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Math Help - How do I verify these identities?

  1. #1
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    How do I verify these identities?

    I was doing an advance reading on trigonometry (identities, specifically) when I encountered these problems on verification:
    <br />
(tan^3 x + sin x sec x -sin x cos x)/(sec x - cos x) = tan x sec x + sin x<br />

    <br /> <br />
(1 + sin 2B + cos 2B)/(1 + sin 2B - cos 2B) = cot B<br /> <br />



    <br /> <br />
(sin x + sin 3x + sin 5x + sin 7x)/(cos x + cos 3x + cos 5x + cos 7x) = tan 4x<br /> <br />

    I'll try to edit this when I come up with answers. Thanks for those who will help!
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  2. #2
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    Quote Originally Posted by azuresonata View Post
    I was doing an advance reading on trigonometry (identities, specifically) when I encountered these problems on verification:
    <br />
(tan^3 x + sin x sec x -sin x cos x)/(sec x - cos x) = tan x sec x + sin x<br />

    Mr F says: At a glance, I'd suggest making the sustitutions {\color{red}\tan x = \frac{\sin x}{\cos x}} and {\color{red}\sec x = \frac{1}{\cos x}} on each side and then simplifying the resulting expressions.

    <br /> <br />
(1 + sin 2B + cos 2B)/(1 + sin 2B - cos 2B) = cot B<br /> <br />

    Mr F says: At a glance I'd suggest substituting from the double angle formulae in the left hand side and simplifying the resulting expression.

    <br /> <br />
(sin x + sin 3x + sin 5x + sin 7x)/(cos x + cos 3x + cos 5x + cos 7x) = tan 4x<br /> <br />

    I'll try to edit this when I come up with answers. Thanks for those who will help!
    For the last, you can start by using the following identities (derivable from the compound angle formulae):

    1. sin(A + B) + sin(A - B) = 2 sin A cos B.

    From which it follows that:

    sin(x) + sin(3x) = 2 sin(2x) cos(x)

    sin(5x) + sin(7x) = 2 sin(6x) cos(x)


    2. cos(A + B) + cos(A - B) = 2 cos A cos B.

    From which it follows that

    cos(x) + cos(3x) = ........

    cos(5x) + cos(7x) = ......
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  3. #3
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    ok for problems such like this knowing the standard identities is VERY useful, a list of them can be found here
    Table of Trigonometric Identities

    for example lets look at the first of your problems

    looking at the denominator we can write it as
    \sec x - \cos x=\frac{1}{\cos x} - \cos x = \frac{1-cos^2 x}{\cos x} = \frac{sin^2x}{\cos x}=\tan x \sin x


    so the LHS becomes

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

    now writing tan x in terms of sine and cosine we have

    \frac{1}{\sin x}(\tan^2 x +1 -\cos^2 x)

    then using  \cos^2 x +\sin^2 x=1

    \frac{1}{\sin x}(tan^2 x + sin^2 x)

    once again writing tan in terms of sine and cosine gives us the RHS

    \tan x \sec x + \sin x

    so all this topic is really is learning identities and practice (hence why this is deathly boring!)
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