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Math Help - Trig help

  1. #1
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    Trig help

    two prolems I really need to get done fast;

    1. sin2(y)= cos4(y) when 0 greater or equal y less than or equal 360 degrees

    2. solve algerbraically

    sin^2(x)/ 1- cos(x) = sec (x) +1/ sec (x)

    help fast please, thanks.
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  2. #2
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    Trigonometry

    Hello alpina
    Quote Originally Posted by alpina View Post
    two prolems I really need to get done fast;

    1. sin2(y)= cos4(y) when 0 greater or equal y less than or equal 360 degrees

    2. solve algerbraically

    sin^2(x)/ 1- cos(x) = sec (x) +1/ sec (x)

    help fast please, thanks.
    \sin 2y = \cos 4y

    \Rightarrow \sin 2y = \sin(90 - 4y)

    Now \sin A = \sin B \Rightarrow A = 360n +B
    or 180(2n+1) -B, where n is any integer

    So you now need to investigate n = 0, \pm 1, \pm 2, \dots, and put A = 2y and B = 90 - 4y.

    For instance: n = 0 gives 2y = 90 - 4y or 2y = 180 - (90 - 4y)

    \Rightarrow 6y = 90
    or -2y = 90

    \Rightarrow y = 15 (or y = -45, which is out-of-range).

    Then try n = +1, then -1, etc until you have all the values of y in the range.

    2 I assume that you need to prove the identity:

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

    So: \frac{\sin^2 x}{1-\cos x} =\frac{1 - \cos^2x}{1 - \cos x}

    = 1 + \cos x

    = \frac{\sec x}{\sec x} + \frac{1}{\sec x}

    = \frac{\sec x + 1}{\sec x}

    Grandad
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  3. #3
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    Quote Originally Posted by alpina View Post
    two prolems I really need to get done fast;

    1. sin2(y)= cos4(y) when 0 greater or equal y less than or equal 360 degrees
    Do you mean sin(2y)= cos(4y) or sin^2(y)= cos^4(y)? Since you use "^" in problem 2 I assume it is the first. cos(4y)= cos(2(2y))= cos^2(2y)- sin^2(2y)= (1- sin^2(2y))- sin^2(2y)= 1- 2sin^2(2y) so you can write the question as sin(2y)= 1- 2sin^2(2y). Let u= sin(2y) and the equation becomes u= 1- 2u^2 or 2u^2+ u- 1= 0. Solve that equation and, once you know u, solve sin(2y)= u for y.

    2. solve algerbraically

    sin^2(x)/ 1- cos(x) = sec (x) +1/ sec (x)

    help fast please, thanks.
    I assume you mean sin^2(x)/(1- cos(x))= (sec(x)+ 1)/sec(x). Multiplying on both sides by 1- cos(x) and sec(x), sin^2(x)sec(x)= (sec(x)+ 1)(1- cos(x)).
    Since sec(x)= 1/cos(x), that is sin^2(x)/cos(x)= (1/cos(x)+ 1)(1- cos(x)).
    Multiplying out the right side gives 1/cos(x)- cos(x)+ 1- 1= 1/cos(x)- cos(x).
    And on the left, sin^2(x)= 1- cos^2(x) so sin^2(x)/cos(x)= (1- cos^2(x)/cos(x)= 1/cos(x)- cos(x).

    Were you asked to "solve" this equation or prove that this is an identity?
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  4. #4
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    Quote Originally Posted by Grandad View Post
    Hello alpina
    \sin 2y = \cos 4y

    \Rightarrow \sin 2y = \sin(90 - 4y)

    Now \sin A = \sin B \Rightarrow A = 360n +B or 180(2n+1) -B, where n is any integer

    So you now need to investigate n = 0, \pm 1, \pm 2, \dots, and put A = 2y and B = 90 - 4y.

    For instance: n = 0 gives 2y = 90 - 4y or 2y = 180 - (90 - 4y)

    \Rightarrow 6y = 90 or -2y = 90

    \Rightarrow y = 15 (or y = -45, which is out-of-range).

    Then try n = +1, then -1, etc until you have all the values of y in the range.

    2 I assume that you need to prove the identity:

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

    So: \frac{\sin^2 x}{1-\cos x} =\frac{1 - \cos^2x}{1 - \cos x}

    = 1 + \cos x

    = \frac{\sec x}{\sec x} + \frac{1}{\sec x}

    = \frac{\sec x + 1}{\sec x}

    Grandad


    Thanks for your responses, and is there any other way to write the 1st problem? because that's not the way I usually solve them, so I am just wondering.
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  5. #5
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    Trigonometry

    Hello alpina
    Quote Originally Posted by alpina View Post
    Thanks for your responses, and is there any other way to write the 1st problem? because that's not the way I usually solve them, so I am just wondering.
    Yes, as HallsOfIvy suggested, you can write \cos 4y as 1 - 2 \sin^2 2y and get a quadratic equation in \sin 2y.

    \sin 2y = 1 - 2\sin^2 2y

    \Rightarrow 2\sin^2 2y + \sin 2y - 1 = 0

    \Rightarrow (2\sin 2y -1)(\sin 2y +1) = 0

    \Rightarrow \sin 2y = \frac{1}{2} or -1

    and take it from there.

    Grandad
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  6. #6
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    thanks a lot. helped me out
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