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Thread: Prove each identity (Trig functions)

  1. #1
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    Default Prove each identity (Trig functions)

    Below is an assignment that I have to do, but I have no idea where to start or what the hell he is asking me to do...any help?


    Prove each identity.

    1. 1+ tan y
    1+ cot y
    = sec y
    csc y

    2. tan x − tan3x
    1+ tan3x tan x
    = 2tan x
    tan2 x −1

    3. cosr + sinr
    cosr
    = 1+ tanr

    4. tan p + cot p
    sec p + csc p
    = 1
    cos p + sin p

    5. sin 4z + sin 6z
    cos 4z − cos6z
    = cot z
    Finished B.S. Program in Radiology!!

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    Make use of the different trigo identities to manipulate the equation. Eg. double angle formula, half angle formula etc.

    I don't understand the equation, is it:

    1+ tan y + 1+ cot y = sec y + csc y ?

    There's no '+' for the next line? So what's the relationship between the first line and the 2nd line?

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    Use trigonometric identities.

    http://en.wikipedia.org/wiki/List_of...ric_identities

    The way you wrote it is also mathematically incorrect. It looks like you're missing a bracket or symbol between line 1 and 2 in each example so I can't help you solve them.

    EDIT: Great..Riwu guy beats me to it. Haha. >_>

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    Corrected***

    1. 1+ tan y/1+ cot y = sec y/ csc y

    2. tan x − tan3x /1+ tan3x tan x = 2tan x /tan2 x −1

    3. cosr + sinr/ cosr = 1+ tanr

    4. tan p + cot p /sec p + csc p = 1/ cos p + sin p

    5. sin 4z + sin 6z/ cos 4z − cos6z = cot z
    Finished B.S. Program in Radiology!!

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    Yea if it looked proper and there were no mistakes or controversies, I would look into it for you and do it on paper first. I like this stuff and just recently did it. You use the list of trig identities to expand and simplify each side of the equals sign (Left side/Right side) to prove that they are indeed equal to one another. You are given the answer, you just have to prove it.
    If you can give me a better format I will help you out a little more as I will have my math book with me and will go over my notes to remember what to do.

    Edit: Thanks for posting in better format Dont worry bout the comment above now really. I will help you out tomorrow if you can wait that long cuz its late and tomorrow I will bring home my math to look over.
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    Please with brackets?

    1. 1+ tan y/1+ cot y = sec y/ csc y

    could be understood as 1. 1+ tan (y/1)+ cot y = sec y/ csc y

    or 1. 1+ (tan y)/1+ cot y = sec y/ csc y

    >>> both of these make no sense since dividing by 1 won't change anything so I'm still wondering if you're actually typing it all out correctly?

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    1.
    1+ tan y/1+ cot y................................................. .......= sec y/ csc y
    1+ (siny/cosy) / 1+ cosy/siny.........................................= 1/cosy / 1/siny
    cosy+siny/cosy / siny+cosy/siny......................................= 1/cosy multiplied by siny/1
    cosy+siny/cosy multiplied by siny/siny+cosy........................=*One's cancel out left with:* siny/cosy
    *cosy+siny cancel out left with:* siny/cosy

    LS=RS
    QED

    Note the invert and multiply step and you can reduce further into tany on both sides*
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    Quote Originally Posted by kevin33 View Post
    1.
    1+ tan y/1+ cot y................................................. .......= sec y/ csc y
    1+ (siny/cosy) / 1+ cosy/siny.........................................= 1/cosy / 1/siny
    cosy+siny/cosy / siny+cosy/siny......................................= 1/cosy multiplied by siny/1
    cosy+siny/cosy multiplied by siny/siny+cosy........................=*One's cancel out left with:* siny/cosy
    *cosy+siny cancel out left with:* siny/cosy

    LS=RS
    QED

    Note the invert and multiply step and you can reduce further into tany on both sides*
    wrong.

    EDIT: The question included 1+ tan y/1+ cot y and you assumed it was (1+ tan y)/(1+ cot y). Then instead of actually stating that you continued to answer without brackets! Whoever posted this question in the first place please include brackets - these are not proper mathematical standards and you don't answer questions by making assumptions.

    Also, when trying to prove an equation to be something else you don't change what you're trying to prove. It defeats the whole purpose. (In this case RHS was changed - why?)
    Last edited by Runehack123; 11-29-2012 at 05:55 AM.

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    Quote Originally Posted by Runehack123 View Post
    wrong.
    How is it wrong?

    OT: Let me know if thats what your looking for. That is an example for you. I will help you with the rest tomorrow after my school.
    Also sorry for it being odd looking. Did it on paper and I cant do everything on the computer I can freely do on paper like cross stuff out. Refer to words within for explanations on each step that seems odd.
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    Read my edit above please.
    There were no brackets in the question and you can only divide tany by 1 since the rest of the equation is not included.
    1+ tan y/1+ cot y <<< Either that is what he wanted answered or

    1+ tan y
    ________
    1+coty

    In which case the two examples I just gave are not one and the same! You answered the question using the latter, but the RHS was changed and as with any proof question you don't change what you're trying to prove. It defeats the whole purpose.

    The approach to the question should be x = ... = .... = ... = ... = ... = y Therefore, x=y
    Not
    x = y
    ...=...
    ...=...
    ...=...

    Aside from missing brackets and a wrong approach to the question I can't follow how you deduced

    1+ (siny/cosy) / 1+ cosy/siny.........................................= 1/cosy / 1/siny
    cosy+siny/cosy / siny+cosy/siny......................................= 1/cosy multiplied by siny/1

    on the LHS of the equation.

    cosY = 1 and sinY = 1 No. That's wrong I'm afraid.

    I will now answer this question correctly!

