Question

Prove the identity.

csc(X)-sin(x)/csc(x)+sin(x)=cos^2x/1+sin^2x

Answer 1

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

Answer 2

its okay, at least you put effort into it

Answer 3

This isn't as hard as it looks :)

Answer 4

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

Answer 5

Do you need it to be explained more?

Answer 6

yes if you don't care to

Answer 7

Is there a specific part?

Answer 8

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

Answer 9

brb

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

Good. What's your other question?

Answer 14

solve this trig equation on the interval 0<x<2pi, express answer for angles as exact value in radians.

2cos^2x+3cosx+1=0

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

Sorry mj

Answer 22

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

Answer 23

Yes it is.

Answer 24

That whole thing is equal to zero, though.

Answer 25

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).

Answer 26

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

Answer 27

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

Answer 28

Okay thanks!

Answer 29

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.

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

*plug

Answer 34

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

Answer 35

Is that equal to zero?

Answer 36

yes sorry i forgot

Answer 37

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?

Answer 38

*'Sot' = 'For'.

Answer 39

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

Answer 40

Yes it is.

Answer 41

3pi/2

Answer 42

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.).

Answer 43

so sinx=3pi/2

Answer 44

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

Answer 45

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

Answer 46

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

Answer 47

Yes.

Answer 48

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

Answer 49

You're welcome, mj :)

Answer 50

So that's it? You're done?

Answer 51

Uhm this one seems simple but i forgot the identity to use

5sin^2x+2cos^2x=5-3cos^2x

Answer 52

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

Answer 53

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.

Answer 54

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

Answer 55

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

Answer 56

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

Answer 57

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

Answer 58

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

Answer 59

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

Answer 60

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

Answer 61

No more questions then?

Answer 62

tan(x/2)sinx=1-cosx
LHS
tan(x/2)sinx=1-cos/sin(sin/1)=1-cosx

did i do that right?

Answer 63

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

Answer 64

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

Answer 65

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

Answer 66

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

Answer 67

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

Answer 68

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

Answer 69

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

Answer 70

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<x<2pi ?

Answer 71

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

Answer 72

What's under the square root?

Answer 73

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

Answer 74

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

Answer 75

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

Answer 76

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

Answer 77

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

Answer 78

oh i didnt know that thank you!

Answer 79

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

Answer 80

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

Answer 81

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

Answer 82

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

Answer 83

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

i used the double angle formula for tan

Answer 84

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?

Answer 85

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

Answer 86

yes thats what i have it equal to

Answer 87

ok...let me see.

Answer 88

okay!

Answer 89

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.

Answer 90

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)

Answer 91

I need you to use the equation editor ::

Answer 92

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

Answer 93

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

Answer 94

yess that is right

Answer 95

Thank you for everything! I appreciate your time!