Prove the identity.
csc(X)-sin(x)/csc(x)+sin(x)=cos^2x/1+sin^2x

- anonymous

Prove the identity.
csc(X)-sin(x)/csc(x)+sin(x)=cos^2x/1+sin^2x

- Stacey Warren - Expert brainly.com

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- chestercat

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- yuki

sorry I tried, but it doesn't seem straight forward.

- anonymous

its okay, at least you put effort into it

- anonymous

This isn't as hard as it looks :)

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## More answers

- anonymous

\[\csc x =\frac{1}{\sin x}\]by definition, so\[\frac{\csc x - \sin x}{\csc x + \sin x}=\frac{\frac{1}{\sin x}-\sin x}{\frac{1}{\sin x}+\sin x}=\frac{\frac{1-\sin^2x}{\sin x}}{\frac{1+\sin^2 x}{\sin x}}=\frac{1-\sin^2 x}{\sin x}\frac{\sin x}{1+\sin^2x}\]\[=\frac{1-\sin^2 x}{1+\sin^2x}=\frac{\cos^2x}{1+\sin^2x}\]

- anonymous

Do you need it to be explained more?

- anonymous

yes if you don't care to

- anonymous

Is there a specific part?

- anonymous

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- anonymous

You can refer to the part by the number on the attachment, if that's easier.

- anonymous

brb

- anonymous

yes that made it alot easier to see thank you! do u care to help me on another one?

- anonymous

Oh, so you're okay with the last one?

- anonymous

Yes i knew how to get it started i just didnt know how to simplify that complex fraction, but i understand now.

- anonymous

Good. What's your other question?

- anonymous

solve this trig equation on the interval 0

- anonymous

Alright...this is a quadratic equation in cos(x). If you 'see' cos(x) as something like, 'u', then you have the equation\[2u^2+3u+1=0\]

- amistre64

2 or 3 more and we can start a club ;)

- anonymous

It can be factored into \[(2u+1)(u+1)=0\]

- anonymous

Yeah, but in 30s you'll be on infinity hero, so... :)

- amistre64

lol.... by then ill just need another account :)

- anonymous

No, by then it's like TRON and you're sucked into the website.

- anonymous

Sorry mj

- anonymous

is it (2cos^2+1)(cosx+1) btw no harm here i aprreciate your help!

- anonymous

Yes it is.

- anonymous

That whole thing is equal to zero, though.

- anonymous

Which means either
2cos^2x+1 = 0
OR
cos(x) + 1 = 0
OR they are BOTH zero.
The solution set is all x's that satisfy those conditions (just like a 'normal' quadratic).

- anonymous

how do i find the angles? thats where i have a problem

- anonymous

Yep...I'll do them on paper first.

- anonymous

Okay thanks!

- anonymous

Okay, if you look at the first factor, you should see something fishy...\[2\cos^2 x + 1 =0 \rightarrow \cos^2x=-\frac{1}{2}\]We have something squared on the left-hand side, and a negative on the right-hand side. If you're only using real numbers (which is what's going on here), this can't happen, so there are no x-values that satisfy this particular equation.

- anonymous

The square of any real number is always 0 or positive.

- anonymous

For the second factor, you have\[\cos x = -1\]On the interval \[0 \le x \lt 2\pi\]this happens for \[x=\pi\]only.

- anonymous

If you plu x=pi into your original equation, you will get 0.

- anonymous

*plug

- anonymous

Thanks! if i had (sin(x)-4)(sin(x)+1) how would i find the angles on the same interval?

- anonymous

Is that equal to zero?

- anonymous

yes sorry i forgot

- anonymous

Okay...it's the same deal...always...
Either one or both of the factors have to be zero, so you set each one to zero and solve for any possible x-values.
Here, in your first factor, you have
sin(x)-4 = 0 --> sin(x)=4
This cannot happen (sin(x) oscillates between -1 and 1...there are no values of x here that will punch out a 4, so you won't be getting any solutions from this factor. Note, if it had been something like 4sin(x)-4 = 0 --> sin(x) =4/4 = 1 and you would have solutions).
Sot the second factor, you have
sin(x)+1=0
which means
sin(x) = -1
When does sin(x) = -1 on this interval?

