Question

Establish the identity
1-(cos^2 Θ/1+sin Θ)=sin Θ

Answer 1

Can you take the LCM and make the denominator common..??

Answer 2

yeah, so..[1(1+sin Θ)/1+sin Θ]-[cos^2/1+sin Θ]?

Answer 3

Yeah right, now can you join these terms because you have denominator like:

\[\frac{4}{5} + \frac{6}{5} = \frac{4 + 6}{10}\]

Like this..

Answer 4

1+sin Θ-cos^2 Θ/1+sin Θ

Answer 5

Yep, good going..

Now use this :

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

Can you use this Identity in place of \(cos^2(x)\) ??

Answer 6

Don't complicate things.
\[\huge \frac{\cos^2x}{1+\sin x}=\frac{1-\sin^2x}{1+\sin x}\]
As you said in the above identity^
And then just use difference of two squares
\[\large x^2-y^2=(x+y)(x-y)\]

Answer 7

Going by another method does not mean that we are going in a complicated way..

Answer 8

Yeah, it doesn't. Complication is something subjective.
However, I think having to get an LCM is more complicated.

Answer 9

Than simply factoring, and cancelling.

Answer 10

\[\frac{1 - \sin(x) - 1 + \sin^2(x)}{1 + \sin(x)} \implies \frac{\sin(x)[1 + \sin(x)]}{1 + \sin(x)}\]

Answer 11

So, you basically had to factor out and cancel, anyway, with the additional step of getting an LCM. Just one step more. However, still a step more.

Answer 12

uh.. i did it by @waterineyes method, & got (1+sin Θ)(1-sin Θ)/1+sin Θ
also cancelled out the two 1+sin Θ so i just have 1-sin Θ left?

Answer 13

Actualy, 1 - (1 - sin Θ)

Answer 14

I think understanding is far more better than Complication, and if you are understanding it in the complicated way, then It is good for you..

And thanks, you just confused the boy, you can give your method after letting me understand my method to him..

Answer 15

her* ._.
& how come it's 1-(1-sin Θ)?

Answer 16

You had a 1 - (cos^2 theta)/(1+sin theta) at the beginning.

Answer 17

You continue first, and if she does not get your method or she still wants to learn my method also, then I will tell her later, you continue..

Answer 18

Do I detect hostility, @waterineyes ?
I'm sorry, I meant no offense.

@MoshiBunny
\[\huge 1-\frac{\cos^2\theta}{1+\sin \theta}\]
And you factored out the fraction bit
\[\huge 1-\frac{1-\sin^2\theta}{1+\sin \theta}=1-\frac{\cancel{(1+\sin \theta)}(1-\sin\theta)}{\cancel{(1+\sin\theta)}}\]

So, don't forget that the "1 - " part was still there, after you factored and cancelled.

Answer 19

you could use the pythageorean identity.
1-cossquared = sinsquared over.
since it's over sin one of your sin's cancel leaving you with the sin you need

Answer 20

pythagorean identity is confusing to me ;/
& yeah i get it, so i get 1-sin^2 Θ

Answer 21

@terenzreignz more to do..

Answer 22

Well, then, if you're having trouble with the Pythagorean Identity, just remember this
\[\large \cos^2x+\sin^2x=1\]
Two more identities relating the other Trigonometric functions can be derived from this.

Or are you having trouble believing the Pythagorean identity itself? ^.^

Answer 23

oh i didn't learn it yet, i just learned trig identities yesterday..

Answer 24

Well, let's take a look at a right triangle

Answer 25

You know the Pythagorean Theorem itself asserts that
x² + y² = r²
right? :)

Answer 26

yeah..

Answer 27

Well, you also note that
cosθ = x/r
sinθ = y/r

Answer 28

So, back to this one...
\[\large x^2+y^2=r^2\]Dividing both sides by r², we get
\[\Large \frac{x^2}{r^2}+\frac{y^2}{r^2}=1\]Which can be written as
\[\Large \left(\frac{x}r\right)^2+\left(\frac{y}r\right)^2=1\]
And since
\[\large \cos\theta=\frac{x}r \ \ \ \ \ \ \ \sin\theta=\frac{x}r\]the above equation may be written as
\[\huge \cos^2\theta+\sin^2\theta=1\]There you have it, the Pythagorean identity, which was derived using the Pythaogorean Theorem, hence its name ^.^

Answer 29

umm... ok...

Answer 30

i think i'm just more confused..

Answer 31

You didn't need to know how to derive it (I hope?)
But I do hope you remember this identity
cos²x + sin²x = 1
It's very useful.

Also note, that if we divide both sides by cos²x, we get
1 + tan²x = sec²x

Again, note, that if we instead divide both sides by sin²x, we get
cot²x + 1 = csc²x

So these three, together, are known as the Pythagorean Identities.
cos²x + sin²x = 1
1 + tan²x = sec²x
cot²x + 1 =csc²x

Answer 32

hmm.. okay i'll keep it in mind, thanks (=