Question

(1-sin^2x)(1+sin^2x) = 2cos^2x - cos^4x

Answer 1

Use
\[\cos^2x=1-\sin^2x\]Then,
\[cos^2x(2+\cos^2x)\]Multiply it and you'll find the way.

Answer 2

for which side?

Answer 3

I begin from the left to reach the right side.

Answer 4

so your saying, (cos^2x)( 1+sin^2x) on the left side?

Answer 5

Exact.

Answer 6

Then what?

Answer 7

I'll write it again because I miss a sign :)

Answer 8

\[(1-\sin^2x)(1+\sin^2x) =\cos^2x(1+1-\cos^2x)=\cos^2x(2-\cos^2x)=\\
=2\cos^2x-\cos^4x\]

Answer 9

That's all ;)

Answer 10

They don't equal though?

Answer 11

Well, the line I wrote you, demonstrate the equality you wrote in the problem. Do you see it? It begins from the left side you wrote and finish in the right side.

Answer 12

ok how did you get (1+sin^2x) to (1+1 - cos^2x)

Answer 13

Well, we can apply this identity,
\[\sin^2x+\cos^2x=1\]Then
\[\sin^2x=1-\cos^2x\]And substitute.

Answer 14

Oh I see it now, I didn't now you can do that with the + 1 with it

Answer 15

Apply it when you need it ;), it is a very tricky identity.

Answer 16

I got another one

Answer 17

secx/tanx - tanx/secx = cosx cotx

Answer 18

Well, we need to use,
\[\sec x=1/\cos x\]\[\tan x=\sin x/\cos x\]\[\cot x=\cos x/\sin x\]Then
\[(1/\cos x)/(\sin x/\cos x)-(\sin x/\cos x)/(1/\cos x)=\frac{1}{\sin x}-\sin x=\\
=\frac{1-\sin^2x}{\sin x}=\frac{\cos^2x}{\sin x}=\cos x\frac{\cos x}{\sin x}=\cos x\cot x\]

Answer 19

alright I think found another way. I times secx/tanx by secx/secx and tanx/secx by tanx/secx, which leads to sec^2x - tan^2x on top which is 1. So 1 over tanx equeal cotx. and 1 over secx equals cosx

Answer 20

I would say you cannot do that, because you are not multiplying by 1 each fraction. So you, finally, would not obtain the equality you are searching for.

Answer 21

It worked out perfectly for me, so Im keeping it

Answer 22

Okay, so when you finally sum the fractions do you obtain cos x* cot x?

Answer 23

If this the case then something wrong is happening in the Universe :)

Answer 24

yeah 1 over secx = cosx and 1over tanx = cotx

Answer 25

I got one more identity, -1/tanx - secx = 1+sinx/cosx

Answer 26

Are you serious?

Answer 27

yes?

Answer 28

\[-\frac{1}{\tan x}-\sec x=-\frac{1}{\sin x/\cos x}-1/\cos x=-\frac{\cos x}{\sin x}-\frac{1}{\cos x}=\\
=-\frac{\sin x+\cos^2x}{\sin x\cos x}\]

Answer 29

Your last relation is impossible. You only need to do x=0 and the relation doesn't hold.

Answer 30

Oh alright, I can do this one on my own. You helped me enough! Helping a stranger to do math, is a great deed. Thank you very much :)

Answer 31

Don't worry. But in case you need, I think wolfram alpha could be very helpful to check this relations. I check the last one with it, using x=1.

http://www.wolframalpha.com/input/?i=-1%2FTan%5B1%5D-1%2FCos%5B1%5D%3D%3D1%2BSin%5B1%5D%2FCos%5B1%5D

Answer 32

ok

Answer 33

You have the identity wrong

Answer 34

Could you please explain me why?

Answer 35

Let me re write it?

Answer 36

Please

Answer 37

-1/tanx - secx = 1 +sin x/cos x

Answer 38

And what is the difference between this and mine?

Answer 39

\[\sec x=\frac{1}{\cos x}\]

Answer 40

Instead of sec x you had cos x

Answer 41

Well, I think you note that,
\[\sec x=\frac{1}{\cos x}\]

Answer 42

But don't worry, Wolfram Alpha also know about Secx, so here is the link with the same result,
http://www.wolframalpha.com/input/?i=-1%2FTan%5B1%5D-Sec%5B1%5D%3D%3D1%2BSin%5B1%5D%2FCos%5B1%5D

Answer 43

Do you see it? The test is false, so the relation doesn't hold.

Answer 44

Maybe my teacher made this one wrong on purpose

Answer 45

I see the relation

Answer 46

May be your teacher put it wrong, it is strange, but it doesn't hold. Sorry.

Answer 47

Alright well your the expert, so it has to be wrong

Answer 48

May be a sign or something. I have tried all the combinations but I didn't find anything that your teacher cpuld put as a problem with this expression.

Answer 49

I got the input to say true when I wrote it

Answer 50

-1/(tanx - secx) = (1+sinx) / cosx

Answer 51

I forgot the parentheses