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Thread: Dreaded identities and double angle formulae

  1. #1
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    Dreaded identities and double angle formulae

    I've always had problems with these. Surely it cant be that hard, but I really cant get my head around them as it stands, even if it is just proving them - Maybe because I cant find many practical uses for it.. Anyway, these are the following coursework questions due for tomorrow. Would be very greatful if you could provide full answers so I can actually start to understand these damned things. Cheers!

    [Y=Theta]

    12) Using double angle formulae rewrite $\displaystyle sin 4Y$ in terms of $\displaystyle sin Y$ and $\displaystyle cos Y$ (Raised to any power).

    13) Prove the following identitys:

    a) $\displaystyle cot^2 x+1 = cosec^2 x$

    b) $\displaystyle (tan x+sec x)/(sec x-cos x+tan x) = cosec x$

    c) $\displaystyle sin^3 x-cos^3 x = (1+sin x*cos x)(sin x-cos x)$

    For c) we were told to go about doing the right hand side first.

    Thanks again!
    Last edited by Backlock; Feb 25th 2009 at 07:21 AM.
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  2. #2
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    Trig Identities

    Hello Backlock
    Quote Originally Posted by Backlock View Post
    [Y=Theta]

    12) Using double angle formulae rewrite $\displaystyle sin 4Y$ in terms of $\displaystyle sin Y and cos Y$ (Raised to any power).
    $\displaystyle \sin 4\theta = \sin (2 \times 2\theta$)

    $\displaystyle = 2\sin 2\theta\cos 2\theta$

    $\displaystyle = 2\times 2\sin\theta\cos\theta \times (2\cos^2\theta - 1)$

    I'll leave you to simplify this.


    Quote Originally Posted by Backlock View Post
    13) Prove the following identitys:

    a) $\displaystyle cot^2 x+1 = cosec^2 x$
    Take $\displaystyle \cos^2 x + \sin^2x=1$ and divide both sides by $\displaystyle \sin^2x$:

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

    $\displaystyle \Rightarrow$ ...?


    Quote Originally Posted by Backlock View Post
    b) $\displaystyle (tan x+sec x)/(sec x-cos x+tan x) = cosec x$
    Note that $\displaystyle \tan x = \frac{\sin x}{\cos x}$ and $\displaystyle \sec x = \frac{1}{\cos x}$ and multiply top and bottom of $\displaystyle \frac{\tan x + \sec x}{\sec x - \cos x + \tan x}$ by $\displaystyle \cos x$ to get:

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

    $\displaystyle = \frac{\sin x + 1}{\sin^2x + \sin x}$

    Can you finish it now? (Take out a common factor in the denominator.)


    Quote Originally Posted by Backlock View Post
    c) $\displaystyle sin^3 x-cos^3 x = (1+sin x*cos x)(sin x-cos x)$

    For c) we were told to go about doing the right hand side first.

    Thanks again!
    OK, but I think it's easier to start with the LHS, noting that $\displaystyle a^3 - b^3$ factorises to $\displaystyle (a-b)(a^2 + ab + b^2)$. So:

    $\displaystyle \sin^3x - \cos^3 x= (\sin x - \cos x) (\sin^2x + \sin x \cos x + \cos^2x)$

    $\displaystyle = (\sin x - \cos x)(1 + \sin x \cos x)$, using $\displaystyle \sin^2x + \cos^2x = 1$

    Grandad
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