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Math Help - Trigonometric identities.

  1. #1
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    [SOLVED]Trigonometric identities.

    Could any kind souls help me with these questions? I am stuck!

    1)Prove the identity: cos(a+b) cos(a-b) = cos^2a -sin^2b
    for this, i am stuck at LHS = cos^2a cos^2b - sin^2a sin^2b .

    2)Prove the identity tanA tan(A-60) + tanA tan(A+60) + tan(A-60) tan(A+60) = -3 ,
    For this,im stuck with LHS = -3tan^2A . Please correct me if this is already wrong cause i assumed A to be 180 degrees.

    3)If A and B are the principal values of arccos(3/sqrt10) and arcsin(1/sqrt5) respectively,show that A+B = pi/4 (I have no idea how to start this,i've never seen this question before!)

    4) Prove that if A+B+C = 180 degrees(sum of angles in a triangle),
    cotAcotB + cotBcotC + cotCcotA =1 holds true.

    (for this, i've tried to use 1/tanAtanB... and also converting it to cosA/sinA... but both of these methods got me into infinite loops and some nice symmetry too. Haha!)

    All help is greatly appreciated!
    Last edited by arccos; November 6th 2010 at 07:08 AM.
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  2. #2
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    Quote Originally Posted by arccos View Post
    Could any kind souls help me with these questions? I am stuck!

    [snip]
    3)If A and B are the principal values of arccos(3/sqrt10) and arcsin(1/sqrt5) respectively,show that A+B = pi/4 (I have no idea how to start this,i've never seen this question before!)

    [snip]
    Consider \cos (A + B) = \cos (A) \cos(B) - \sin(A) \sin(B) where \cos(A) = \frac{3}{\sqrt{10}} \Rightarrow \sin(A) = .... and \sin(B) = \frac{1}{\sqrt{5}} \Rightarrow \cos(B) = .....

    Once you substitute these values you should get \cos(A + B) = \frac{1}{\sqrt{2}} \Rightarrow A + B = \frac{\pi}{4}.
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  3. #3
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    Quote Originally Posted by arccos View Post
    1)Prove the identity: cos(a+b) cos(a-b) = cos^2a -sin^2b
    for this, i am stuck at LHS = cos^2a cos^2b - sin^2a sin^2b .
    Good start. Now change \cos^2{b} with [LaTeX ERROR: Convert failed] .
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  4. #4
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    Thanks for the inputs! I have solved qn 1 and 3 with your help. Now i'm left with 2 & 4.
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  5. #5
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by arccos View Post
    Thanks for the inputs! I have solved qn 1 and 3 with your help. Now i'm left with 2 & 4.
    Q2:

    Use the following identitites:

     \tan(a+b) = \dfrac{\tan(a)+ \tan(b)}{1- \tan(a) \; \tan(b)}

    \tan(a-b) = \dfrac{\tan(a)-\tan(b)}{1+\tan(a) \; \tan(b)}

    tan(60) = tan(\frac{\pi}{3})=\sqrt{3}


    and then keep simplifying to get -3.
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  6. #6
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    Quote Originally Posted by harish21 View Post
    Q2:

    Use the following identitites:

     \tan(a+b) = \dfrac{\tan(a)+ \tan(b)}{1- \tan(a) \; \tan(b)}

    \tan(a-b) = \dfrac{\tan(a)-\tan(b)}{1+\tan(a) \; \tan(b)}

    tan(60) = tan(\frac{\pi}{3})=\sqrt{3}


    and then keep simplifying to get -3.

    How can i use the addition formulaes for tangent in the question? i cant see any relation. would you please enlighten me?
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  7. #7
    MHF Contributor harish21's Avatar
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    tanA tan(A-60) + tanA tan(A+60) + tan(A-60) tan(A+60)

    =\bigg(\tan(A) \times \dfrac{\tan(A)-\tan(60)}{1+\tan(A)\;\tan(60)}\bigg) \;+ \; \bigg(\tan(A) \times \dfrac{\tan(A)+\tan(60)}{1-\tan(A)\;\tan(60)}\bigg)+ \bigg(\dfrac{\tan(A)-\tan(60)}{1+\tan(A)\;\tan(60)} \times \dfrac{\tan(A)+\tan(60)}{1-\tan(A)\;\tan(60)}\bigg)

