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    I need help with the question labeled 'e' and '3'

    I dont even know were to start

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    (Original post by User32432432)
    I need help with the question labeled 'e' and '3'

    I dont even know were to start

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    For e. I would suggest multiplying by 2 to get rid of the fraction.

    for 3. Use the trig identitiy  Sin^2 + Cos^2 = 1 and rearranging gives:  Sin^2 = 1-Cos^2
    Now sub that in  Sin^2 and you should be okay
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    (Original post by User32432432)
    I need help with the question labeled 'e' and '3'

    I dont even know were to start

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    You know that \sin^2 \theta + \cos^2 \theta=1 so from there:

    \sin^2 \theta = 1 - \cos^2 \theta

    you now have: 1 - \cos^2 \theta = 4 \cos \theta

    rearrange to get: 4 \cos \theta + \cos^2 \theta -1 = 0

    You can see this is in the form of a quadratic if you let \cos \theta = y

    y^2 + 4y - 1 = 0

    factorise to get  y = -2 + \sqrt 5 , -2 - \sqrt 5 (use quadratic formula or complete the square for this)

    because you let \cos \theta = y, now \cos \theta = -2 + \sqrt 5 , -2 - \sqrt 5

    (HINT: -2 - \sqrt 5 is less than -1 so it is not a valid solution as \cos \theta must be between -1 and 1)

    for e, I would first let \frac{\theta}{2} = y, then solve for y in \cos y

    when you have the solutions for y, get solutions of y = \frac{\theta}{2} , and times all answers by 2 to get \theta

    Hope I helped!
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    for e. use the formula for cos(2x) and let x = theta / 2
    for 3. use the identity sin^2(x) + cos^2(x) = 1 , rearrange and you can get a quadratic of cos(theta)
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    (Original post by Anythingoo1)
    You know that \sin^2 \theta + \cos^2 \theta=1 so from there:

    \sin^2 \theta = 1 - \cos^2 \theta

    you now have: 1 - \cos^2 \theta = 4 \cos \theta

    rearrange to get: 4 \cos \theta + \cos^2 \theta -1 = 0

    You can see this is in the form of a quadratic if you let \cos \theta = y

    y^2 + 4y - 1 = 0

    factorise to get  y = -2 + \sqrt 5 , -2 - \sqrt 5 (use quadratic formula or complete the square for this)

    because you let \cos \theta = y, now \cos \theta = -2 + \sqrt 5 , -2 - \sqrt 5

    (HINT: -2 - \sqrt 5 is less than -1 so it is not a valid solution as \cos \theta must be between -1 and 1)

    for e, I would first let \frac{\theta}{2} = y, then solve for y in \cos y

    when you have the solutions for y, get solutions of y = \frac{\theta}{2} , and times all answers by 2 to get [latex]theta[\latex]

    Hope I helped!
    Wow! Yes you did help Thanks sooo much <3
    To + rep do I just like your posts ? if so comment and I'll like more and if you could look at this

    http://www.thestudentroom.co.uk/show....php?t=2297725
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    (Original post by User32432432)
    Wow! Yes you did help Thanks sooo much <3
    To + rep do I just like your posts ? if so comment and I'll like more and if you could look at this

    http://www.thestudentroom.co.uk/show....php?t=2297725
    Yeah but there's a limit to how many you can +rep per day I think, it's no problem at all it's helpful for me because I'm trying to familiarise myself with latex, and with rep I'm not doing this for rep I just like helping people, but I won't say no if you're offering

    If you have any more questions I'm happy to help.
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    (Original post by Anythingoo1)
    Yeah but there's a limit to how many you can +rep per day I think, it's no problem at all it's helpful for me because I'm trying to familiarise myself with latex, and with rep I'm not doing this for rep I just like helping people, but I won't say no if you're offering

    If you have any more questions I'm happy to help.
    for part e can you confirm my answers ?

    I got 135,-135,225,-225


    Thanks

    EDIT: for part 3 I have Sin^2=4Cos(-2+5^1/2)

    What do I do from there ?
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    (Original post by User32432432)
    for part e can you confirm my answers ?

    I got 135,-135,225,-225


    Thanks

    EDIT: for part 3 I have Sin^2=4Cos(-2+5^1/2)

    What do I do from there ?
    Yes solutions are fine for part e, just remember that \cos -\theta = \cos \theta

    Erm, not really sure what you've done there, see my above working guide...
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    (Original post by Anythingoo1)
    Yes solutions are fine for part e, just remember that \cos -\theta = \cos \theta

    Erm, not really sure what you've done there, see my above working guide...
    I checked what I did..... I though you just sub in the value of CosX into the equation ?

    So I got Sin^2X=4Cos(-2+5^1/2)

    But I'm not sure what to do from there ?
 
 
 
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