    Now, let's assume
    (The way it was stated this is not the case, because brackets were missing! You HAVE to include brackets otherwise this question is NOT solvable, as 1+ tan y/1+ cot y is NOT equal to secY/cscY , which defeats the whole purpose of proving something if there is no logic to it in the first place...)
    prove that....
    1+ tan y
    ________ = (secy)/(cscy)
    1+coty

    was the question.
    RHS needs to remain unchanged as already explained above so we will start off with

    (1+tanY)/(1+cotY)

    = (1+tanY) / ((tanY+1) / (tanY))

    = (1+tanY)((tanY) / (tanY+1))

    = tanY

    = (sinY) / (cosY)

    = (1/(cosY)) / (1/sinY)

    = (secY)/(cosecY)

    Therefore, LHS = RHS.
    Last edited by Runehack123; 11-29-2012 at 06:52 AM.

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    I learned never to use tan always switch to sin and cos unless the answer (what your solving for) includes a tan that needs to be solved. Also, you are able to work with both sides, that is how I have learned it the past 2 years. Most universities and things however, prefer you work only on one side. That's how things go down where I'm from anyways so it is rude to question our education system and state I am wrong when truly I am right.
    I also did not write it down wrong as you see by my answer, I did it the same way you did there. Its hard to show it completely on a computer and because no brackets were used(even though they arent needed).

    side from missing brackets and a wrong approach to the question I can't follow how you deduced

    1+ (siny/cosy) / 1+ cosy/siny.........................................= 1/cosy / 1/siny
    cosy+siny/cosy / siny+cosy/siny......................................= 1/cosy multiplied by siny/1

    on the LHS of the equation.

    cosY = 1 and sinY = 1 No. That's wrong I'm afraid.

    1+ siny/cosy
    _______________

    1+ cosy/siny

    You cant have a fraction in a fraction so you simply find a common denominator for the top then the bottom (top is cosy, bottom is siny) then change it to 1 term (change 1). With 1 term on top and 1 term on the bottom, you then, instead of divide, invert and multiply. The numerator and denominators are the same (cosysiny) so they cancel out leaving you with the answer which matches the RHS.
    To sum it all up, cosy = 1 and siny = 1 is clearly not what I did so you have interpreted it wrong. cosy/cosy and siny/siny is what I have and each equal the previous 1, they are just combined with the 2nd term
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    Quote Originally Posted by kevin33 View Post
    You cant have a fraction in a fraction so you simply find a common denominator for the top then the bottom (top is cosy, bottom is siny) then change it to 1 term (change 1). With 1 term on top and 1 term on the bottom, you then, instead of divide, invert and multiply. The numerator and denominators are the same (cosysiny) so they cancel out leaving you with the answer which matches the RHS.
    To sum it all up, cosy = 1 and siny = 1 is clearly not what I did so you have interpreted it wrong. cosy/cosy and siny/siny is what I have and each equal the previous 1, they are just combined with the 2nd term
    The way you approached the question was very confusing for me since you changed the right-hand side, but if that's how you do it in your country then I take back whatever I said against that.

    Ok so I see what you did there

    sinY(cosY+sinY)
    _______________
    cosY(sinY+cosY) etc.

    Nevertheless, you still need brackets!
    Take this as an example...

    1 + 2 x 3 / 4 (=2.5) is not the same as (1+2x3)/4 (=1.75)

    Type it in your calculator if you don't believe me.

    Based on that,
    1+ (siny/cosy) / 1+ cosy/siny
    is not the same as
    (1+(siny)/(cosy))/(1+(cosy)/(siny))

    And yes, you can have a fraction within a fraction?
    Here's what I found on 'complex fractions':

    HTML Code:
    http://www.sosmath.com/algebra/fraction/frac4/frac4.html
    HTML Code:
    http://www.purplemath.com/modules/compfrac.htm
    Thanks though for sharing - didn't mean to offend you.
    We're probably the only 2 people following this thread haha...

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    Simplify each to sin and cos then rationalize
    Last edited by tehq; 11-29-2012 at 03:49 PM.

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    Quote Originally Posted by tehq View Post
    You can't just cross out the two sinycosy in

    sinycosy + (siny)^2
    _________________
    sinycosy + (cosy)^2

    Because they are included in a bracket:

    (sinycosy + (siny)^2)
    ____________________
    (sinycosy + (cosy)^2)

    And you can only cancel common factors found in the numerator and denominator.

    In this case a common factor can be found by factorisation:

    sinycosy + (siny)^2
    _________________
    sinycosy + (cosy)^2

    =

    siny(cosy+siny)
    ______________
    cosy(cosy+siny)

    Then you can cancel out (cosy+siny) since it is a common factor.

    I'm sorry to say that this part of your thought process is incorrect.
    Allow me to prove this to you by replacing the cosy, siny values with integers.

    2+ 5
    _____ is not equal to 5/10 since its 7/12.
    2+ 10

    5/10 is achieved by cancelling out the 2's, which is not possible since there is an 'invisible' bracket:

    (2+5)
    _____
    (2+10)

    This can further be proven by using a value for y. Let's say 60 degrees!
    With my calculator in degrees mode I get:

    sin60cos60 + (sin60)^2
    ______________________ = 1.732... Rounded to 4 s.f.
    sin60cos60 + (cos60)^2

    Therefore, based on your calculation,

    (sin60)^2
    __________ = 3
    (cos60)^2

    should get me exactly the same answer. However, this is in fact equal to 3!

    From (siny)^2/(cosy)^2 you then deduced siny/cosy, which is again *sorry* incorrect since
    (siny)^2/(cosy)^2

    = (sinysiny)/(cosycosy)

    And as you can see from that there is no common factor.

    Using an integer example e.g. 2^2/3^2 you can quickly see that it does not equal to 2/3.

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