Expert: Josh - 12/1/2003

Question
Hi

I am really stuck on this question:

For each of the points show what P(x), cos X, sin x, tan x cot x, sec x & cosec x are using exact notation

a) 11 pye divided by 6
b) 11 pye divided by 4
c) pye divided by 8
d) pye divided by 12
e) -7 pye divided by 12
f) 5 pye divided by 8

I'm hopeless at trig so if you could help that would be great.

Thanks

Clair

Answer
Hi Clair,

The first thing I got to say to you is don't ever give up on something because you think you are hopeless, you are not. No matter where you are at right now, if you persevere and get the help that you need, you can always and only get better.

For this question, we work with angles, X, where X is measured in radians.

Part I- Degrees versus Radians
There are two ways of measuring angle. We can divide a circle, (one complete revolution) into 360 degrees. Or we can say that a complete revolution, ie., 360 degrees is equivalent to 2*pi.
Get use to these, right now.
360 degrees=2*pi
180 degrees=pi
90 degrees=pi/2
60 degrees=pi/3
45 degrees=pi/4
30 degrees=pi/6
15 degrees=pi/12
You need to be able to convert between these mentally, as they come up everywhere.
Note: If you ever need to use a calculator to work out nasty angles (which are none of the above), make sure you switch it into the right mode. On Casio, this is commonly given by "D", "R", which stands for degrees and radians respectively.

Part II- Meaning of Trigonometry
Trigonometry is all about relating the ratio between any two sides of a triangle, to an angle subtended by the triangle.
Trigonometry is useful in virtually all branches of science. Did you know that investigators work out the hiding place of the assessin of JF Kennedy by back projection, using basic trigonometry and geometry?

Look at this diagram.
A
|
|
C-------B

Let angle ^ABC be "X". The hypotenuse is the longest side, |AB|, which is opposite to the right angle ^ACB.
By definition, (you must familiarize with this, there are only three fundamental identities, all the rest can be derived, so please commit these to memory.)
sin(X)=opposite/hypotenuse=|AC|/|AB|
cos(X)=adjacent/hypotenuse=|BC|/|AB|
tan(X)=opposite/adjacent==|AC|/|BC|
Remember, we are taking angle ^ABC as X, for the moment.

Part III
Remember the sign of the trigs in each quadrant.
S|A
---
T|C
In quadrant 1, 0<=X<90 OR 0<=X<(pi/2)
sin(X),cos(X),tan(X) are "ALL" positive quantities.

In quadrant 2, 90<=X<180 OR (pi/2)<=X<pi
Only sin(X), "S" is positive. ie., cos(X),tan(X) negative.
Specifically, sin(X)=sin(pi-X),
cos(X)=-cos(pi-X),tan(X)=-tan(pi-X)

In quadrant 3, 180<=X<270 OR pi<=X<(3*pi/2)
Only tan(X), "T" is positive. ie., sin(X),cos(X) negative.
Specifically, tan(X)=tan(X-pi)
sin(X)=-sin(X-pi),cos(X)=-cos(X-pi)

In quadrant 4, 270<=X<360 OR (3pi/2)<=X<(2*pi)
Only cos(X), "C" is positive. ie., sin(X),tan(X) negative.
Specifically, cos(X)=cos(2*pi-X)
sin(X)=-sin(2*pi-X),tan(X)=-tan(2*pi-X)

What does this all means?
eg., say X=135 degrees, sin(135)=sin(45) according to the observations above. ie.,sin(3*pi/4)=sin(pi/4)
but, cos(135)=-cos(45), ie, cos(3*pi/4)=-cos(pi/4)
...get the drift?

What happens when X>2*pi, subtract by multiples of 2*pi until angle is reduced to fundamental interval between 0 and 2*pi.
eg., sin(495)=sin(360+135)=sin(135) and the rest is as before.

Answers to your questions.

a) (11/6)*pi=2*pi-(pi/6).
First, identify that angle is between 3*pi/2 and 2*pi.
ie., it's in quadrant 4.
Let X=(11/6)*pi
from above, when (3*pi/2)<=X<2*pi
cos(X)=cos(2*pi-X),
sin(X)=-sin(2*pi-X),
tan(X)=-tan(2*pi-X),

Applying these results,
note: make sure you use the brackets, otherwise, it's wrong.
We are talking about (11/6)*pi NOT 11/6*pi.
*recognize that
cos((11/6)*pi)=cos(2*pi-(11/6)*pi)=cos((2-11/6)*pi)
*then, the rest follows,
cos((2-11/6)*pi)=cos(pi/6)=sqrt(3)/2
note: we are implicitly quoting the result,
cos(X)=cos(2*pi-X),

sin((11/6)*pi)=sin(2*pi-(11/6)*pi)=sin((2-11/6)*pi)
Therefore, sin((2-11/6)*pi)=-sin(pi/6)=-sin(30)= -1/2
since, sin(X)=-sin(2*pi-X), when (3*pi/2)<=X<2*pi ...

