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Thread: trigo >find angles

  1. #1
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    trigo >find angles

    0* to 350*, find A
    1. sin5A=sin3A
    2.cos4A+cos2A=cos3A

    Inverse Trigonometric Functions
    Show, without using a calculator, that 〖Tan〗^(-1)4 -〖 Tan〗^(-1) 3/5 = π/4
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    Quote Originally Posted by sanikui View Post
    0* to 350*, find A
    1. sin5A=sin3A
    [snip]
    General solution:

    Case 1: $\displaystyle 5A = 3A + 2n \pi$ where n is an integer. Therefore $\displaystyle A = \, .... $

    Case 2: $\displaystyle 5A = (\pi - 3A) + 2n\pi$ where n is an integer. Therefore $\displaystyle A = \, .... $

    In each case substitute values of n that give values of A that lie in the given domain
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    Quote Originally Posted by sanikui View Post
    0* to 350*, find A
    [snip]
    2.cos4A+cos2A=cos3A

    [snip]
    You know that $\displaystyle \cos (x + y) + \cos (x - y) = 2 \cos x \cos y$. Substitute x = 3A and y = A: $\displaystyle \cos (4A) + \cos (2A) = 2 \cos (3A) \cos (A)$. Therefore the equation can be re-written as:

    $\displaystyle 2 \cos (3A) \cos (A) = \cos (3A) \Rightarrow \cos (3A) ( 2 \cos (A) - 1) = 0$.

    I'm sure you can continue the solution from here.
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    Quote Originally Posted by sanikui View Post
    [snip]
    Inverse Trigonometric Functions
    Show, without using a calculator, that 〖Tan〗^(-1)4 -〖 Tan〗^(-1) 3/5 = π/4
    Let $\displaystyle \tan^{-1} (4) = \alpha \Rightarrow \tan \alpha = 4$.

    Let $\displaystyle \tan^{-1} \left( \frac{3}{5} \right) = \beta \Rightarrow \tan \beta = \frac{3}{5}$.

    Then you have to show that $\displaystyle \alpha - \beta = \frac{\pi}{4}$.

    Now note that $\displaystyle \tan (\alpha -\beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} = \frac{4 - \frac{3}{5}}{1 + (4)\left( \frac{3}{5}\right)} = \frac{20 - 3}{5 + 12} = 1$.

    Therefore $\displaystyle \alpha - \beta = \tan^{-1} (1)$.
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    Quote Originally Posted by mr fantastic View Post
    General solution:

    Case 1: $\displaystyle 5A = 3A + 2n \pi$ where n is an integer. Therefore $\displaystyle A = \, .... $

    Case 2: $\displaystyle 5A = (\pi - 3A) + 2n\pi$ where n is an integer. Therefore $\displaystyle A = \, .... $

    In each case substitute values of n that give values of A that lie in the given domain
    This i till cant understand...can explain more clearly?
    Next 2 question i can get ur mean.Thx.
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  6. #6
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    Quote Originally Posted by mr fantastic View Post
    General solution:

    Case 1: $\displaystyle 5A = 3A + 2n \pi$ where n is an integer. Therefore $\displaystyle A = \, .... $

    Case 2: $\displaystyle 5A = (\pi - 3A) + 2n\pi$ where n is an integer. Therefore $\displaystyle A = \, .... $

    In each case substitute values of n that give values of A that lie in the given domain
    My mistake. I was using radians instead of degrees.

    Case 1: It should be obvious that if $\displaystyle \sin (5A) = \sin (3A)$ then either 5A = 3A or the angles 5A and 3A differ by an integer multiple of $\displaystyle 360^0$.

    So $\displaystyle 5A = 3A + 360^0 n$ where n is an integer.


    Case 2: You should know that $\displaystyle \sin x = \sin (180^0 - x)$. Therefore $\displaystyle \sin (3A) = \sin (180^0 - 3A)$.

    So the equation can be written $\displaystyle \sin (5A) = \sin (180^0 - 3A)$.

    Using the same logic that gave Case 1, it should be obvious that either $\displaystyle 5A = (180^0 - 3A)$ or the angles $\displaystyle 5A$ and $\displaystyle (180^0 - 3A)$ differ by an integer multiple of $\displaystyle 360^0$.

    So $\displaystyle 5A = (180^0 - 3A) + 360^0 n$ where n is an integer.


    In each case, substitute the values of n into the general solutions that give values of A lying in the domain $\displaystyle 0^0 \leq A \leq 350^0$.
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  7. #7
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    Oic...know liaw....thx Mr.fantastic~!!
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