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    Is it possible to write ( 1-sin²θ/sinθ ) as 1/sinθ-sinθ

    Also I am really confused on what to do to prove the following:
    1] ( sin2θ+sinθ ) / ( 1+cos2θ+cosθ ) ≡tanθ
    2] ( 1-cos2θ+sin2θ ) / ( 1+cos2θ+sin2θ ) ≡tanθ

    I've started with LHS = and then i get stuck

    Any answers and help??
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    You need to get some brackets in there or use latex, what you have written is ambiguous without.
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    You should put parenthesis, it is quite confusing...
    If it is (1-sin²)/(sin), then it is equal to (1/sin)-sin because sin²/sin just gives you sin
    If it is 1-(sin²/sin), then it gives 1-sin

    1) and 2) => put parenthesis, and someone might help
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    1) It is very easy, just replace the ones with "2x" using the duplication formulae

    [sin(2x)+sin(x)]/(1+cos(2x)+cos(x)) = (2sin(x)cos(x)+sin(x))/(1+(cos²x-sin²x)+cos(x)
    = [(sin(x)(2cos(x)+1)]/[(cos(x)(2cos(x)+1)]
    = sin(x)/cos(x)
    = tan(x)

    2) [(1-cos(2x)+sin(2x)]/[(1-cos(2x)+sin(2x)] = 1...you should have made a mistake here
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    2nd one should have a + after the 1 not a minus
    my mistake

    should be ( 1-cos2θ+sin2θ ) / ( 1+cos2θ+sin2θ ) ≡tanθ
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    then you have
    2) LHS = 1 - (cos²x - sin²x) + 2sin(x)cos(x)
    =2sin²x+2sin(x)cos(x)
    = 2sin(x)(sin(x)+cos(x)

    RHS=1+cos²x-sin²x +2sin(x)cos(x)
    = 2cos²(x)+2sin(x)cos(x)
    =2cos(x)(cos(x)+sin(x)

    LHS/RHS = sin(x)/cos(x) = tan(x)
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    Thanks

    do you know how to get the maximum and minimum of acos(x)+bsin(x)?
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    Well, if you do not want to use calculus, you have
    -1<cos(x)<1 (it can also be EQUAL, but I do not know how to write it without latex)
    Therefore, for a>0, -a<acos(x)<a
    For a<0, then it would change the sign, but basically since you have a on the LHS and RHS, it would not change much

    Then , you do the same for b, you take into account the signs of a and of b, and here you are
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    for example i've done cosx- root3 sinx=2
    in the for Rcos(x-a) gives 2cos(x- [root3/1]

    how would i get the max/min for this?
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    I do not really understand...for f(x)=acos(x)+bsin(x), do you want to solve f(x) = k, where k is a constant, or just find the variations of the function f?
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    i dont know! my teacher taught us this but i didnt get any of it!
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    (Original post by notan123)
    for example i've done cosx- root3 sinx=2
    in the for Rcos(x-a) gives 2cos(x- [root3/1]

    how would i get the max/min for this?
    Can you state the maximum and minimum of f(x) = cos x ?

    What about y = f (x - k) ?

    What about y = 2 f (x - k) ?
 
 
 
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