Question

TRUE OR FALSE, The equation csc^2x-1=cot^2x is an identity

Answer 1

\[\csc^2x-1 =\cot^2x\]Let's replace everything with their equivalents in the basic trig functions:
\[\csc x = \frac{1}{\sin x}\]\[\cot x = \frac{1}{\tan x} = \frac{1}{\frac{\sin x}{\cos x}} = \frac{\cos x}{\sin x}\]

\[\csc^2 x -1 = \cot^2x\]\[(\frac{1}{\sin x})^2 - 1 = (\frac{\cos x}{\sin x}) ^2\]Multiply both sides by \(\sin^2x\) to clear the fractions. What do you get?

Answer 2

Do i plug in a random number and see if it works?
I was going through my notes but i didn't see anything that could help me out on this.

Answer 3

if you do what I suggested, what is the result? it's going to be an equation, not a number.

Answer 4

Oh, ok. 1^2-1=(cosx)^2
1-1=0
0=(cosx)^2

Answer 5

No...
\[(\frac{1}{\sin x})^2 - 1 = (\frac{\cos x}{\sin x})^2\]Multiply both sides by \(\sin^2x\)
\[\sin^2x *(\frac{1}{\sin x})^2 - \sin^2x*1 = \sin^2x * (\frac{\cos x}{\sin x})^2\]\[\cancel{sin^2x }*(\frac{1}{\cancel{\sin x}})^2 - \sin^2x*1 = \cancel{\sin^2x} * (\frac{\cos x}{\cancel{\sin x}})^2\]\[1-\sin^2x = \cos^2x\]Right? Does that resemble any trig identities you know?

Answer 6

ooo oo I know but I can't just jump in... I'll just watch

Answer 7

You can jump in help me out too lol

Answer 8

But where did you get the extra sin^2x?

Answer 9

which extra sin^2x? I multiplied both sides of the equation by sin^2x to get rid of the fractions. The first term becomes 1. The -1 that was already there becomes - sin^2x. And the term on the right hand side becomes cos^2x.

Answer 10

ok, i think i get it. So if it comes out to be a basic trig it's an identity?

Answer 11

@UsukiDoll what does that mean?

Answer 12

it's a trig identity XD

Answer 13

alright, wait is the answer true?

Answer 14

@UsukiDoll

Answer 15

yes/...

Answer 16

thanks

Answer 17

what about sinx=(sqrt2)/2. is that an identity?

Answer 18

:/

Answer 19

Well, what does \(\sin 0 =\)

Answer 20

0

Answer 21

How about \[\sin \frac{\pi}{2}=\]

Answer 22

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