Question

I really need help on a question for my test corrections, I don't have a clue. Any help is appreciated!
4cos(x)
------ + 15
tan(x)

=

4sin^2(x) - 15sin(x) - 4

Answer 1

so.. .what are you supposed to do there anyway?

Answer 2

\[4\cos(x) / \tan(x) + 15 = 4\sin^2(x) - 15\sin(x) - 4\]

Answer 3

Show that the equation (first one) can be expressed as the (second one).

Answer 4

They are both equal to zero, which is why they are equal to each other.

Answer 5

hmmm have you covered trigonometric identities yet?

Answer 6

Yes, I can get my notebook, if you want me to.
It's only for my test corrections, I am just beyond confused.
Hold on a second, please.

Answer 7

hmmm lemme check about

Answer 8

Like, I know how to do the second part of the question, where you have to find the numbers between 0 < x < 360. I just have trouble with converting them to equal each other. I have my notebook now.

Answer 9

I have the information to do it, I just don't know what to do in the case of this problem.

Answer 10

tis \(\bf \cfrac{4cos(x)}{tan(x)}+15=4sin^2(x)-15sin(x)-4\ right?\)

is there a -4 on the RHS?

Answer 11

Yes.

Answer 12

i put a value for x angle, evaluating it, the equality is not the same

Answer 13

hmm lemme gimme a few seconds lemme post what I have

Answer 14

In this case, we are not finding the value of x. We are converting the one equation into the other.

Answer 15

With identities. I just do not know how to convert it correctly to equal the other one.

Answer 16

i picked a random test angle, the left and right sides of the equality give different values

Answer 17

Danjs, the second part of the problem has two different values, yes. Can you tell me what you got, because I have the correct values for that part.

Answer 18

\(\bf \cfrac{4cos(x)}{tan(x)}+15=4sin^2(x)-15sin(x)-4
\\ \quad \\ \quad \\
\cfrac{4cos(x)}{tan(x)}+15\implies \cfrac{4cos(x)+15tan(x)}{tan(x)}
\\ \quad \\
\cfrac{4cos(x)+15{\color{brown}{ \frac{sin(x)}{cos(x)}}}}{tan(x)}\implies \cfrac{4cos(x)+\frac{15sin(x)}{cos(x)}}{tan(x)}\implies
\cfrac{\frac{4cos^2(x)+15sin(x)}{cos(x)}}{\frac{sin(x)}{cos(x)}}
\\ \quad \\
\cfrac{4cos^2(x)+15sin(x)}{\cancel{cos(x)}}\cdot \cfrac{\cancel{cos(x)}}{sin(x)}\implies \cfrac{4cos^2(x)+15sin(x)}{sin(x)}
\\ \quad \\
\cfrac{4[{\color{brown}{ 1-sin^2(x)}}]+15sin(x)}{sin(x)}\implies \cfrac{4-4sin^2(x)+15sin(x)}{sin(x)}
\)

which doesn't quite equate the RHS btw

Answer 19

the numerator comes close, but it has the wrong sign
but doesn't quite equate the RHS though

Answer 20

The issue is the sin(x) on the bottom?

Answer 21

The numerator rearranged equals the other equation, yes?

Answer 22

well, it does, but it doesn't, the sign is different

now
take a peek at their graph

http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiIoNGNvcyh4KSkvKHRhbih4KSkrMTUiLCJjb2xvciI6IiMyNzJDRDYifSx7InR5cGUiOjAsImVxIjoiNChzaW4oeCkpXjItMTVzaW4oeCktNCIsImNvbG9yIjoiI0RFMEQwRCJ9LHsidHlwZSI6MTAwMCwid2luZG93IjpbIi00OC40Mjg3NzM4ODAwMDQ4MiIsIjQ4LjQyODc3Mzg4MDAwNDgyIiwiLTI0LjQ4OTgyMjM4NzY5NTI3NyIsIjM1LjExNDgyMjM4NzY5NTI4Il19XQ--

they do differ, thus, they're indeed, NOT equal

Answer 23

anyhow.. need to dash
but they're not equal

Answer 24

Mhm, I see that.
Logically for the question, they are supposed to equal each other in some way, or form.
I believe you have given me enough information to help me finish it myself. I am greatly pleased with your help, as an old member of Open Study, I am grateful for the help.
You have an wonderful day, or night.
You aswell, DanJS.