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Thread: Developing a double angle formula for Sin4x

  1. #1
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    Question Developing a double angle formula for Sin4x

    Hi,

    I hope someone can help. I'm just trying to understand how Sin4x can be placed into a double-angle formula. One of the ways that I came across is as follows:

    Developing a double angle formula for Sin4x-18986423_1426484374074319_1207453464_o.jpg

    Can someone please tell me the thought-process behind the first-step? I know that 2Sinx= 2SinxCosx, but where does the (2) and the (Cos2x) come into play?


    Sincerely,
    Olivia
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    Re: Developing a double angle formula for Sin4x

    $\begin{align*}\sin (4x) & = \sin (2 (2x)) \\ & = 2\sin (2x)\cos (2x) \\ & = 2 (2\sin x \cos x)\cos (2x)\end{align*} $
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    Re: Developing a double angle formula for Sin4x

    Ohh okay I understand what you mean. The steps are pretty cool Thanks for your help.
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    Re: Developing a double angle formula for Sin4x

    If I may, I will add two more lines:
    $\begin{align*}\sin (4x) & = \sin (2 (2x)) \\ & = 2\sin (2x)\cos (2x) \\ & = 2 (2\sin x \cos x)\cos (2x)\\&=2(2\sin x \cos x)(\cos^3(x)-sin^2(x))\\&=4(\sin(x)\cos^3(x)-\cos(x)\sin^3(x)\end{align*}$

    Doing that makes all functions of a single $x$
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    Re: Developing a double angle formula for Sin4x

    Quote Originally Posted by Plato View Post
    If I may, I will add two more lines:
    $\begin{align*}\sin (4x) & = \sin (2 (2x)) \\ & = 2\sin (2x)\cos (2x) \\ & = 2 (2\sin x \cos x)\cos (2x)\\&=2(2\sin x \cos x)(\cos^3(x)-sin^2(x))$
    Typo: This should be $cos^2(x)$ not $cos^3(x)$

    $\\&=4(\sin(x)\cos^3(x)-\cos(x)\sin^3(x)\end{align*}$

    Doing that makes all functions of a single $x$
    Thanks from Plato
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    Re: Developing a double angle formula for Sin4x

    Thanks for responding! I have two questions to ask:

    1. I understand your equation above too some extent. However, the image which I attached in my original post has a solution where it is 4x in terms of x. Is your solution sin4x = 4sin(x)cos^3(x) - cos(x)sin^3(x) just an alternative?

    2. Couldn't I theoretically use the equation Sin2x=2CosxSinx and let x = 4x/2 which would then result in the equation: Sin4x = 2Cos2xSin2x. Would this equation for Sin4x be as equally correct to the other equations of Sin4x? And if they are equivalent, I'm curious under what conditions it would be more advantageous to use one equation of Sin4x over the other?
    Last edited by otownsend; Jun 4th 2017 at 10:40 AM.
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    Re: Developing a double angle formula for Sin4x

    1) yes, it is another alternative
    2) you should use whatever is easiest to use.
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    Re: Developing a double angle formula for Sin4x

    Quote Originally Posted by otownsend View Post
    1. I understand your equation above too some extent. However, the image which I attached in my original post has a solution where it is 4x in terms of x. Is your solution sin4x = 4sin(x)cos^3(x) - cos(x)sin^3(x) just an alternative?
    I am not sure the source of your confusion?
    The expression $\sin(4x)$ is a function of $x$. But the question asks you to find an expression is terms of a single $x$.
    $ \sin(4x) = 4\sin(x)\cos^3(x) - \cos(x)\sin^3(x)$.
    The advantage?: suppose you are given that $\cos(\theta)=A$ and told to find $\sin(4\theta)$ do you see how easy it would be?

    Quote Originally Posted by otownsend View Post
    2. Couldn't I theoretically use the equation Sin2x=2CosxSinx and let x = 4x/2 which would then result in the equation: Sin4x = 2Cos2xSin2x. Would this equation for Sin4x be as equally correct to the other equations of Sin4x? And if they are equivalent, I'm curious under what conditions it would be more advantageous to use one equation of Sin4x over the other?
    The example above answers your question.
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    Re: Developing a double angle formula for Sin4x

    I don't see how 4SinxCosx-8Sin^3Cosx is not expressing Sin4x in terms of x. Is this what you're suggesting?

    And could you further explain what you meant by "suppose you are given that cos(x) = A and told to find sin(4x) do you see how easy it would be?" Could you show what you mean?

    Also... thank you for your help so far.
    Last edited by otownsend; Jun 5th 2017 at 08:38 AM.
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  10. #10
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    Re: Developing a double angle formula for Sin4x

    Example: for some angle $x$ we have $\cos x = \dfrac{3}{5}$

    By the Pythagorean Theorem, we know $\sin x =\dfrac{4}{5}$

    Calculate $\sin (4x) $
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