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    part of a question can someone help me express 1+sinX in terms of half angles.


    part fo the queston says

    by using the identity for 1+sinx in terms of half angles.....

    its for finding arc length its review exercise Q67

    thanks for any help
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    (Original post by david cool)
    part of a question can someone help me express 1+sinX in terms of half angles.


    part fo the queston says

    by using the identity for 1+sinx in terms of half angles.....

    its for finding arc length its review exercise Q67

    thanks for any help
    wot is the identity of 1+sinx for half angles?
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    dunno wot question is, but off top of my head id do
    1+sinx = 1+2sin(x/2)cos(x/2) maybe?
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    Good point since sin x = sin 2(x/2)
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    √ cos²x - cos2x = sinx
    √ cos²x - (2cos²x -1) = sinx
    √1 - cos²x = sinx

    na no idea
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    i've got a question as well guys,... its on page34 of the p5 book... it says

    a curve C is given by x=2cos t, y=sin t, 0<=t<2pi, wher t is a parameter. find the area enclosed by C.

    so far i've don this...

    using the identity sin^2x + cos^2x =1 we get

    <x/2>^2 + y^2 = 1

    y = sq root <1 - 1/4x>

    then i dunno how to integrate it or wot limits i need...

    n e help wud B great....

    Dake
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    (Original post by david cool)
    part of a question can someone help me express 1+sinX in terms of half angles.


    part fo the queston says

    by using the identity for 1+sinx in terms of half angles.....

    its for finding arc length its review exercise Q67

    thanks for any help
    I remember this question and thought it was quite clever.
    1+sinx=[(cos0.5x)^2+(sin0.5x)^2]+2sin(0.5x)cos(0.5x)
    1+sinx=(sin0.5x)^2+2sin(0.5x)cos (0.5x)+(cos0.5x)^2
    What do you now notice about the expression on the right that will help you to perfrom the integration?
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    (Original post by DaKe)
    i've got a question as well guys,... its on page34 of the p5 book... it says

    a curve C is given by x=2cos t, y=sin t, 0<=t<2pi, wher t is a parameter. find the area enclosed by C.

    so far i've don this...

    using the identity sin^2x + cos^2x =1 we get

    <x/2>^2 + y^2 = 1

    y = sq root <1 - 1/4x>

    then i dunno how to integrate it or wot limits i need...

    n e help wud B great....

    Dake
    If you want to do it in cartesian form then you'll notice that the extreme values of x are +-2, and the curve is similar in all 4 quadrants. Thus, you can use 4INTy dx with limits 0,2.
    Alternatively you could have used INT y.dx/dt dt
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    (Original post by Gaz031)
    I remember this question and thought it was quite clever.
    1+sinx=[(cos0.5x)^2+(sin0.5x)^2]+2sin(0.5x)cos(0.5x)
    1+sinx=(sin0.5x)^2+2sin(0.5x)cos (0.5x)+(cos0.5x)^2
    What do you now notice about the expression on the right that will help you to perfrom the integration?
    HAHA! Quadratic, anyway wouldnt this be simpler by intergrating parametrically?

    INT y dt/dx dx?
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    (Original post by Womble548)
    HAHA! Quadratic, anyway wouldnt this be simpler by intergrating parametrically?
    It's not just a quadratic. It's also a perfect ....
 
 
 
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