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Math Help - Wrong answers on test, need help

  1. #1
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    Wrong answers on test, need help

    Here are a few problems I missed on the test. The professor doesn't show where you missed the problem or anything. Would like to know how to work these:

    1) Solve for x: tan(pi(x)/2)=0

    2) Solve for x: 1-2cos(x)=0

    3) Calculate (give exact answers. i.e. no decimals)
    a: sin(105)
    b: cos(7pi/12)

    4)Solve for x: 2sin^2x=1+cos^2x


    Thanks for any help!
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  2. #2
    Senior Member DeMath's Avatar
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    Quote Originally Posted by pinnacle2009 View Post
    Here are a few problems I missed on the test. The professor doesn't show where you missed the problem or anything. Would like to know how to work these:

    1) Solve for x: tan(pi(x)/2)=0

    2) Solve for x: 1-2cos(x)=0

    3) Calculate (give exact answers. i.e. no decimals)
    a: sin(105)
    b: cos(7pi/12)

    4)Solve for x: 2sin^2x=1+cos^2x


    Thanks for any help!
    For 1)

    \tan \left( {\frac{\pi }{2}x} \right) = 0 \Leftrightarrow \frac{\pi }{2}x = \underbrace {\arctan 0}_0 + \pi k \Leftrightarrow x = 2k{\text{ }}\left( {k \in \mathbb{Z}} \right).

    For 2)

    1 - 2\cos x = 0 \Leftrightarrow \cos x = \frac{1}{2} \Leftrightarrow x =  \pm \arccos \frac{1}{2} + 2\pi k \Leftrightarrow x =  \pm \frac{\pi }{3} + 2\pi k{\text{ }}\left( {k \in \mathbb{Z}} \right).
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  3. #3
    MHF Contributor red_dog's Avatar
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    3)

    \sin 105^{\circ}=\sin(60^{\circ}+45^{\circ})=\sin 60^{\circ}\cos 45^{\circ}+\sin 45^{\circ}\cos 60^{\circ}=

    =\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2}+\frac{\  sqrt{2}}{2}\cdot\frac{1}{2}=\frac{\sqrt{6}+\sqrt{2  }}{4}
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  4. #4
    MHF Contributor red_dog's Avatar
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    3) b)

    \cos\frac{7\pi}{12}=\cos\left(\frac{\pi}{3}+\frac{  \pi}{4}\right)=

    =\cos\frac{\pi}{3}\cos\frac{\pi}{4}-\sin\frac{\pi}{3}\sin\frac{\pi}{4}=

    =\frac{1}{2}\cdot\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2}=\frac{\s  qrt{2}-\sqrt{6}}{4}
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  5. #5
    MHF Contributor red_dog's Avatar
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    4)

    2\sin^2x=1+\cos^2x\Rightarrow 2\sin^2x=2-\sin^2x\Rightarrow3\sin^2x=2\Rightarrow\sin^2x=\fr  ac{2}{3}\Rightarrow

    \Rightarrow\sin x=\pm\sqrt{\frac{2}{3}}\Rightarrow x=(-1)^k\arcsin\left(\pm\sqrt{\frac{2}{3}}\right)+k\pi  , \ k\in\mathbb{Z}
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  6. #6
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    Thanks very much!
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  7. #7
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    Quote Originally Posted by pinnacle2009 View Post
    Here are a few problems I missed on the test. The professor doesn't show where you missed the problem or anything.
    [snip]
    Then you should arrange an appointment with your professor and ask him/her.
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  8. #8
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    Somehow 2 and 4 he still says are wrong. I think on 4 he wanted us to use the power reducing formulas and on 2, I have no idea, unless he wanted us to use another formula.

    I tried working 4 using the power reducing formula and am still having trouble.

    Could someone look at these problems again please?

    I may be taking pre-calculus again lol
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  9. #9
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    Quote Originally Posted by pinnacle2009 View Post
    Somehow 2 and 4 he still says are wrong. I think on 4 he wanted us to use the power reducing formulas and on 2, I have no idea, unless he wanted us to use another formula.

    I tried working 4 using the power reducing formula and am still having trouble.

    Could someone look at these problems again please?

    I may be taking pre-calculus again lol
    There's no formula to use in #2, per se. You have to recognize the trig functions of special angles. The answer provided by DeMath is correct, but you could also state your answers as
    x = \frac{\pi}{3} + 2\pi k and
    x = \frac{5\pi}{3} + 2\pi k, where k is an integer.
    (Remember that \frac{5\pi}{3} is coterminal to -\frac{\pi}{3}).

    If the answers are supposed to be in 0 \le x < 2\pi, then shame on you for not stating that in the first place. The answers would then be x = \frac{\pi}{3} and x = \frac{5\pi}{3}.


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  10. #10
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    Quote Originally Posted by pinnacle2009 View Post
    4)Solve for x: 2sin^2x=1+cos^2x
    Use the power-reducing identities? Ok.

    \begin{aligned}<br />
2\sin^2 x &= 1 + \cos^2 x \\<br />
2\left(\frac{1 - \cos 2x}{2}\right) &= 1 + \frac{1 + \cos 2x}{2} \\<br />
1 - \cos 2x &= 1 + \frac{1 + \cos 2x}{2} \\<br />
2 - 2\cos 2x &= 2 + 1 + \cos 2x \\<br />
3\cos 2x &= -1 \\<br />
\cos 2x &= -\frac{1}{3}<br />
\end{aligned}

    Does the professor want answers approximated in decimal? Then,
    2x \approx 1.911 + 2\pi k\;\Rightarrow\;x \approx 0.955 + k\pi
    2x \approx 4.372 + 2\pi k\;\Rightarrow\;x \approx 2.186 + k\pi
    2x \approx 8.194 + 2\pi k\;\Rightarrow\;x \approx 4.097 + k\pi
    2x \approx 10.656 + 2\pi k\;\Rightarrow\;x \approx 5.328 + k\pi


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  11. #11
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    Quote Originally Posted by yeongil View Post
    There's no formula to use in #2, per se. You have to recognize the trig functions of special angles. The answer provided by DeMath is correct, but you could also state your answers as
    x = \frac{\pi}{3} + 2\pi k and
    x = \frac{5\pi}{3} + 2\pi k, where k is an integer.
    (Remember that \frac{5\pi}{3} is coterminal to -\frac{\pi}{3}).

    If the answers are supposed to be in 0 \le x < 2\pi, then shame on you for not stating that in the first place. The answers would then be x = \frac{\pi}{3} and x = \frac{5\pi}{3}.


    01
    No range was stated on the test.


    Thanks Yeongil. I had a friend work this one for me (she's a math major with the same professor). She got that same answer.
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