    =\bigg(\tan(A) \times \dfrac{\tan(A)-\sqrt{3}}{1+\sqrt{3}\tan(A)}\bigg) \;+ \; \bigg(\tan(A) \times \dfrac{\tan(A)+\sqrt{3}}{1-\sqrt{3}\tan(A)}\bigg)+ \bigg(\dfrac{\tan(A)-\sqrt{3}}{1+\sqrt{3}\tan(A)} \times \dfrac{\tan(A)+ \sqrt{3}}{1-\sqrt{3}\tan(A)}\bigg)

    =\bigg(\dfrac{\tan^{2}(A)-\sqrt{3} \tan(A)}{1+\sqrt{3}\tan(A)}\bigg) + \bigg( \dfrac{\tan^{2}(A)+\sqrt{3}\tan(A)}{1-\sqrt{3}\tan(A)}\bigg)+\bigg(\dfrac{\tan^{2}(A)-3}{(1+\sqrt{3}\tan(A))(1-\sqrt{3}\tan(A))}\bigg)

    now finish it..
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  8. #8
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    Quote Originally Posted by harish21 View Post
    tanA tan(A-60) + tanA tan(A+60) + tan(A-60) tan(A+60)

    =\bigg(\tan(A) \times \dfrac{\tan(A)-\tan(60)}{1+\tan(A)\;\tan(60)}\bigg) \;+ \; \bigg(\tan(A) \times \dfrac{\tan(A)+\tan(60)}{1-\tan(A)\;\tan(60)}\bigg)+ \bigg(\dfrac{\tan(A)-\tan(60)}{1+\tan(A)\;\tan(60)} \times \dfrac{\tan(A)+\tan(60)}{1-\tan(A)\;\tan(60)}\bigg)

    =\bigg(\tan(A) \times \dfrac{\tan(A)-\sqrt{3}}{1+\sqrt{3}\tan(A)}\bigg) \;+ \; \bigg(\tan(A) \times \dfrac{\tan(A)+\sqrt{3}}{1-\sqrt{3}\tan(A)}\bigg)+ \bigg(\dfrac{\tan(A)-\sqrt{3}}{1+\sqrt{3}\tan(A)} \times \dfrac{\tan(A)+ \sqrt{3}}{1-\sqrt{3}\tan(A)}\bigg)

    =\bigg(\dfrac{\tan^{2}(A)-\sqrt{3} \tan(A)}{1+\sqrt{3}\tan(A)}\bigg) + \bigg( \dfrac{\tan^{2}(A)+\sqrt{3}\tan(A)}{1-\sqrt{3}\tan(A)}\bigg)+\bigg(\dfrac{\tan^{2}(A)-3}{(1+\sqrt{3}\tan(A))(1-\sqrt{3}\tan(A))}\bigg)

    now finish it..

    I simplified it to 8tan^2A - 3 / 1 -3tan^2A and this was as much as I could go. I got to here by multiplying the first 2 brackets of fractions together and i found that it had the same denominator as the last bracket.Expanded(really long) and simplified to that. Am i doing it wrong?
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  9. #9
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by arccos View Post
    I simplified it to 8tan^2A - 3 / 1 -3tan^2A and this was as much as I could go. I got to here by multiplying the first 2 brackets of fractions together and i found that it had the same denominator as the last bracket.Expanded(really long) and simplified to that. Am i doing it wrong?
    Check your calculation again..you should be getting:

    \dfrac{9\tan^{2}(A)-3}{1-3\tan^{2}(A)}
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  10. #10
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    Quote Originally Posted by harish21 View Post
    Check your calculation again..you should be getting:

    \dfrac{9\tan^{2}(A)-3}{1-3\tan^{2}(A)}

    Testing latex: (tan\^{}2A-sqrt3tanA)(1-sqrt3tanA)+ (tan\^{}2A + sqrt3tanA)(1+sqrt3tanA)<br />
=tan\^{}2A -sqrt3tan\^{}3A -sqrt3tanA +3tan\^{}2A + tan\^{}2A +sqrt3tan\^{}3A + sqrt3tanA +3tan\^{}2A

