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    \sin^4 \theta + 2sin^2 \theta cos^2 \theta + cos^4 \theta
    What is the first step to consider when doing these questions? I try using sin^2 theta = 1 - cos^2 theta etc but I seem to get no where. I tried writing \sin^4 \theta as (\sin^2 \theta)^2 and tried stuff from there to no avail.

    Thanks
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    (Original post by UKBrah)
    \sin^4 \theta + 2sin^2 \theta cos^2 \theta + cos^4 \theta
    What is the first step to consider when doing these questions? I try using sin^2 theta = 1 - cos^2 theta etc but I seem to get no where. I tried writing \sin^4 \theta as (\sin^2 \theta)^2 and tried stuff from there to no avail.

    Thanks

    It's a perfect square

    (a+b)^2 = a^2 + 2ab + b^2

    and in fact, for all theta, it equals one.
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    It sure does, thanks mate.
    Are there any other algebraic identities I should be aware of, like (a+b)^2 = a^2 + 2ab + b^2?
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    (Original post by claret_n_blue)
     (a - b)^2 = a^2 + b^2
    :curious: Are you sure?
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    (Original post by Farhan.Hanif93)
    :curious: Are you sure?
    (a-b)^2 = (a-b)(a+b) is that the one implied?
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    (Original post by UKBrah)
    It sure does, thanks mate.
    Are there any other algebraic identities I should be aware of, like (a+b)^2 = a^2 + 2ab + b^2?
    Well, it would help to know the binomial theorem.

    \text{For any}\  n \geq 1,

    (x+y)^n = \displaystyle \sum_{r=0}^{n} x^r y^{n-r} \binom{n}{r} = y^n x^0 \binom{n}{0} + \binom{n}{1}x y^{n-1} + ... + x^n y^0 \binom{n}{0}

    where

    \displaystyle \binom{n}{r} = \dfrac{n!}{r!(n-r)!}
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    (Original post by Farhan.Hanif93)
    :curious: Are you sure?
    Oops.
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    (Original post by UKBrah)
    (a-b)^2 = (a-b)(a+b) is that the one implied?
    Not quite; the RHS is equivalent to a^2-b^2 but (a-b)^2 \not\equiv a^2  - b^2.

    (Original post by Indeterminate)
    Well, it would help to know the binomial theorem.

    \text{For any}\  n \geq 1,

    (x+y)^n = \displaystyle \sum_{r=0}^{n} x^r y^{n-r} \binom{n}{r} = y^n x^0 \binom{n}{0} + \binom{n}{1}x y^{n-1} + ... + x^n y^0 \binom{n}{0}

    where

    \displaystyle \binom{n}{r} = \dfrac{n!}{r!(n-r)!}
    (with n\in \mathbb{Z})
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    (Original post by Farhan.Hanif93)
    Not quite; the RHS is equivalent to a^2-b^2 but (a-b)^2 \not\equiv a^2  - b^2.


    (with n\in \mathbb{Z})

    Yep, but since they don't go into sets at A-level, I'd tell an A-level student that n is an integer.
 
 
 
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