sin&cos 8/16/2007 1

1

x sin(

)x

cos( )x P

O

-1

-0.5

0

0.5

1

-1

-0.5

0

0.5

1

6π

6π

2π

2π

23π

23π

25π

25π

27π

27π

29π

29π

1

π/6

π/6 2π

2π

5π

5π

π

π

3π

3π

4π

4π

y= xsin( )

y= xcos( ))6cos(π

)6sin(π

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).

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: sin(x) the distance from P to the horizontal axis cos(x) the distance from P to the vertical axis 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? 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.

sin&cos 8/16/2007 2

xπ−x

−xπ+x

sin(

)x

cos( )x

O

1

x sin(

)x

cos( )xO

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

0

1

2π

23π

25π

27π2ππ−π 3π 4π

2π−

y= xsiny= xcos

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)

The internal symmetries of each of the graphs give us:

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. 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 sin-graph π/2 units left we get the cos-graph can be written:

cos(x) = sin( 2π +x)

and since cos(x) = cos(–x), this can be also written:

cos(x) = sin( 2π –x).

The second relationship between sin and cos comes from the Pythagorean Theorem applied to the bold triangle at the right:

sin2(x) + cos2(x) = 1.

sin&cos 8/16/2007 3

⎪⎭

⎪⎬

⎫

⎭⎬⎫

1

x y

y

cosx

sinx

cos sinx y

sin cosx y

⎪⎭

⎪⎬

⎫sin( )x+y

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( 2π ) + sin( 2

π )

0 = 1 + 1 !!! 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)

cos(x+y) = cos(x)cos(y) – sin(x)sin(y) cos(x–y) = cos(x)cos(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 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.

Some time ago a student submitted an

assignment with the assertion sin(4π ) =

½. When questioned, he replied that

since sin( 2π ) = 1, should not sin(

4π ) be

half that? 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” mis-statements such as: sin(2x) = 2 sin(x) sin(x+y) = sin(x) + sin(y) neither of which are true. See below.

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.

sin&cos 8/16/2007 4

1 1

22 3

π/6 π/6

π/3 π/3

1

1

π/4

π/4

2

Double and half angle. Setting x=y in the sum formulae, we get:

sin(2x) = 2sin(x)cos(x)

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).

2cos1

2cos xx +

=

2cos1

2sin xx −

=

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. We summarize the results in the following table. sin 0 = 0 sin π/6 = 1/2 sin π/3 = 3 /2

sin π/4 = 1/ 2 sin π/2 = 1

cos 0 = 0 cos π/6 = 3 /2 cos π/3 = 1/2 cos π/4 = 1/ 2 cos π/2 = 1

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.

The cos(2x) formula has three forms, the 2nd and 3rd obtained from the 1st with a substitution from sin2 + cos2 = 1. 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.

sin&cos 8/16/2007 5

Example 1. Here are two different ways to write 15° in terms of the special angles

15º = 60º–45º.

15° = 30°/2 In radians:

4312πππ

−=

621

12ππ

=

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:

4sin

3cos

4cos

3sin)

43sin(

12sin πππππππ

−=−=

21

21

21

23

−=

2213 −

=

The half-angle formula gives us:

2)2/3(1

2)6/cos(1

12sin

−=

−=

ππ

432 −

=

232 −

=

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 .

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 .

sin&cos 8/16/2007 6

Problems. 1. Calculate the following. Give exact answers. Work with a picture, either the cos and sin graphs, or the circle diagram (or both!) (a) sin(2π/3) (b) cos(3π/4) (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 = sin9π . 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

sinx + 3cosx = 1 in the interval –2π ≤ x ≤ 2π.

sin&cos 8/16/2007 7

-1

0

1

2π

23π

25π

27π2ππ−π 3π 4π

2π−

y= xsiny= xcos

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( 2π +x) to draw, on the same set of axes, the graph

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: (a) Based on this diagram, sketch the graph of the average of the two functions:

f(x) = 2

cossin xx + .

(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.]