- anonymous

*'Sot' = 'For'.

- anonymous

is it -pi/2 +2kpi =3pi/2

- anonymous

Yes it is.

- anonymous

3pi/2

- anonymous

Be careful when using formulas like that; you need to ensure you give the angle that lies in the interval they want (even though -pi/2 is the same as 3pi/2 when considered in the context of plugging into sine, cosine, etc.).

- anonymous

so sinx=3pi/2

- anonymous

No, \[\sin \frac{3\pi}{2}=-1\]

- anonymous

You're trying to find the x's that, when plugged into sine, give -1.

- anonymous

oh so x=3pi/2 i made a mistake

- anonymous

Yes.

- anonymous

you have been a lifesaver, that helped me out alot, especially with the identities.

- anonymous

You're welcome, mj :)

- anonymous

So that's it? You're done?

- anonymous

Uhm this one seems simple but i forgot the identity to use
5sin^2x+2cos^2x=5-3cos^2x

- anonymous

Okay...this might be more than you have to do, but technically, this is how you would prove it 'properly'.

- anonymous

Assume that the statement is not true. Then,\[5\sin^2 x + 2\cos^2 x \ne 5-3 \cos^2 x\]But then adding 3cos^2x to both sides gives,\[5\sin^2 x + 5 \cos^2 x \ne 5\]Now, the left-hand side is just\[5\sin^2 x + 5 \cos^2 x=5(\sin^2 x + \cos^2 x) = 5\]which would mean,\[5=5\sin^2 x + 5 \cos^2 x \ne 5\]That is,\[5 \ne 5\]which is false.

- anonymous

Our assumption that the original statement was false led to a contradiction, so we must take the original statement as true.

- anonymous

The statement is either true or false. If it's not false, it must me true.

- anonymous

If you're not happy with that way, I have another.

- anonymous

i dont think thats what im lookin for my teacher makes us make one side the same as the other side :/

- anonymous

Yeah...I'll do it another way.

- anonymous

Take the right-hand side first. You have\[5-3\cos^2 x\]which equals,\[5(1)-3\cos^2 x\]\[=5(\sin^2x + \cos^2x)-3\cos^2x\]\[=5\sin^2x + 5\cos^2x-3\cos^2 x\]\[=5\sin^2 x+2\cos^2x\]

- anonymous

ahh thats exactly what i was looking for i was thinking you would use sin^2+cos^2=1 for this

- anonymous

No more questions then?

- anonymous

tan(x/2)sinx=1-cosx
LHS
tan(x/2)sinx=1-cos/sin(sin/1)=1-cosx
did i do that right?

- anonymous

I'm not sure what you did. This is what I would do:

- anonymous

\[\tan \frac{x}{2}\sin x = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}.2\sin \frac{x}{2}\cos \frac{x}{2}=2\sin^2 \frac{x}{2}=1-\cos x\]

- anonymous

Since by the double angle formula for cosine,\[\cos x = \cos ^2 \frac{x}{2}-\sin^2 \frac{x}{2}=1-2\sin^2 \frac{x}{2} \rightarrow 2\sin^2 \frac{x}{2}=1-\cos x\]

- anonymous

cos(alpha+beta)/sin(alpha)cos(beta)=cot(alpha)-tan(beta)
cos(alpha)cos(beta)-sin(alpha)sin(beta)/.5(sin(alpha+beta)+sin(alpha-beta)
how would i finish that

- anonymous

\[\frac{\cos(\alpha + \beta)}{\sin \alpha \cos \beta}=\frac{\cos \alpha \cos \beta - \sin \alpha \sin \beta}{\sin \alpha \cos \beta}\]\[=\frac{\cos \alpha \cos \beta}{\sin \alpha \cos \beta }-\frac{\sin \alpha \sin \beta }{\sin \alpha \cos \beta}\]\[=\frac{\cos \alpha}{\sin \alpha}-\frac{\sin \beta }{\cos \beta}\]\[=\cot \alpha - \tan \beta\]

- anonymous

That was a much easier way of doing it, identities must take alot of practice to get use to.

