Expert: Josh - 1/29/2010

Question
Josh,
as requested.

6cos(x)-5sec(x) = tan(x)   for -180 < x <180
| how do i do this?   looked at formulas and they are all squared.

I've quickly scanned over your reply,and will do a full read up tonight.One thing that did catch my eye was the explanation for my question on where does the 1 come from? How did you get the other side of triangle?
Is there a calculation you did, or am i reading this wrong,usually you have 2 things on a triangle to get either a angle,or a measurement. sin x is saying = 0.3,,
 quote
RATIO obtained by dividing the length of the opposite side (to angle x) by the hypotenuse.  I have no angle, just one measurement of sin x=0.3   am i missing something?  

thanks for replying so quickly
richard.

Answer
Richard,

Okay, let's deal with the outstanding issue first. I believe I have already addressed your concern "regarding where the 1 comes from". I think yesterday's explanation will make sense when you read my response more closely tonight.

Re: "How did you get the other side of triangle? given sin(x)=0.3"
Ans: Don't forget, we identified angle ^ACB as "x" right at the beginning, without any loss of generality. This setup does not rely on any assumption, okay? Then, after labeling angle ACB as x, we use the definition of sin(x)=|AB|/|AC| (i.e., opposite/hypotenuse) and set |AB|/|AC|=0.3 because this is what we have been told. It is true that we do not know the EXACT dimensions of |AB| and |AC|; but we do not care. Understand this: To work out cos(x) or tan(x) [these are likewise ratios], we only need to know the RATIO between two sides of a right angle triangle. If the question tells us sin(x)=0.3, then |AB|/|AC| must be 0.3 (by definition). In other words, |AB|=0.3*|AC|. We do not need to know the exact length of |AC|. Any choice of |AB| and |AC| satisfying sin(x)=|AB|/|AC|=0.3 will do. The simplest choice being |AC|=1, |AB|=0.3. [We can set |AC|=2.5 if we like, but then |AB|=0.3*2.5=0.75; both not very nice numbers]

A
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B-------------C

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A quick review of today's problem:
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| Solve 6*cos(x)-5*sec(x) = tan(x) for -180 < x <180
|

We can rewrite the equation as 6*cos(x) - 5/cos(x) = sin(x)/cos(x). At first sight, we seem to run into a dead end, it is difficult to solve this (using techniques we already know) because there are two unknowns "sin(x) and cos(x)" that we need to determine. To overcome this, we need to exploit the underlying relationship between tan(x) and cos(x), and reduce this to an equation with ONE unknown. Here is a technique for achieving this.

Introduction to the t-substitution Technique

A
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B----------C

Let |AB|=t, |BC|=1.
Let angle ^ACB=(x/2).

As usual, the reason for picking "x/2" is not immediately obvious, at least not until you see the computational advantage when we exploit the trigonometric formula for "double angle".

Before we get there, we need to develop the ground work using Pythagoras and fundamental definitions.

tan(x/2)=|AB|/|BC|=t              ...[#1]
sin(x/2)=|AB|/|AC|=t/sqrt(1+t^2)  ...[#2]
cos(x/2)=|BC|/|AC|=1/sqrt(1+t^2)  ...[#3]

Comment: Remember this idea, and commit these to memory if necessary. The rest will follow.

Using well-known identities:

sin(x) = 2*sin(x/2)*cos(x/2) ...[#A] and
cos(x) = [cos(x/2)]^2-[sin(x/2)]^2 = 2[cos(x/2)]^2-1 ...[#B],
we can express sin(x), cos(x) and tan(x) in terms of "t". This is extremely powerful, because we no longer need to view sin(x), cos(x) and tan(x) as independent entities whenever a combination of sin/cos/tan(x) appear in the same equation. In particular, using [#2] and [#3] in [#A] and [#B],

sin(x) = 2t/(1+t^2)                    ....[#4]
cos(x) = 2/(1+t^2)-1 = (1-t^2)/(1+t^2) ....[#5]
tan(x) = sin(x)/cos(x) = 2t/(1-t^2)    ....[#6]

Development of this powerful idea is now complete. We are good to go.
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Returning to 6*cos(x)-5*sec(x)=tan(x), this is clearly equivalent to 6*cos(x)-5/cos(x)=tan(x).

Using [#5] and [#6], we transform this into an equation with one variable. Here is the magic.

6*(1-t^2)/(1+t^2) - 5*(1+t^2)/(1-t^2) = 2t/(1-t^2)
Multiply by (1+t^2)(1-t^2) on both sides,
6*(1-t^2)(1-t^2) - 5*(1+t^2)(1+t^2) = 2t(1+t^2)
6*(1-2*t^2+t^4) - 5*(1+2*t^2+t^4) = 2t+2*t^3
6-12*t^2+6*t^4 -5-10*t^2-5*t^4 = 2t+2*t^3
t^4-2*t^3-22*t^2-2t+1 = 0

Comments: There is actually an explicit formula for finding the zeros of a quartic polynomial. But it is complicated looking, and not worth going into. Derivation of the "quartic formula" involves pages of algebra (take a look at "quartic_function" on wikipedia if you are curious. Warning: you may suffer a heart attack). So, appreciate the fact that we have accomplished the following:

1. Successfully transformed an equation that contains a mixture of sin(x)/cos(x)/tan(x) into an equation with only 1 unknown [expressed in terms of t=tan(x/2)];
2. This problem is now solvable [even though finding the answers still takes a bit of work....with LOTS of algebra manipulation, we can still find the answers by hand].

Ending: Using a numerical package (computer), this equation (with only one unknown in "t") has solutions t = 5.8284, -3.7321, -0.2679 and 0.1716.

Recall that t=tan(x/2), so x=2*tan^-1(t).
x = 2.8018, -2.6180, -0.5236, 0.3398 in radian; OR
 160.5288, -150, -30, 19.4712 in degrees.
 
These solutions have been verified to be correct.

Comments: Referring to the attached figure, you can identify all four solutions at x=-150, -30, 19.4712 and 160.5288. I have zoomed into the graph to emphasize the zero crossings. Apparently, there is also a (fake) zero crossing near -90; but this resulted due to a graphing artefact, there is actually no solution at this point. In fact, there is an asymptote (discontinuity) at that point. Approaching from the left, the curve shoots off to +infinity, approaching from the right, the curve shoots off to -infinity. You can guess what happens when the program tried to join the dots between these two extremes. It produces a spurious zero-crossing which does not exist in reality. The function is actually forbidden to go there (value is undefined).