# Row reduction with trig

• EV33
In summary, the conversation is about row reducing a matrix containing trigonometric functions. The person is unsure if they can multiply by a function and asks for help with trig identities. The expert explains the process of row reduction and uses a trig identity to simplify the problem.

## Homework Statement

I am having trouble row reducing...

5cos(t) 5sin(t) | -cos(t)

2cos(t)+sin(t) 2sin(t)-cos(t) | sin(t)

## The Attempt at a Solution

I know I am allowed to multiply a row by a constant but I was wondering if I am allowed to multiply by a function?

I have row reduced it to

5cos(t) 5sin(t)|-cos(t)
sin(t) -cos(t)|sin(t)+(2/5)cos(t)

I was unable to reduce this any further so I tried solving the first equation for x1 and then plugging it into the second equation but it didn't come out pretty.

So I am curious if there are any trig identities which would make this problem easier?

Any help would be much appreciated. Thank you.

"Cosine" is a function. "Cos(t)" is a number- the value of the cosine function at the number, t. So, yes, you can multiply by such a thing or divide by it if you are careful about values of t that make that number 0.

Here, I would row reduce that in the "standard" way. Since the first row of the first column is "5 cos(t)", divided entire first row by it to make that entry "1". The first row becomes:
$$\left[\begin{array}{cc}1 & \frac{sin(t)}{cos(t)}\end{array}\right| -\frac{1}{5}\right]$$
Now, to change that "sin(t)" in the second row of the first column to "0", subtract sin(t) times the new first row from that second row. The number in the second column of the second row becomes
$$-cos(t)- sin(t)\frac{sin(t)}{cos(t)}= -cos- \frac{sin^2(t)}{cos(t)}= \frac{-cos^2(t)- sin^2(t)}{cos(t)}$$
and we certainly can use a "trig identity": $sin^2(xt)+ cos^2(t)= 1$
to get
$$-\frac{1}{cos(t)}$$
(moral- you can't expect to see trig identities from the start- go ahead, do the algebra and they will come!)

For the third column, then, we will need to subtract (-1/5) times sin(t) from sin(t)+ (2/5)cos(t): sin(t)+ (2/5)cos(t)+ (1/5)sin(t)= (4/5)sin(t)+ (6/5)sin(t)+ (2/5)cos(t) so that after reducing the first column you have
$$\left[\begin{array}{cc}1 & \frac{sin(t)}{cos(t)} \\ 0 & \frac{1}{cos(t)}\end{array}\right|\left|\begin{array}{c}-\frac{1}{5} \\ \frac{6}{5}sin(t)+ \frac{2}{5}cos(t)\end{array}\right]=\left[\begin{array}{cc}1 & tan(t) \\ 0 & sec(t)\end{array}\right|\left|\begin{array}{c}-\frac{1}{5} \\ \frac{6}{5}sin(t)+ \frac{2}{5}cos(t)\end{array}\right]$$