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Math Help - Multiple choice trigonometry help pls?

  1. #1
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    Multiple choice trigonometry help pls?

    I'm terrible at trig, mind if I get some explanation on the following questions?

    1) What is the exact value of sin2(pi/6) - 2sin(pi/6)cos(pi/6) + cos2(-pi/6)

    a)2 - SqRt(3)
    b)(2- SqRt(3))/2
    c)(2+SqRt(3))/2
    d)2+SqRt(3)

    2) If cos(x)= -sin(x), then sin2(x) is

    a)1/4
    b)1/2
    c)SqRt(2)/2
    d)-SqRt(2)/2

    3)For the equation 4cos(x) - SqRt(12) = 0, where 0<x<(or equal to)2pi, determine the measure of x.

    a)pi/3, 5pi/3
    b)2pi/3, 4pi/3
    c)5pi/6 , 7pi/6
    d)pi/6 , 11pi/6

    4)If cos2(x) - 6sin(2x)=0, then one value of x is

    a)0 Degrees
    b)90 Degrees
    c)180 Degrees
    d)255 Degrees

    5) For the graph of y=2sin(3x), which of the following is not an x-intercept?

    a)60 Degrees
    b)90 Degrees
    c)240 Degrees
    d)2pi Rad

    6)What is another form of Sin(A+B)sin(A-B)/cos2(A)cos2(B)

    a)cos2(A) - cos2(B)
    b)Tan2(A) + Tan2(B)
    c)Tan2(A) - Tan2(B)
    d)sin2(A) - sin2(B)
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  2. #2
    MHF Contributor Amer's Avatar
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    Jordan
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    Quote Originally Posted by Blahdkm View Post
    I'm terrible at trig, mind if I get some explanation on the following questions?

    1) What is the exact value of sin2(pi/6) - 2sin(pi/6)cos(pi/6) + cos2(-pi/6)

    a)2 - SqRt(3)
    b)(2- SqRt(3))/2
    c)(2+SqRt(3))/2
    d)2+SqRt(3)

    2) If cos(x)= -sin(x), then sin2(x) is

    a)1/4
    b)1/2
    c)SqRt(2)/2
    d)-SqRt(2)/2

    3)For the equation 4cos(x) - SqRt(12) = 0, where 0<x<(or equal to)2pi, determine the measure of x.

    a)pi/3, 5pi/3
    b)2pi/3, 4pi/3
    c)5pi/6 , 7pi/6
    d)pi/6 , 11pi/6

    4)If cos2(x) - 6sin(2x)=0, then one value of x is

    a)0 Degrees
    b)90 Degrees
    c)180 Degrees
    d)255 Degrees

    5) For the graph of y=2sin(3x), which of the following is not an x-intercept?

    a)60 Degrees
    b)90 Degrees
    c)240 Degrees
    d)2pi Rad

    6)What is another form of Sin(A+B)sin(A-B)/cos2(A)cos2(B)

    a)cos2(A) - cos2(B)
    b)Tan2(A) + Tan2(B)
    c)Tan2(A) - Tan2(B)
    d)sin2(A) - sin2(B)
    all your questions need to know these

    sin2x=2sin(x)cos(x),cos2(x)=1-2sin^2(x), cos2(x)=2cos^2(x)-1,cos2(x)=cos^2x-sin^2x

    sin(A+B)=sinAcosB+cosAsinB,sin(A-B)=sinAcosB-cosAsinB

    cos\left(\frac{\pi}{6}+2n\pi\right)=\frac{\sqrt{3}  }{2}

    sin\frac{\pi}{4}=\frac{1}{\sqrt{2}},cos\frac{\pi}{  4}=\frac{1}{\sqrt{2}}

    you should know that
    lats take

    3) 4cosx-\sqrt{12}=0...4cosx=\sqrt{12}....4cosx=\sqrt{4(3)}

    4cosx= 2\sqrt{(3)}...cosx=\frac{\sqrt{3}}{2} what is angle when cosx = squre root of 3 divded by 2 and it is positive

    2)when -sinx = cosx , sin and cosx had a defferent signs this happened in the foruth quarter and in the second quarter and they have the same value that mean that the angle is pi/4 right now pi\4 in the second quarter equal ... and pi/4 in the fourth quarter equal afer you find x find sin2(x)



    Multiple choice trigonometry help pls?-unitcircle.gif
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  3. #3
    Senior Member Twig's Avatar
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    From
    Gothenburg
    Posts
    396
    hi

    1)  \frac{2-\sqrt{3}}{2}

    Fot this one you need to know the 'standard' angles.  sin(\frac{\pi}{6})=\frac{1}{2} etc ... These can be found on internet easily, or in any decent calculus book.

    2)  \frac{1}{2}

     cos(x)=-sin(x) when they have the same magnitude, which is only for  x = \frac{\pi}{4} . So here we get  sin(\frac{3\pi}{4}) = \frac{1}{\sqrt{2}} \, \Rightarrow sin^{2}(\frac{3\pi}{4}) = \frac{1}{2}
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  4. #4
    Member Awsom Guy's Avatar
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    You could try to put them in the calculator. If it doesn't work put them in seperatly and check brackets and things.
    I don't really know the calculations but you could give it a try.
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  5. #5
    Super Member
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    Quote Originally Posted by Blahdkm View Post
    5) For the graph of y=2sin(3x), which of the following is not an x-intercept?

    a)60 Degrees
    b)90 Degrees
    c)240 Degrees
    d)2pi Rad
    The x-intercepts of y = sin x are x = n\pi, where n is an integer. The x-intercepts of y = 2 sin x do not change because the only thing that changes in the graph is the amplitude. However, the graph of y = 2 sin 3x is a horizontal shrink by a factor of 1/3, so the x-intercepts are x = \frac{n\pi}{3}, where n is an integer. In other words, the x-intercepts are multiples of 60 degrees. The answer is (b) because 90 isn't a multiple of 60.


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  6. #6
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    Quote Originally Posted by Blahdkm View Post
    6)What is another form of Sin(A+B)sin(A-B)/cos2(A)cos2(B)

    a)cos2(A) - cos2(B)
    b)Tan2(A) + Tan2(B)
    c)Tan2(A) - Tan2(B)
    d)sin2(A) - sin2(B)
    \frac{\sin (A + B) \sin (A - B)}{\cos^2 A \cos^2 B}

    = \frac{(\sin A \cos B + \cos A \sin B)(\sin A \cos B - \cos A \sin B)}{\cos^2 A \cos^2 B}

    = \frac{\sin^2 A \cos^2 B - \cos^2 A \sin^2 B}{\cos^2 A \cos^2 B}

    = \frac{\sin^2 A \cos^2 B}{\cos^2 A \cos^2 B} - \frac{\cos^2 A \sin^2 B}{\cos^2 A \cos^2 B}

    = \frac{\sin^2 A}{\cos^2 A} - \frac{\sin^2 B}{\cos^2 B}

    = \tan^2 A - \tan^2 B .

    The answer is (c).


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