Hints for other parts-
note: sec(X)=1/cos(X), cosec(X)=1/sin(X), cot(X)=1/tan(X).
b) (11*pi/4)=2*pi+(3/4)*pi.
As we have discussed earlier, if X>2*pi, the strategy is to keep reducing it by 2*pi until it is in the range 0 to 2*pi.
So, the problem is same as considering X=(3/4)*pi, which sits in the second quadrant.

c) X=pi/8 or 22.5 degrees. We know very well that sin(pi/4)=cos(pi/4)=1/sqrt(2) and tan(pi/4)=1.
Quoting identities:
sin(2*theta)=2*sin(theta)*cos(theta) ...[#A]
cos(2*theta)=2*[cos(theta)]^2 -1 ...[#B]
tan(2*theta)=2*tan(theta)/[1-[tan(theta)]^2] ...[#C]

If we let theta=X=(pi/8), 2X=(pi/4),
we can firstly, work out cos(X)=cos(pi/8) using [#B]
cos(2*pi/8)=2*[cos(pi/8)]^2 -1  ...[*]
Remember the left hand side (LHS) is known.
we know that cos(2*pi/8)=cos(pi/4)=1/sqrt(2).
Rearranging equation [*],
cos(pi/8)=sqrt((cos(pi/4)+1)/2) ...[**]
Okay, we can do this in one line.
[**] simplifies to,
cos(pi/8)=sqrt(((1/sqrt(2))+1)/2) in exact form.

Then, you can work out sin(X)=sin(pi/8) using [#A]
sin(2*pi/8)=2*sin(pi/8)*cos(pi/8)
By, this time, tan(X)=sin(X)/cos(X) is a given.
you can reciprocate the fractions to obtain sec,cosec,cot...

d) pi/12=pi*(4/12-3/12)=pi*(1/3-1/4)=pi/3-pi/4.
Read "Part II (continued)" below for pointers on how to proceed. Just use [#1],[#2] and [#3].

e) -(7*pi/12) ~ -(7*pi/12)+2*pi = (5*pi/12)

f) (5/8)*pi = (4/8+1/8)*pi=pi/2+pi/8. Use result obtained in part (c).

You can do the rest, same ideas.

If you have got time, read the following afterwards.

Good luck.
====================================================
Part II (continued)...
If we call angle ^BAC as alpha, we know that "alpha" must be (pi/2)-X. Why? Cos there are 180 degrees in a triangle, which equals the sum of all angles. But ^ACB is 90 degrees.
So, 180=^ABC+^BCA+^BAC. Thus, 180=X+90+alpha.
Putting these into radians, [see Part I]
alpha=pi-(pi/2)-X=(pi/2)-X as claimed.

Exercise: work out the expressions for sin(alpha),cos(alpha),tan(alpha). Attempt these before reading the answers, don't cheat yourself.
Solution:
sin(alpha)=opposite/hypotenuse=|BC|/|AB|
cos(alpha)=adjacent/hypotenuse=|AC|/|AB|
tan(alpha)=opposite/adjacent==|BC|/|AC|

What do you notice?
Something you would have learnt or will soon learn are the trig. identities with the sum of two angles as the argument.

Consider sin(alpha)=sin(pi/2-X)
It can be shown that in general,
sin(A+B)=sinA*cosB+cosA*sinB ...[#1]
cos(A+B)=cosA*cosB-sinA*sinB ...[#2]
tan(A+B)=[tan(A)+tan(B)]/[1-tan(A)tan(B)] ...[#3]

Fact: sin(pi/2)=sin(90degrees)=1 & cos(pi/2)=0
Draw the oscillatory curves to see this.

So, simplifying [#1] & [#2], with A=pi/2, B=-X
sin(pi/2-X)=sin(pi/2)*cos(-X)+cos(pi/2)*sin(-X) ...[#1A]
=1*cos(X)+0*[-sin(X)]
=cos(X)
using the fact that cos(-X)=cos(X), sin(-X)=-sin(X)
cos(pi/2-X)=cos(pi/2)*cos(-X)-sin(pi/2)*sin(-X) ...[#2A]
=0*cos(X)+0*[-sin(X)]
=-sin(X)