Expert: Josh - 1/21/2010

Question
QUESTION: Hi Josh,, can you help me?

ive attached an image for the following question. the question is the first line, and both formulas are above,i chose the sin+cos formula. My question is where am i going wrong?
If cos 1/2 = 60 degrees

why are the solutions
-135,-45,45,and 135

have i used the formula wrong?
Thanks for all your help

Richard

ANSWER: Hi Richard,

The question reads:
5*[cos(x)]^2 = 4 - 3*[sin(x)]^2

Your working:
5*[cos(x)]^2 -4 = -3*[sin(x)]^2
5*[cos(x)]^2 -4 = -3*(1-[cos(x)]^2) ....using [sin(x)]^2+[cos(x)]^2=1
5*[cos(x)]^2 -4 = -3 +3*[cos(x)]^2
2*[cos(x)]^2 -1 = 0          ....[#1]

Everything is fine up till this point.
But the following factorization does not follow from the previous line.

(2*cos(x)-1)(cos(x)+1) = 0  is where the mistake is made.

Note: LHS equals 2*[cos(x)]^2 + cos(x) -1, NOT 2*[cos(x)]^2 -1.

I would rewrite [#1] as 2*[cos(x)]^2 = 1.
Then, [cos(x)]^2 = 1/2.
Taking square root on both sides, cos(x) = +1/sqrt(2) or -1/sqrt(2).
Always remember the plus and minus (two separate cases to consider).

From the cosine waveform,
Solutions for cos(x)=1/sqrt(2) consist of 45 and -45 degrees.
Solutions for cos(x)=-1/sqrt(2) consist of (180-45) and 180+45 degrees.
Note: 180+45=225 is same as 225-360 = -135 degrees.

Hence, we have x = +/- 45, +/- 135.

Josh

---------- FOLLOW-UP ----------

QUESTION: Hi Josh,

this line    2*[cos(x)]^2 -1 = 0          ....[#1]

the examples i have done so far have required me to make into a quadratic.
How do i know when and when not to make the line into a quadratic? What i mean is, when im rearranging how do i know if something should equal 0 or a number?
As you seen my working out, i subtracted to get 0, the solution was to to keep the 1. i seem to be very confused by this , as sometimes i get it, and sometimes i dont!

thanks a lot,

Richard


Answer
Hi Richard,

I'm a little surprise that you are already fully engaged with mathematics, so early in the year. I thought the school holidays aren't over yet ;)

Unless the question asks you to express [#1] as a product of linear factors, i.e., something like (cos(x)-a)(cos(x)-b)=0, you can solve it in anyway as you see fit.

When I see an unknown term (either cos(x) OR [cos(x)]^2, but NOT both) isolated on one side of an equation, I usually proceed by separating the known and unknown. i.e., writing 2*[cos(x)]^2-1=0 as [cos(x)]^2=1/2, followed by taking the square root.

| COMMENT:
| We can perform this separation only when the LHS polynomial
| a*[cos(x)]^3 + b*[cos(x)]^2 + c*[cos(x)] + d = 0
| contains one power term that involves the unknown variable.
| e.g., if "a" and "b" are 0 [both cos(x)^3 and cos(x)^2 are missing] or
| "a" and "c" are 0 [both cos(x)^3 and cos(x) are missing] or
| "b" and "c" are 0 [both cos(x)^2 and cos(x) are missing].

This is easier to do than finding a quadratic factorization; although the latter can also be done.
Here, we can view 2*[cos(x)]^2 + 0*cos(x) -1 =0 as a quadratic, with the middle term missing.
| Using factorization,
| 2*[cos(x)]^2-1
| = 2*{[cos(x)]^2-1/2}  ....inside curly bracket is a square of difference
| = 2*{(z+a)(z-a)}
|   where z=cos(x), a=1/sqrt(2)
| Hence, 2*{(z+a)(z-a)} = 0 when cos(x)=-1/sqrt(2),+1/sqrt(2).
|
When we see multiple unknowns (e.g., "cos(x)" and "[cos(x)]^2" both present in an equation), we can no longer write the equation in the form of [cos(x)]^n = constant. We have to use something like quadratic factorization. I avoid this when I can (see earlier COMMENT), because this route is longer and it is easier to make mistakes.

Summary:
a) Direct method - expresses the unknown "cos(x)" in terms of the known quantities. Once you obtain [cos(x)]^n=some_constant, taking the nth root gives you the answers. We can do this only if the equation contains a single powered term of the unknown.

b) Root finding approach - Equation is organized in such a way that the LHS is described by a polynomial, and equated to zero on the RHS. If the LHS is a quadratic, you can use factorization, or the quadratic formula, to solve for z^2-(a+b)z+ab = (z-a)(z-b) = 0.

Both methods essentially amount to doing the same thing. The objective is still to find z (in this question, z represents cos(x)) and there is more than one way to get there, depending on what steps you take. There is no need to worry which method you choose. As long as you do everything right, you will get the answer.

Homework questions are meant to reinforce certain ideas. But just because you have done a dozen questions in a certain way, it doesn't mean it is always best to follow that approach. So, keep an open mind as better alternatives might be available.