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Math Help - Double / Half & Area Trig Help

  1. #1
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    Double / Half & Area Trig Help

    I have two questions:
    1. Given that sin 3pi/10 = sqrt(5)+1/4 find an exact expression for cos (3pi/5)
    I used cos(2theta) = 1 - 2sin^2(theta)
    cos(3pi/5) = 1- 2sin^2(3pi/10)
    = 1 - 2(sqrt(5)+1/4)^2
    = 1- 2(3+sqrt(5)/8) = 4-3+sqrt(5)/4 = 1+sqrt(5)/4 <- i got that as the final answer but the answer key says that it's 1-sqrt(5)/4

    2. What is the area of the triangle whose sides all have length r
    i used A = 1/2absin(theta)
    = 1/2 r^2sin(theta)
    i got to that part and now i'm confused as to what to do next. the answer is sqrt(3)r^2/4
    would it be right if i said that sin(theta) = 1/2 = pi/6 and then cos(pi/6) = sqrt(3)/2 and then use that to get sqrt(3)r^2/4?
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  2. #2
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    Do you mean \displaystyle \sin{\frac{3\pi}{10}} = \frac{\sqrt{5} + 1}{4}?

    You should know that \displaystyle \cos{2\theta}= 1 - 2\sin^2{\theta}. Here \displaystyle \theta = \frac{3\pi}{10}.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    Do you mean \displaystyle \sin{\frac{3\pi}{10}} = \frac{\sqrt{5} + 1}{4}?

    You should know that \displaystyle \cos{2\theta}= 1 - 2\sin^2{\theta}. Here \displaystyle \theta = \frac{3\pi}{10}.
    yes i know but i still can't figure out what i did wrong to get 1+sqrt(5) instead of 1 - sqrt(5)
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  4. #4
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    Let's see...

    \displaystyle 1 - 2\sin^2{\frac{3\pi}{10}} = 1 - 2\left(\frac{\sqrt{5} + 1}{4}\right)^2

    \displaystyle = 1 - 2\left(\frac{5 + 2\sqrt{5} + 1}{16}\right)

    \displaystyle = 1 - \frac{6 + 2\sqrt{5}}{8}

    \displaystyle = \frac{8}{8} - \frac{6 + 2\sqrt{5}}{8}

    \displaystyle = \frac{8 - (6 + 2\sqrt{5})}{8}

    \displaystyle = \frac{8 - 6 - 2\sqrt{5}}{8}. Go from here.

    It appears you did not subtract ALL of the second term...
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  5. #5
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    Quote Originally Posted by Prove It View Post
    Let's see...

    \displaystyle 1 - 2\sin^2{\frac{3\pi}{10}} = 1 - 2\left(\frac{\sqrt{5} + 1}{4}\right)^2

    \displaystyle = 1 - 2\left(\frac{5 + 2\sqrt{5} + 1}{16}\right)

    \displaystyle = 1 - \frac{6 + 2\sqrt{5}}{8}

    \displaystyle = \frac{8}{8} - \frac{6 + 2\sqrt{5}}{8}

    \displaystyle = \frac{8 - (6 + 2\sqrt{5})}{8}

    \displaystyle = \frac{8 - 6 - 2\sqrt{5}}{8}. Go from here.

    It appears you did not subtract ALL of the second term...
    yay i got it~ wow, i didn't multiply the 8 to get 8-(6+2sqrt(5))
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  6. #6
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    Hello, dondonlouie!


    \text{2. What is the area of the triangle whose sides all have length } r\,\text{?}

    \text{i used: }\:A \:=\: \frac{1}{2}ab\sin\theta \:=\:\frac{1}{2}r^2\sin\theta

    \text{i got to that part and now i'm confused as to what to do next.}

    Since the triangle is equilateral, . \theta \,=\,\frac{\pi}{3}

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