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Math Help - Calculate tan^2(36) tan^2(72)

  1. #1
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    Calculate tan^2(36) tan^2(72)

    tan^{2}36\cdot tan^{2}72=
    Last edited by mr fantastic; September 24th 2011 at 03:52 AM. Reason: Re-titled.
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  2. #2
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    Re: Calculate tan^2(36) tan^2(72)

    The Problem is very easy.

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  3. #3
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    Re: Calculate tan^2(36) tan^2(72)

    no tan^2(18)tan^2(72) I know this is easy

    but
    tan^2(36)tan^2(72)
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  4. #4
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    Re: Calculate tan^2(36) tan^2(72)

    Okay sorry I read \tan^2(18)

    \tan^2{(36)}=5-2 \sqrt{5}

    also, \tan^2{(72)}=5+2 \sqrt{5}

    Therefore \tan^2{(72)}.\tan^2{(36)}=25-20=5
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  5. #5
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    Re: Calculate tan^2(36) tan^2(72)

    hahah very quickly I dont understad where you get worth of tan^2(36) tan^(72) can you explain
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  6. #6
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    Re: Calculate tan^2(36) tan^2(72)

    He multiplied! (tan^2(36))(tan^2(72))= (5- 2\sqrt{5})(5+ 2\sqrt{5})= 25- 4(5)= 5.

    Now, if you are asking how sbhatnagar determined that tan^2(36)= 5- 2\sqrt{5}, that's a really good question! Come on, sbhatnagar, let us in on the secret.
    Last edited by HallsofIvy; September 24th 2011 at 11:42 AM.
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    Re: Calculate tan^2(36) tan^2(72)

    yes
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  8. #8
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    Re: Calculate tan^2(36) tan^2(72)

    triple angle formula can be used to find the exact value of tan 36 and tan 72.
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  9. #9
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    Re: Calculate tan^2(36) tan^2(72)

    Quote Originally Posted by HallsofIvy View Post
    He multiplied! (tan^2(36))(tan^2(72))= (5- 2\sqrt{5})(5+ 2\sqrt{5})= 25- 4(5)= 5.

    Now, if you are asking how sbhatnagar determined that tan^2(36)= 5- 2\sqrt{5}, that's a really good question! Come on, sbhatnagar, let us in on the secret.
    To calculate tan(36 degrees), I'd consider a regular pentagon ....
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    Exclamation Re: Calculate tan^2(36) tan^2(72)

    To find sin(72):

    Let x = 18

    5x=90

    \Rightarrow 3x+2x=90

    \Rightarrow 2x=90-3x

    \Rightarrow \sin{2x}=\sin{(90-3x)}

    \Rightarrow 2\sin{x}\cos{x}=\cos{3x}

    \Rightarrow 2\sin{x}\cos{x}=4\cos^3{x}-3\cos{x}

    \Rightarrow 4\sin^2{x}+2\sin{x}-1=0

    So now using quadratic formula, \Rightarrow \sin{x}=\frac{\sqrt{5}-1}{4}

    Now use sin^(2)x+cos^(2)x=1 to find cos(x)

    \cos{x}=\frac{\sqrt{10+2\sqrt{5}}}{4}

    Now you have, \cos{18}=\frac{\sqrt{10+2\sqrt{5}}}{4}

    Therefore \sin{(90-18)}=\frac{\sqrt{10+2\sqrt{5}}}{4}

    \Rightarrow \sin{72}=\frac{\sqrt{10+2\sqrt{5}}}{4}
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  11. #11
    Member sbhatnagar's Avatar
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    Re: Calculate tan^2(36) tan^2(72)

    If \sin{72}=\frac{\sqrt{10+2\sqrt{5}}}{4} then \cos{72}=\frac{\sqrt{5}-1}{4} and \tan{72}=\frac{\sqrt{10+2\sqrt{5}}}{\sqrt{5}-1}}

    Therefore \tan^2{72}=5+2\sqrt{5}

    The value of sin(36 degrees) can be found by using the formula of sin(2x).

    \sin{36}=2\sin{18}\cos{18}
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  12. #12
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    Re: Calculate tan^2(36) tan^2(72)

    ha ha I searched a lot of similar crazy expressions. It seems difficult to learn all of them.
    Exact trigonometric constants - Wikipedia, the free encyclopedia
    Trigonometry Angles -- from Wolfram MathWorld
    Last edited by BookEnquiry; October 11th 2011 at 10:48 PM.
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  13. #13
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    wrong reply

    Rcos2\theta+iRsin2\theta=(rcos\theta+irsin\theta)^  2
    R=rr=r^2
    R+Rcos2\theta=2rrcos^2\theta
    R-Rcos2\theta=2rrsin^2\theta
    Last edited by BookEnquiry; October 11th 2011 at 08:04 AM.
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