- anonymous

Yeah. It's mostly intuition as to how to start and how to close. That comes from practice!

- anonymous

I can tell the preparation and follow through is the hardest part!
cos(x/2)
cos(x/2)=square root of 1+4/5/2
=square root of 9/5/2 =square root of 9/10
is that right if cosx=4/5 with 3pi/2

- anonymous

Um, it's hard for me to tell since I'm not sure of your order of operations.

- anonymous

What's under the square root?

- anonymous

\[\frac{\sqrt{1+4/5}}{2}\]?

- anonymous

i used the half angle formula for cos YES that is correct

- anonymous

Yes, and it's positive root if you're in that interval.

- anonymous

yes that is what i used how bout tan(2x) on same interval
if i typed out what i have it would be hard for you to understand

- anonymous

Do you know how to use the equation editor? The button below will take you there.

- anonymous

oh i didnt know that thank you!

- anonymous

\[\tan2(\theta)=1-\cos(2\theta)\div2\div1+\cos(2\theta)\div2 \]

- anonymous

lol...it's not clear...is it...\[\tan 2 \theta =\frac{\frac{1-\cos 2 \theta}{2}}{\frac{1+ \cos 2 \theta}{2}}\]?

- anonymous

yes thats what i was trying to do i got -3/13 for the answer but im not sure about it

- anonymous

Are you trying to find tan(2 theta) and the right-hand side is what you came up with?

- anonymous

yes im trin to find tan(2theta) on the interval 3pi/2

- anonymous

For you to get a number, tan(2theta) would have to have been set equal to something in the first place. Is that what you have?

- anonymous

Because tan(2theta) takes an infinity of values on that interval...unless tan(2theta)=(something) to begin with.

- anonymous

yes thats what i have it equal to

- anonymous

ok...let me see.

- anonymous

okay!

- anonymous

I'll tell you what I'm doing (because I'm being distracted)...since cos2theta is all the rage in this equation, try to get everything in terms of cos(2theta) and determine a polynomial in cos(2theta),\[P(\cos 2 \theta)=0\]Then solve for the roots of the polynomial.

- anonymous

okay i will do that!
can you tell me how to simplify real quick
(-1/2)(4/5)-square root of 3/2(3/5)

- anonymous

I need you to use the equation editor ::

- anonymous

\[-1/2(4/5)-(\sqrt{3/2})(3/5)\]

- anonymous

\[-\frac{1}{2}\frac{4}{5}-\frac{3}{5}\sqrt{\frac{3}{2}}\]do you mean this?

- anonymous

yess that is right

- anonymous

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- anonymous

Thank you for everything! I appreciate your time!

- anonymous

welcome

- anonymous

did you finish the last one?

- anonymous

nope :/

- anonymous

Is this due online or something?

- anonymous

no i have to turn in next week

- anonymous

Okay. I have to go sort some things out (like my life). I can look at the last problem and post back here. You should get an e-mail from the site saying it's been updated. I'll probably be back during the day to do it...probably :)

- anonymous

Alright sounds good! I'll be sure to check back on here for it! Have a good day kind sir

- anonymous

You too ;)

- anonymous

Here you are...remember to check it.

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- anonymous

So, in summary, you have a quartic polynomial in tan(theta), and so, four possible roots. Two of these roots will be complex conjugates of each other, so are not included.
The roots are:\[\tan \theta = 0\]\[\tan \theta \approx -1.52138\]which means that your angles are (for this interval)\[\theta = 2\pi\](from the first root)\[\theta \approx 5.294^c\](from the second root). The 'c' superscript means, 'radians'.

- anonymous

The plot is for
f(t)=t^3-t+2

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