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Math Help - simplify trig

  1. #1
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    simplify trig

    A.2sinx+1=0
    B.2cos^2x-1=0
    C.3tan^3x-tanx=0
    D.2sin^2x=2+cosx

    I really need help.. I have my answers here but im not sure if they are correct:

    A.x=7/6pi, 11/6pi
    B.x=pi/4 7/4pi
    C.x=pi/6 7/6pi 2pi pi
    D.x=pi/2, 3/2pi pi/3 5/3pi


    I also need these checked
    Using a sum of difference identity find exact values for the following:

    A.Cos75(degrees)
    B.Sin15(degrees)
    C.tan345(degrees)

    I got
    A.*squareroot*6-2/4
    B.*squareroot*6-2/4
    C.3-*squareroot*3/3

    For problems 2 and 3 find the following.
    A.cos(x-y): sin(x-y); tan (x-y); the quadrant (x-y) is in.
    B.cos(x+y); sin(x+y) tan(x+y) the quadrant (x+y) is in.

    2.If (pi/2)<x<pi and pi<y<(3pi/2) and cosx=-5/13 and tany=3/4

    3.If pi<x< or equal too(3pi/2) and (3pi/2)<y<2pi, and secx=-41/9 and tany=-12/5

    If you could show me how to do them it would be appreciated.
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  2. #2
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    A.x=7/6pi, 11/6pi
    right

    B.x=pi/4 7/4pi
    C.x=pi/6 7/6pi 2pi pi
    these solutions work but there are others as well. Remember that if x^2 = a then x = \pm \sqrt {a}

    D.x=pi/2, 3/2pi pi/3 5/3pi
    You have lost a minus sign somewhere here, you should have ended up with cos x = 0 or cos x = -1/2, but the good news is that you appear to know what you are doing.

    A.Cos75(degrees)
    B.Sin15(degrees)
    C.tan345(degrees)
    cos (75^o) = cos(45^o+30^o)
    = cos(45^o)cos(30^o)-sin(45^o)sin(30^o)
    = \frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\frac{1}{2}
    = \frac{\sqrt{3}-1}{2\sqrt{2}}

    sin (15^o) = cos (75^o), as you probably noticed.

    tan (345^o) = -tan (15^o)
    = -\frac {sin(15^o)}{cos(15^o)}
    = -\frac{sin(15^o)}{cos (45^o+-30^o)}
    = -\frac{sin(15^o)}{cos (45^o)cos(-30^o)-sin(45^o)sin(-30^o)}
    = -\frac{\frac{\sqrt{3}-1}{2\sqrt{2}}}{\frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2  }+\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}}
    = -\frac{\sqrt{3}-1}{\sqrt{3}+1}
    = \frac{1-\sqrt{3}}{1+\sqrt{3}}
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  3. #3
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    For problems 2 and 3 find the following.
    A.cos(x-y): sin(x-y); tan (x-y); the quadrant (x-y) is in.
    B.cos(x+y); sin(x+y) tan(x+y) the quadrant (x+y) is in.

    2.If (pi/2)<x<pi and pi<y<(3pi/2) and cosx=-5/13 and tany=3/4
    The first thing you need to do is draw up the triangles for x and y, using the definition of cos as adj/hyp and tan as opp/adj and pythagoras theorem.
    simplify trig-tris.jpg

    Now you can read off the absolute values of anything else you may need and combine it with the quadrants of x and y to get the actual values. Then its just a matter of using the formulae for addition of angles in trigonometric functions.

    To find the quadrant of x-y, you will need to combine the restrictions imposed by the range for x and y and the signs of the results you get for sin, cos, tan etc of x-y
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  4. #4
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    wouldnt cos(-5/13+4/5) be:

    (-5/13.-4/5)-(12/14.-3/5)

    Because cos in the third quadrant is always negative.
    And sin in the third quadrant is always negative.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ninelivesmeow View Post
    wouldnt cos(-5/13+4/5) be:

    (-5/13.-4/5)-(12/14.-3/5)

    Because cos in the third quadrant is always negative.
    And sin in the third quadrant is always negative.
    I presume that you are trying to do a difference of angles formula for cosine, but I can't identify which (or even what) problem you are working on.

    I think you are trying to do cos(x + y) where cos(x) = -\frac{5}{13} and sin(y) = \frac{4}{5}?

    I need you to confirm this before I go any further.

    -Dan
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