ≤ x ≤ 2π.
sin&cos
sinx + 3cosx = 1
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9. Refer to Example 1.
(a) Show that an expression of the form a + b 3 has a square root that is of the same form (if it is
positive).
(b) Calculate the square root of 2 – 3 . Use this to show that the two answers we obtained in Example 1
are equal.
10.(a) The sin graph is drawn below. Sketch on this set of axes what you think the graph of sin2(x) ought
to look like. Justify the principle features of your graph. Use one of the formulae of this section to obtain a
very revealing alternative formula for sin2(x). [Don’t use your graphing calculator for this problem!]
(b) Now use the translation property that cos(x) = sin( π +x) to draw, on the same set of axes, the graph
2
of cos2(x). Finally make a sketch of the curve that lies exactly halfway between your two graphs. Use the
Pythagorean relation:
sin2(x) + cos2(x) = 1
to explain an unexpected property of this graph.
11. Study the graphs of sinx and cosx drawn below:
1
y=sinx
y=cosx
−π
−π
2
0
π
2
π
3π
2
2π
5π
2
3π
7π
2
4π
-1
(a) Based on this diagram, sketch the graph of the average of the two functions:
f(x) =
sin x + cos x
.
2
(b) Argue that the graph of f(x) is periodic with period 2π and bilateral symmetric about the line x = π/4.
There are different ways to handle the bilateral symmetry, some quite elegant.
(c) Now find a simplified expression for f(x) which makes it much clearer what the graph of f(x) really is.
[Hint: try to adapt the sin-sum formula.]
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