    Normal text:
    (tan^2A-sqrt3tanA)(1-sqrt3tanA)+ (tan^2A + sqrt3tanA)(1+sqrt3tanA)
    =tan^2A -sqrt3tan^3A -sqrt3tanA +3tan^2A + tan^2A +sqrt3tan^3A + sqrt3tanA +3tan^2A (just the numerator)

    Its late here. I still see 8 only. :O
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  11. #11
    MHF Contributor Unknown008's Avatar
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    For the last one:

    A + B + C = 180

    A + B = 180 - C

    \tan(A+B) = \tan(180 - C)

    \dfrac{\tan A + \tan B}{1 - \tan A\tan B} = \dfrac{\tan 180 - \tan C}{1 + \tan 180 \tan C}

    but tan 180 = 0;

    \dfrac{\tan A + \tan B}{1 - \tan A\tan B} =  - \tan C

    \tan A + \tan B =  - \tan C + \tan A\tan B\tan C

    \tan A + \tan B + \tan C = \tan A\tan B\tan C

    \dfrac{1}{\tan A + \tan B + \tan C} = \dfrac{1}{\tan A\tan B\tan C}

    \dfrac{1}{\tan A + \tan B + \tan C} = \cot A\cot B\cot C}

    now cross multiply, keeping in mind that:

    \cot A = \dfrac{1}{\tan A}
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  12. #12
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    Unknown008,thanks so much! I totally overlooked the fact that tan180(i wrote it as tan(pi) radians) = 0. Looks like I haven't got the special values at my fingertips yet.
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  13. #13
    MHF Contributor harish21's Avatar
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    You should have:

    \dfrac{(\tan^{2}(A)-\sqrt{3}\tan(A))(1-\sqrt{3}\tan(A))+(\tan^{2}(A)+\sqrt{3}\tan(A))(1+\  sqrt{3}\tan(A))+\tan^{2}(A)-3}{(1-\sqrt{3}\tan(A))(1-\sqrt{3}\tan(A))}

    the numerator would be:

    \tan^{2}(A)-\sqrt{3}\tan^{3}(A)-\sqrt{3}\tan(A)+3\tan^{2}(A)+\tan^{2}(A)+\sqrt{3}\  tan^{3}(A)+\sqrt{3}\tan(A)+3tan^{2}(A)+tan^{2}(A)-3

    =\tan^{2}(A)+3\tan^{2}(A)+\tan^{2}(A)+3\tan^{2}(A)  +\tan^{2}(A)-3

    =9tan^{2}(A)-3
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  14. #14
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    Quote Originally Posted by arccos View Post
    Unknown008,thanks so much! I totally overlooked the fact that tan180(i wrote it as tan(pi) radians) = 0. Looks like I haven't got the special values at my fingertips yet.
    Well, I had a similar proof some time ago which stopped at the tanAtanBtanC = tanA + tanB + tanC, so, it was easier to get this right

    And it's always good to check the special values now and again.
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  15. #15
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    Quote Originally Posted by harish21 View Post
    You should have:

    \dfrac{(\tan^{2}(A)-\sqrt{3}\tan(A))(1-\sqrt{3}\tan(A))+(\tan^{2}(A)+\sqrt{3}\tan(A))(1+\  sqrt{3}\tan(A))+\tan^{2}(A)-3}{(1-\sqrt{3}\tan(A))(1-\sqrt{3}\tan(A))}

    the numerator would be:

    \tan^{2}(A)-\sqrt{3}\tan^{3}(A)-\sqrt{3}\tan(A)+3\tan^{2}(A)+\tan^{2}(A)+\sqrt{3}\  tan^{3}(A)+\sqrt{3}\tan(A)+3tan^{2}(A)+tan^{2}(A)-3

    =\tan^{2}(A)+3\tan^{2}(A)+\tan^{2}(A)+3\tan^{2}(A)  +\tan^{2}(A)-3

    =9tan^{2}(A)-3
    Opps! I forgot about the last bracket.
    Can you hint me on what to do with the simplified fraction?

    EDIT: Oh i got it! I factored out the 3 for the numerator and multipled the fraction by -1 and the 3tan^2A -1 cancels out. Thanks so much!!!
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