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Math Help - Trigonometry.

  1. #1
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    Trigonometry.

    Alevel (module 2/3 standard). Any help at all would be hugely appriciated.

    32. a) Solve in radians



    sin (x + pi/6) = 2cosx



    giving all solutions in range of 0 < x < 2pi



    b) Use the formular tan (a - b) = (tana - tanb)/(1 + tana tanb)



    to show that tan(x - pi/4) = (tanx - 1)/(tan x+1)



    c) Hence or otherwise solve in radians tan(x - pi/4) = 6tanx



    67. Betweem -180 < x < 180 solve

    cosec^2 x = 1 + 2 sec^2 x



    68. For all values of x prove that



    8 sin^4 x = 3 - 4 cos 2 x + cos 4 x



    Hence find in radians in terms of pi the values of x in pi and -pi for which cos 4x - 4cos 2x + 1 = 0



    70. Find the values of x for 0-360 for which tan^2 x + 2 sec x = 5



    85. Find values to which



    3 cos 2x + 7 cos x = 0 and 0 < x < 2pi



    Thank you for reading. xx
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    Quote Originally Posted by Sarah. View Post
    Alevel (module 2/3 standard). Any help at all would be hugely appriciated.

    32. a) Solve in radians



    sin (x + pi/6) = 2cosx



    giving all solutions in range of 0 < x < 2pi
    Hint: use the identity \sin (A + B) = \sin A \cos B + \sin B \cos A and see where you get
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    Quote Originally Posted by Sarah. View Post

    b) Use the formular tan (a - b) = (tana - tanb)/(1 + tana tanb)



    to show that tan(x - pi/4) = (tanx - 1)/(tan x+1)
    just use the formula:

    \tan \left( x - \frac {\pi}4 \right) = \frac {\tan x - \tan \frac {\pi}4 }{1 + \tan x \tan \frac {\pi}4 } .........i trust you know what \tan \frac {\pi}4 is


    c) Hence or otherwise solve in radians tan(x - pi/4) = 6tanx
    from above, we know that \frac {\tan x - 1}{\tan x + 1} = 6 \tan x

    begin by solving for tanx, then take the arctan (that is, the inverse tangent) of both sides to find the required x-values
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    Quote Originally Posted by Sarah. View Post

    67. Betweem -180 < x < 180 solve

    cosec^2 x = 1 + 2 sec^2 x
    use the fact that:

    \csc^2 x = 1 + 2 \sec^2 x

    \Rightarrow \frac 1{\sin^2 x} = 1 + \frac 2{\cos^2 x}

    combine the fractions on the right and see if you can take it from there
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    Quote Originally Posted by Sarah. View Post
    68. For all values of x prove that

    8 sin^4 x = 3 - 4 cos 2 x + cos 4 x

    usually i'd tell you to start working on the right, but maybe it's better to start on the left here

    note,
    8 \sin^4 x = 8 \left( \sin^2 x \right)^2 = 8 \left( 1 - \cos^2 x \right)^2

    now expand and simplify to get the right hand side

    (remember your formulas for the double angle of cosine)
    Hence find in radians in terms of pi the values of x in pi and -pi for which cos 4x - 4cos 2x + 1 = 0
    in light of the above, we have 4 - 8 \sin^4 x = 0

    can you continue?
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    32 was fine thanks.

    67. [tex]
    1/(sin^2x) = 1 + 2/(cos^2x)

    After this, I got a bit stuck. Another hint would be useful please.

    68.
    Quote:
    Hence find in radians in terms of pi the values of x in pi and -pi for which cos 4x - 4cos 2x + 1 = 0
    in light of the above, we have

    can you continue?

    In short, no. I'm not 100% sure where you got the first line from, and from there how to solve the final equation.

    Thanks for the help so far.
    Last edited by Sarah.; December 2nd 2007 at 01:13 PM.
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    Quote Originally Posted by Sarah. View Post
    32 was fine thanks.

