Question

Simplify the Equation; Trigonometric Solving

Answer 1

\[1-\sin^2x/\sin(x)-\csc(x)\]

Ok, so I know that I should simplify the csc first into maybe "1/sin(x)", right? Then what..?

Answer 2

@terenzreignz Hey, could you help me out? =)

Answer 3

Ambiguity kills... is it like this~
\[\Large \frac{1-\sin^2(x)}{\sin(x)-\csc(x)}\]?

Answer 4

Uhuh, that's it.

Answer 5

Well the top... it should simplify into something... something elusive :D
Remember the pythagorean identity?
\[\large \cos^2(x) + \sin^2(x) = 1\]

Answer 6

Yup, I do.

Answer 7

Hang on...

Answer 8

If we subtract \(\large \sin^2(x)\) from both sides of the pythagorean identity, we should end up with...
\[\Large \cos^2(x) +\sin^2(x) \color{red}{-\sin^2(x)}=1\color{red}{-\sin^2(x)}\]

Simplifying into
\[\Large \cos^2(x) = 1-\sin^2(x)\]

right?

Answer 9

I believe so, yes =o

Answer 10

the numerator then becomes...?

Answer 11

cos^2(x)?

Answer 12

Yup...
\[\Large \frac{\cos^2(x)}{\sin(x) -\csc(x)}\]

Answer 13

Now, the denominator could be written, such as you said, like this...
\[\Large \frac{\cos^2(x)}{\sin(x) -\frac1{\sin(x)}}\]

Answer 14

So... let's manipulate the denominator some more...
\[\Large \frac{\cos^2(x)}{\frac{\sin^2(x)}{\sin(x)}-\frac1{\sin(x)}}= \frac{\cos^2(x)}{\frac{\sin^2(x) -1}{\sin(x)}}\]

Answer 15

Wouldn't the "sin^2(x)" then become negative while the second denominator becomes sin^2(x)?

Answer 16

Why would the \(\large \sin^2(x)\) be negative?

Answer 17

Nevermind got confused. Please proceed =)

Answer 18

remember this identity...
\[\large \cos^2(x) + \sin^2(x) = 1\]

It can also be rearranged into this...
\[\Large \color{green}{\sin^2(x) -1 = \color{black}{-}\cos^2(x)}\]
right?

Answer 19

Right. Ok, so \[\cos^2(x)/-\cos^2(x)/\sin(x)\] . What next?

Answer 20

Would that leave use -1 + sin(x)...?

Answer 21

No, the \(\Large \sin(x)\) at the denominator can be brought to the numerator, like so...
\[\Large \Large \frac{\sin(x)\cos^2(x)}{-\cos^2(x)}\]
Now just simplify :P

Answer 22

sin(x) - 1

Answer 23

Nope... this happens...
\[\Large \frac{\sin(x)\cancel{\cos^2(x)}}{-\cancel{\cos^2(x)}}=\frac{\sin(x)}{-1}\]

Answer 24

Oh, well then it should be -sin(x)..?

Answer 25

That's much better :P

Answer 26

^_^ ok, thank you so much. I may need your help in a second, but it's just for verifying my answers if you don't mind.

Answer 27

right, let's do it :P