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anonymous
 5 years ago
Prove the identity.
csc(X)sin(x)/csc(x)+sin(x)=cos^2x/1+sin^2x
anonymous
 5 years ago
Prove the identity. csc(X)sin(x)/csc(x)+sin(x)=cos^2x/1+sin^2x

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Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0sorry I tried, but it doesn't seem straight forward.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its okay, at least you put effort into it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This isn't as hard as it looks :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\csc x =\frac{1}{\sin x}\]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
 5 years ago
Best ResponseYou've already chosen the best response.0Do you need it to be explained more?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes if you don't care to

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is there a specific part?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can refer to the part by the number on the attachment, if that's easier.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes that made it alot easier to see thank you! do u care to help me on another one?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, so you're okay with the last one?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes i knew how to get it started i just didnt know how to simplify that complex fraction, but i understand now.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Good. What's your other question?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0solve this trig equation on the interval 0<x<2pi, express answer for angles as exact value in radians. 2cos^2x+3cosx+1=0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Alright...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
 5 years ago
Best ResponseYou've already chosen the best response.02 or 3 more and we can start a club ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It can be factored into \[(2u+1)(u+1)=0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, but in 30s you'll be on infinity hero, so... :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0lol.... by then ill just need another account :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, by then it's like TRON and you're sucked into the website.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is it (2cos^2+1)(cosx+1) btw no harm here i aprreciate your help!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That whole thing is equal to zero, though.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Which 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
 5 years ago
Best ResponseYou've already chosen the best response.0how do i find the angles? thats where i have a problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yep...I'll do them on paper first.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, 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 lefthand side, and a negative on the righthand side. If you're only using real numbers (which is what's going on here), this can't happen, so there are no xvalues that satisfy this particular equation.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The square of any real number is always 0 or positive.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0For the second factor, you have\[\cos x = 1\]On the interval \[0 \le x \lt 2\pi\]this happens for \[x=\pi\]only.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you plu x=pi into your original equation, you will get 0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks! if i had (sin(x)4)(sin(x)+1) how would i find the angles on the same interval?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is that equal to zero?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay...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 xvalues. 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
 5 years ago
Best ResponseYou've already chosen the best response.0is it pi/2 +2kpi =3pi/2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Be 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
 5 years ago
Best ResponseYou've already chosen the best response.0No, \[\sin \frac{3\pi}{2}=1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You're trying to find the x's that, when plugged into sine, give 1.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh so x=3pi/2 i made a mistake

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you have been a lifesaver, that helped me out alot, especially with the identities.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You're welcome, mj :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So that's it? You're done?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Uhm this one seems simple but i forgot the identity to use 5sin^2x+2cos^2x=53cos^2x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay...this might be more than you have to do, but technically, this is how you would prove it 'properly'.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Assume that the statement is not true. Then,\[5\sin^2 x + 2\cos^2 x \ne 53 \cos^2 x\]But then adding 3cos^2x to both sides gives,\[5\sin^2 x + 5 \cos^2 x \ne 5\]Now, the lefthand 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
 5 years ago
Best ResponseYou've already chosen the best response.0Our assumption that the original statement was false led to a contradiction, so we must take the original statement as true.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The statement is either true or false. If it's not false, it must me true.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you're not happy with that way, I have another.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i dont think thats what im lookin for my teacher makes us make one side the same as the other side :/

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah...I'll do it another way.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Take the righthand side first. You have\[53\cos^2 x\]which equals,\[5(1)3\cos^2 x\]\[=5(\sin^2x + \cos^2x)3\cos^2x\]\[=5\sin^2x + 5\cos^2x3\cos^2 x\]\[=5\sin^2 x+2\cos^2x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ahh thats exactly what i was looking for i was thinking you would use sin^2+cos^2=1 for this

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No more questions then?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0tan(x/2)sinx=1cosx LHS tan(x/2)sinx=1cos/sin(sin/1)=1cosx did i do that right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm not sure what you did. This is what I would do:

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\tan \frac{x}{2}\sin x = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}.2\sin \frac{x}{2}\cos \frac{x}{2}=2\sin^2 \frac{x}{2}=1\cos x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Since by the double angle formula for cosine,\[\cos x = \cos ^2 \frac{x}{2}\sin^2 \frac{x}{2}=12\sin^2 \frac{x}{2} \rightarrow 2\sin^2 \frac{x}{2}=1\cos x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cos(alpha+beta)/sin(alpha)cos(beta)=cot(alpha)tan(beta) cos(alpha)cos(beta)sin(alpha)sin(beta)/.5(sin(alpha+beta)+sin(alphabeta) how would i finish that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{\cos(\alpha + \beta)}{\sin \alpha \cos \beta}=\frac{\cos \alpha \cos \beta  \sin \alpha \sin \beta}{\sin \alpha \cos \beta}\]\[=\frac{\cos \alpha \cos \beta}{\sin \alpha \cos \beta }\frac{\sin \alpha \sin \beta }{\sin \alpha \cos \beta}\]\[=\frac{\cos \alpha}{\sin \alpha}\frac{\sin \beta }{\cos \beta}\]\[=\cot \alpha  \tan \beta\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That was a much easier way of doing it, identities must take alot of practice to get use to.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah. It's mostly intuition as to how to start and how to close. That comes from practice!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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<x<2pi ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Um, it's hard for me to tell since I'm not sure of your order of operations.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What's under the square root?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{\sqrt{1+4/5}}{2}\]?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i used the half angle formula for cos YES that is correct

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, and it's positive root if you're in that interval.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes 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
 5 years ago
Best ResponseYou've already chosen the best response.0Do you know how to use the equation editor? The button below will take you there.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh i didnt know that thank you!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\tan2(\theta)=1\cos(2\theta)\div2\div1+\cos(2\theta)\div2 \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol...it's not clear...is it...\[\tan 2 \theta =\frac{\frac{1\cos 2 \theta}{2}}{\frac{1+ \cos 2 \theta}{2}}\]?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes thats what i was trying to do i got 3/13 for the answer but im not sure about it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Are you trying to find tan(2 theta) and the righthand side is what you came up with?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes im trin to find tan(2theta) on the interval 3pi/2<x<2pi i used the double angle formula for tan

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0For 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
 5 years ago
Best ResponseYou've already chosen the best response.0Because tan(2theta) takes an infinity of values on that interval...unless tan(2theta)=(something) to begin with.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes thats what i have it equal to

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'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
 5 years ago
Best ResponseYou've already chosen the best response.0okay 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
 5 years ago
Best ResponseYou've already chosen the best response.0I need you to use the equation editor ::

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[1/2(4/5)(\sqrt{3/2})(3/5)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{2}\frac{4}{5}\frac{3}{5}\sqrt{\frac{3}{2}}\]do you mean this?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thank you for everything! I appreciate your time!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0did you finish the last one?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is this due online or something?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no i have to turn in next week

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay. 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 email from the site saying it's been updated. I'll probably be back during the day to do it...probably :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Alright sounds good! I'll be sure to check back on here for it! Have a good day kind sir

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Here you are...remember to check it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So, 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
 5 years ago
Best ResponseYou've already chosen the best response.0The plot is for f(t)=t^3t+2
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