    67. [tex]
    1/(sin^2x) = 1 + 2/(cos^2x)

    After this, I got a bit stuck. Another hint would be useful please.
    that's where i left off. you did nothing after that? do you know how to combine fractions?

    you would have:
    \frac 1{\sin^2 x} = \frac {\cos^2 x + 2}{\cos^2 x}

    do you think you can continue now?

    68.
    Quote:
    Hence find in radians in terms of pi the values of x in pi and -pi for which cos 4x - 4cos 2x + 1 = 0
    in light of the above, we have

    can you continue?

    In short, no. I'm not 100% sure where you got the first line from, and from there how to solve the final equation.

    Thanks for the help so far.
    remember, at this point we showed that 8 \sin^4 x = 3 - 4 \cos 2x + \cos 4x

    \Rightarrow -4 \cos 2x + \cos 4x = 8 \sin^4 x - 3 (i had a typo before, i thought it was - cos (4x) and + 4cos(2x))

    add 1 to both sides, we get:

    \cos 4x - 4 \cos 2x + 1 = 8 \sin^4 x - 2

    now we are told that \cos 4x - 4 \cos 2x + 1 = 0

    thus we have:

    0 = 8 \sin^4 x - 2

    now continue
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    Quote Originally Posted by Sarah. View Post

    70. Find the values of x for 0-360 for which tan^2 x + 2 sec x = 5


    remember, \tan^2 x = \sec^2 x - 1

    thus,

    \tan^2 x + 2 \sec x = 5

    \Rightarrow \sec^2 x - 1 + 2 \sec x = 5

    bring everything to one side, we have:

    \sec^2 x + 2 \sec x - 6 = 0

    this is quadratic in \sec x, if we were to replace \sec x with y, for instance, we would have

    y^2 + 2y - 6 = 0

    i think you can solve that for y. when done, replace y with \sec x and solve for x
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    Quote Originally Posted by Sarah. View Post
    85. Find values to which



    3 cos 2x + 7 cos x = 0 and 0 < x < 2pi
    recall that \cos 2x = 2 \cos^2 x - 1, thus we have:

    6 \cos^2 x - 3 + 7 \cos x = 0

    \Rightarrow 6 \cos^2 x + 7 \cos x - 3 = 0

    and it is a problem like #70 from there
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    67. So you end up with:

    cos^2x = (sin^2x)(cos^2x + 2)

    So you divide by cos^2 x?

    And get

    tan^2 x + 2 tan^2 x -1

    3 tan^2 x -1?

    Or have I done something wrong?

    What do you do here next?

    68. How do you simplify/factorise this?
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    Quote Originally Posted by Sarah. View Post
    67. So you end up with:

    cos^2x = (sin^2x)(cos^2x + 2)

    So you divide by cos^2 x?

    And get

    tan^2 x + 2 tan^2 x -1

    3 tan^2 x -1?

    Or have I done something wrong?

    What do you do here next?
    i do not recommend dividing by cos^2 x. if a solution of x ends up being one that causes cos(x) = 0, it would mean you have divided by 0, hence making your solution invalid. bring everything to one side and factor out the cos^2 x instead


    68. How do you simplify/factorise this?
    no factorization necessary, just solve for sin^4 x and take it from there
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    Doesn't it end up being
    = sin^2 x cos^2 x + 2sin^2x - cos^2x

    So cos^2x isn't common through out the equation and so cannot take out as a factor?
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    Oh wait. for 67, the original problem has sec^2 x. that means cos^2 x in not zero, since sec^2 x would be undefined then. thus it was ok to divide by cos^2 x
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    Quote Originally Posted by Sarah. View Post
    Doesn't it end up being
    = sin^2 x cos^2 x + 2sin^2x - cos^2x

    So cos^2x isn't common through out the equation and so cannot take out as a factor?
    you could have taken out cos^2 x out of the terms where it's common. but i made a post saying why it is ok to divide by cos^2 x, so don't worry about it

    EDIT: i'm kind of busy with other stuff right now, and as a result, i cannot concentrate on your problems as much as i would like to. i'm going to log off. hopefully someone else will help you. if not, i'll pick up where i left off when i come back on again
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    Thank you very much for sticking through this with me, I know how tedious it must be. Enjoy the rest of your night.
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