Expert: Josh - 6/28/2004

Question
Thanks, it took me an hour or so to work through that, but you explained it well and I can see its potential. However, I'm guessing from the fact that we've never studied any of that, and that the question was only worth 3 marks, that perhaps there is a 'watered-down' method for solving it? If you can, I'd be interested if you could show me that. Thanks,

Sam

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Followup To
Question -
Hi, this question doesn't bear much importance anymore but it's really getting on my nerves. On my additional maths exam yesterday, there was a question:

'Find all values of x in the interval 0 <= x <= 360 that fit the expression sin x = -2 cos x'

Okay so it hardly makes any difference to me any more, but I can't even start to think how you might go about solving this. I've tried using a spreadsheet to find the two answers of 116.5 degrees and 296.6 degrees. However, I know that there is a better way of doing this and I can't think how! Thanks in advance,
Sam
Answer -
Hi Sam,

I know how you're feeling. It's always good clearing things up.

With trigonometry, there are usually many ways for solving a problem. The technique that I am about to show you may seem an "over-kill" initially, but you will find this algebraic substitution method useful also for harder  problems, where the solutions cannot be easily spotted by inspection.

BACKGROUND:
Think of (x/2) as an angle. If we use a new variable "t" to denote the quantity tan(x/2)...[i.e., let t=tan(x/2)], it can be shown that, sin(x/2)=t/sqrt(1+t^2) and cos(x/2)=1/sqrt(1+t^2).
========================================================
Proof: Refer to the right angle triangle (RAT) below.
A
|
|
C-----B
Label angle ^ABC=x/2.
Since tan(x/2)=t/1=|AC|/|CB|, we can let |AC|=t be the length of side AC, similarly, let |CB|=1.
Using Pythagoras, |AB|=square_root_of(1+t^2). Thus,
sin(x/2)=|AC|/|AB|=t/sqrt(1+t^2)
cos(x/2)=|CB|/|AB|=1/sqrt(1+t^2)

We can also derive expressions for sin(x),cos(x) and tan(x) using the standard identities,
sin(a+b)=sin(a)cos(b)+cos(a)sin(b) ...[#1a]
cos(a+b)=cos(a)cos(b)-sin(a)sin(b) ...[#1b]

sin(x)=sin((x/2)+(x/2))
=2*sin(x/2)cos(x/2)
=(2*t)/(1+t^2) ...[Result A]
cos(x)=[cos(x/2)]^2-[sin(x/2)]^2
=[1/(1+t^2)]-[(t^2)/(1+t^2)]
=(1-t^2)/(1+t^2) ...[Result B]
=========================================================
CONVENTIONS: In what follows,
[1] I will use "atan()" to represent the inverse tan function, it is usually hand-written as tan^(-1), where ^ denotes superscript. The w=atan(c) function seeks an angle "w" which subtends a ratio "c" between the opposite and adjacent sides of the triangle. So, for instance, atan(1) returns 45 degrees.
[2] All angles are given in radians, unless otherwise stated. Conversion beteen "degrees" and "radians" is quite easy, PI radian (3.141592654) corresponds to 180 degrees.
deg      rad
180  <-> pi
90   <-> pi/2
60   <-> pi/3
45   <-> pi/4
30   <-> pi/6
22.5 <-> pi/8 and so forth.

NOW, SOLVING THE PROBLEM:
Using t-substitution,
sin(x)=-2*cos(x) is now nothing more than
(2t)/(1+t^2)=-2*[(1-t^2)/(1+t^2)], where t=tan(x/2)...[#2]
We managed to convert the original problem into a quadratic expression, in terms of a new variable t=tan(x/2).
Simplifying equation [#2], we obtain
2t=-2(1-t^2)
t=-1+t^2
t^2-t-1=0
Using the quadratic formula, t=[-b+or-sqrt(b^2-4*a*c)]/(2*a), where coefficients a=1,b=-1,c=-1,
t=[-1+sqrt(5)]/2 and [-1-sqrt(5)]/2.
That is, tan(x/2)=[-1+sqrt(5)]/2, [-1-sqrt(5)]/2.
(The approximate values for tan(x/2) are 1.618033989 and -0.618033989 respectively.)

There are four feasible solutions here in theory. But first,
we worry about getting the 2 obvious solutions:
(x/2)=atan(1.618033989)=1.017221968 radian or 58.28252559 degrees, ...[#3a]
(x/2)=atan(-0.618033989)=-0.553574359 radian or -31.71747441 degrees. ...[#3b]
(I used the calculator to obtain these by-the-way)

Since the "tangent(x/2)" function is positive in both the first and third quadrant, viz., 0<(x/2)<90 & 180<(x/2)<270.
From [#3a], (x/2)=180+58.28252559 must also be a solution...[#3c]
(If you feel unsure, draw some pictures to see this, using a protractor perhaps)

Since the "tangent(x/2)" function negative in both the second and fourth quadrant, viz., 90<(x/2)<180 & 270<(x/2)<360.
From [#3b], (x/2)=180-31.71747441 must also be a solution...[#3d]

In summary, from [#3(a-d)]
x=2*(58.28252559),
x=2*(-31.71747441),
x=2*(238.28252559),
x=2*(148.2825256) must all be solutions to sin(x)=-2cos(x).

That is, x=116.5650512,-63.43494882,476.5650512,296.5650512.
Wrapping this inside the [0,360) interval, the two solutions x={-63.43494882,476.5650512} are redundant.

So, in the end, x={116.5650512,296.5650512} degrees.

Please get back to me if there's anything that you don't get. This is a very powerful technique and is certainly worth learning in my opinion.

Cheers,
Josh

Answer
Yeah, I agree. For a 3 mark question, your teacher can't expect you to suddenly become an inspired mathematician and work everything out from scratch...under the stress of an exam.
Having said that, as long as you remember the identities for sin(x/2), cos(x/2) you can basically derive the rest without too much difficulty. The t-substitution method is not actually that tedious.

Here is another approach.

Rewrite sin(x)=-2cos(x) in the form,
A*sin(x)+B*cos(x)=C. ...[#1]
The trick is to pick A=cos(y) and B=sin(y).
This selection allows you to isolate "y", thus you are not making this into a bigger problem.

Notice that [#1] is equivalent to
sin(y+x)=cos(y)*sin(x)+sin(y)*cos(x)=C.

For your question,
1*sin(x)+2*cos(x)=0, ...[#2]
where A=cos(y)=1, although the second term cannot be strictly defined in terms of a cosine function, you can pretend that B=sin(y)=2.

Now, isolate "y".
B/A=tan(y)=2,
y=atan(2)=1.1071..rad or 63.4349 deg (1st quadrant)
      or 4.2487..rad or 243.4349 deg (3rd quadrant),
for -pi<y<=pi.

Thus,
Now, we can express [#2] in terms of an auxillary angle,
and solve "sin(x+y)=0" for "x".

case 1: with y=63.4349 deg
Taking inverse cosine,
(x+63.4349)=0,180
possible values are x=-63.4349 deg, 116.5650 deg.

case 2: with y=243.4349 deg
Taking inverse sine,
(x+243.4349)=360,540 [N.B. can pick 360 instead of 0 deg]
possible values are x=116.5650 deg, 296.5651 deg

Remeber that you need to always check for all possibilities.
These amount to the same results as before.

Certainly no harm done seeing a few different techniques.
I guess you now have an edge over your class mates:)

How's your teacher by the way? Don't you guys talk about the questions after the paper has been marked?

Hey, where are you from?
guess where I'm from?
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Josh