Question

How to prove identity: sin x-1/sin x+1 = -cos^2 x / (sin x+1)^2?

Answer 1

\[\sin x-1divsin x+1 = -\cos ^{2}x \div(\sin x+1)^{?}\]
\[\sin x-1/\sin x+1 = -1-\sin ^{2}x/(sinx+1)(sinx+1)\]

Answer 2

that's all I've got up to and I'm stuck D:

Answer 3

\[\LARGE {\sin x-1\over \sin x+1}={-\cos^2x\over (\sin x+1)^2}\]
is this the given? :)

Answer 4

yes!

Answer 5

ok... on the right side you have the numerator \(-\cos^2 x\) from identity \(\sin^2x+\cos^2x=1\)
you have:

\[\LARGE \cos^2x=1-\sin^2x\]
so in your case you have:
\[\LARGE -1\cdot \cos^2x\]
\[\LARGE -1\cdot (1-\sin^2x)\]
so that becomes :
\[\LARGE \frac{\sin x-1}{\sin x+1}=\frac{-1(1-\sin^2x)}{(\sin x+1)^2}\]
are we good till here :)

Answer 6

noo :( Do i switch the spots -1(1-sin^2x)/(sinx+1)^2 to (sinx+1)^2/-1(1-sinx^2) .. oh I'm confused D:

Answer 7

no you don't ....
from this:
\[\large \frac{{\sin x - 1}}{{\sin x + 1}} = \frac{{ -\cos^2x }}{{{{\left( {\sin x + 1} \right)}^2}}}\]


we've came up with...
\[\large \frac{{\sin x - 1}}{{\sin x + 1}} = \frac{{ - 1 \cdot \left( {1 - {{\sin }^2}x} \right)}}{{{{\left( {\sin x + 1} \right)}^2}}}\]

Answer 8

have you ever heard about this identity

\[\LARGE \sin^2x+\cos^2x=1\] ?

Answer 9

yes I have

Answer 10

do you understand how this came up?

\[\LARGE \frac{{\sin x - 1}}{{\sin x + 1}} = \frac{{ - 1 \cdot \left( {1 - {{\sin }^2}x} \right)}}{{{{\left( {\sin x + 1} \right)}^2}}}\]

Answer 11

@milliex51

Answer 12

yes

Answer 13

I understand

Answer 14

ok... nice.

Now we multiply the numerator of right side which from:

\[\LARGE \frac{{\sin x - 1}}{{\sin x + 1}} = \frac{{ - 1 \cdot \left( {1 - {{\sin }^2}x} \right)}}{{{{\left( {\sin x + 1} \right)}^2}}}\]
becomes:
\[\LARGE \frac{{\sin x - 1}}{{\sin x + 1}} = \frac{{{{\sin }^2}x - 1}}{{{{\left( {\sin x + 1} \right)}^2}}}\]
are we good till here? :)

Answer 15

um, bah. I guess so, do I: cancel sin^2x-1 from one of sinx+1? O.O

Answer 16

don't hurry... now we can solve this in two ways...

one of them is using cross multiply ...

\[\LARGE \frac{{\sin x - 1}}{{\sin x + 1}} = \frac{{{{\sin }^2}x - 1}}{{{{\left( {\sin x + 1} \right)}^2}}}\]
we get:
\[\large \left( {{{\sin }^2}x - 1} \right)\left( {\sin x + 1} \right) = \left( {\sin x - 1} \right){\left( {\sin x + 1} \right)^2}\]

\[\large \left( {{{\sin }^2}x - 1} \right)\cancel{\left( {\sin x + 1} \right)} = \left( {\sin x - 1} \right){\left( {\sin x + 1} \right)^{\cancel{2}}}\]

\[\large \left( {{{\sin }^2}x - 1} \right) = \left( {\sin x - 1} \right){\left( {\sin x + 1} \right) }\]
and now multiplying the right side by each other... we get:
\[\left( {{{\sin }^2}x - 1} \right) = \left( {{{\sin }^2}x - 1} \right)\]


the second way...

\[\LARGE \frac{{\sin x - 1}}{{\sin x + 1}} = \frac{{{{\sin }^2}x - 1}}{{{{\left( {\sin x + 1} \right)}^2}}}\]

\[\LARGE \frac{{\sin x - 1}}{{\sin x + 1}} = \frac{{{{\sin }^2}x - 1^2 }}{{{{\left( {\sin x + 1} \right)}^2}}}\]
\[\LARGE \frac{{\sin x - 1}}{{\sin x + 1}} = \frac{{{{(\sin x+1)(\sin x-1 )}}}}{{{{\left( {\sin x + 1} \right)}^2}}}\]
\[\LARGE \frac{{\sin x - 1}}{{\sin x + 1}} = \frac{{{{\cancel{(\sin x+1)}(\sin x-1 )}}}}{{{{\left( {\sin x + 1} \right)}^{\cancel{2}}}}}\]

...

Answer 17

why did you square -1 from sin^2x-1?

Answer 18

to show you that we can use this formula:

\[\LARGE a^2-b^2=(a+b)(a-b)\]

but you don't have to square it... :)

Answer 19

really? ohh, thank you very much!

Answer 20

My pleasure xD