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

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

Mathematics
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sorry I tried, but it doesn't seem straight forward.
its okay, at least you put effort into it
This isn't as hard as it looks :)

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\[\csc x =\frac{1}{\sin x}\]by definition, so\[\frac{\csc x - \sin x}{\csc x + \sin x}=\frac{\frac{1}{\sin x}-\sin x}{\frac{1}{\sin x}+\sin x}=\frac{\frac{1-\sin^2x}{\sin x}}{\frac{1+\sin^2 x}{\sin x}}=\frac{1-\sin^2 x}{\sin x}\frac{\sin x}{1+\sin^2x}\]\[=\frac{1-\sin^2 x}{1+\sin^2x}=\frac{\cos^2x}{1+\sin^2x}\]
Do you need it to be explained more?
yes if you don't care to
Is there a specific part?
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You can refer to the part by the number on the attachment, if that's easier.
brb
yes that made it alot easier to see thank you! do u care to help me on another one?
Oh, so you're okay with the last one?
Yes i knew how to get it started i just didnt know how to simplify that complex fraction, but i understand now.
Good. What's your other question?
solve this trig equation on the interval 0
Alright...this is a quadratic equation in cos(x). If you 'see' cos(x) as something like, 'u', then you have the equation\[2u^2+3u+1=0\]
2 or 3 more and we can start a club ;)
It can be factored into \[(2u+1)(u+1)=0\]
Yeah, but in 30s you'll be on infinity hero, so... :)
lol.... by then ill just need another account :)
No, by then it's like TRON and you're sucked into the website.
Sorry mj
is it (2cos^2+1)(cosx+1) btw no harm here i aprreciate your help!
Yes it is.
That whole thing is equal to zero, though.
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).
how do i find the angles? thats where i have a problem
Yep...I'll do them on paper first.
Okay thanks!
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.
The square of any real number is always 0 or positive.
For the second factor, you have\[\cos x = -1\]On the interval \[0 \le x \lt 2\pi\]this happens for \[x=\pi\]only.
If you plu x=pi into your original equation, you will get 0.
*plug
Thanks! if i had (sin(x)-4)(sin(x)+1) how would i find the angles on the same interval?
Is that equal to zero?
yes sorry i forgot
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?
*'Sot' = 'For'.
is it -pi/2 +2kpi =3pi/2
Yes it is.
3pi/2
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.).
so sinx=3pi/2
No, \[\sin \frac{3\pi}{2}=-1\]
You're trying to find the x's that, when plugged into sine, give -1.
oh so x=3pi/2 i made a mistake
Yes.
you have been a lifesaver, that helped me out alot, especially with the identities.
You're welcome, mj :)
So that's it? You're done?
Uhm this one seems simple but i forgot the identity to use 5sin^2x+2cos^2x=5-3cos^2x
Okay...this might be more than you have to do, but technically, this is how you would prove it 'properly'.
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.
Our assumption that the original statement was false led to a contradiction, so we must take the original statement as true.
The statement is either true or false. If it's not false, it must me true.
If you're not happy with that way, I have another.
i dont think thats what im lookin for my teacher makes us make one side the same as the other side :/
Yeah...I'll do it another way.
Take the right-hand side first. You have\[5-3\cos^2 x\]which equals,\[5(1)-3\cos^2 x\]\[=5(\sin^2x + \cos^2x)-3\cos^2x\]\[=5\sin^2x + 5\cos^2x-3\cos^2 x\]\[=5\sin^2 x+2\cos^2x\]
ahh thats exactly what i was looking for i was thinking you would use sin^2+cos^2=1 for this
No more questions then?
tan(x/2)sinx=1-cosx LHS tan(x/2)sinx=1-cos/sin(sin/1)=1-cosx did i do that right?
I'm not sure what you did. This is what I would do:
\[\tan \frac{x}{2}\sin x = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}.2\sin \frac{x}{2}\cos \frac{x}{2}=2\sin^2 \frac{x}{2}=1-\cos x\]
Since by the double angle formula for cosine,\[\cos x = \cos ^2 \frac{x}{2}-\sin^2 \frac{x}{2}=1-2\sin^2 \frac{x}{2} \rightarrow 2\sin^2 \frac{x}{2}=1-\cos x\]
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
\[\frac{\cos(\alpha + \beta)}{\sin \alpha \cos \beta}=\frac{\cos \alpha \cos \beta - \sin \alpha \sin \beta}{\sin \alpha \cos \beta}\]\[=\frac{\cos \alpha \cos \beta}{\sin \alpha \cos \beta }-\frac{\sin \alpha \sin \beta }{\sin \alpha \cos \beta}\]\[=\frac{\cos \alpha}{\sin \alpha}-\frac{\sin \beta }{\cos \beta}\]\[=\cot \alpha - \tan \beta\]
That was a much easier way of doing it, identities must take alot of practice to get use to.
Yeah. It's mostly intuition as to how to start and how to close. That comes from practice!
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
Um, it's hard for me to tell since I'm not sure of your order of operations.
What's under the square root?
\[\frac{\sqrt{1+4/5}}{2}\]?
i used the half angle formula for cos YES that is correct
Yes, and it's positive root if you're in that interval.
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
Do you know how to use the equation editor? The button below will take you there.
oh i didnt know that thank you!
\[\tan2(\theta)=1-\cos(2\theta)\div2\div1+\cos(2\theta)\div2 \]
lol...it's not clear...is it...\[\tan 2 \theta =\frac{\frac{1-\cos 2 \theta}{2}}{\frac{1+ \cos 2 \theta}{2}}\]?
yes thats what i was trying to do i got -3/13 for the answer but im not sure about it
Are you trying to find tan(2 theta) and the right-hand side is what you came up with?
yes im trin to find tan(2theta) on the interval 3pi/2
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?
Because tan(2theta) takes an infinity of values on that interval...unless tan(2theta)=(something) to begin with.
yes thats what i have it equal to
ok...let me see.
okay!
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.
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)
I need you to use the equation editor ::
\[-1/2(4/5)-(\sqrt{3/2})(3/5)\]
\[-\frac{1}{2}\frac{4}{5}-\frac{3}{5}\sqrt{\frac{3}{2}}\]do you mean this?
yess that is right
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Thank you for everything! I appreciate your time!
welcome
did you finish the last one?
nope :/
Is this due online or something?
no i have to turn in next week
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 :)
Alright sounds good! I'll be sure to check back on here for it! Have a good day kind sir
You too ;)
Here you are...remember to check it.
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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'.
The plot is for f(t)=t^3-t+2

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