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    Hi,

    Can you please tell me how to solve the following problems. a) is three expressions I do not know how to simplify and b) is a trigonometric equation whose solution I cannot find. I'd be very grateful if you could tell me the solution process, not just the solution. Thanks so much!

    a) Simplify: sin(pi/2-alpha) cos(pi+alpha) tan(2.pi-alpha)
    What principle do you use do simplify these terms?

    b) cos3x=1
    This one is really tricky. After getting rid of the 3x I get the following equation, which I cannot solve: 4cos³x-3cosx=1.
    Normally I solve these equations through substitution or zero products, but here I just don't know what to do. Any advice is greatly appreciated!
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    Someone else could probably explain a) better than myself, but you should probably know how to do b):

    Cos3x=1
    So, 3x=0

    If you're solving for x, where 0<x<360, then you will be solving for 3x where 0<3x<1080 (which is 360*3)

    Does this help at all? When you've found all the solutions for 3x for the values 0<3x<1080, then you can divide all these answers by 3 to give you answers for x. In cases like this imo get rid of the 3x last (Unless of course you are using identities to rewrite the equation.)
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    (Original post by Chelle-belle)
    Someone else could probably explain a) better than myself, but you should probably know how to do b):

    Cos3x=1
    So, 3x=0

    If you're solving for x, where 0<x<360, then you will be solving for 3x where 0<3x<1080 (which is 360*3)

    Does this help at all? When you've found all the solutions for 3x for the values 0<3x<1080, then you can divide all these answers by 3 to give you answers for x. In cases like this imo get rid of the 3x last (Unless of course you are using identities to rewrite the equation.)
    Hi,
    Thanks for your reply. Further to b), it says on my assignment sheet that I should find a value of x between -pi and pi. I don't quite understand this interval... On the unit circle, pi is 180°, and minus pi would be minus 180°, which is the same as plus 180°. Do you know what this interval means? Does it mean the same as the interval between 0° and 360°?

    Thanks
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    do you know the compound angle formulae?
    sin(A+B)=sinAcosB+cosAsinB
    cos(A+B)=cosAcosB-sinAsinB
    tan(A+B)=(tanA+tanB)/(1-tanAtanB)

    and opposites for opposite signs
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    (Original post by Toneh)
    do you know the compound angle formulae?
    sin(A+B)=sinAcosB+cosAsinB
    cos(A+B)=cosAcosB-sinAsinB
    tan(A+B)=(tanA+tanB)/(1-tanAtanB)

    and opposites for opposite signs
    Yes, I know them and I actually used the second one in my attempt to solve ex. b). I still do not really know how to solve it, though.
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    Do you know the double-angle formulae?

    If not, learn them and how to use them first, there's no point anybody explaining if you don't know what's going on.

    Quote me with whether you do and what you've attempted / got an idea of so far.
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    \sin(\frac{\pi}{2}-\alpha) = \sin(\frac{\pi}{2}) \cos(\alpha) - \cos(\frac{\pi}{2}) \sin(\alpha)
    now you should be able to work out the values of \sin(\frac{\pi}{2}) and \cos(\frac{\pi}{2})
    thus:
    \sin(\frac{\pi}{2}-\alpha) =  \cos(\alpha)

    I think that should be sufficient

    Can you do the rest?
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    (Original post by Toneh)
    now you should be able to work out the values of \sin(\frac{\pi}{2}) and \cos(\frac{\pi}{2})
    =  \frac{1}{\sqrt{2}}
    :facepalm2:

    \dfrac{\pi}{2} is 90 degrees.
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    (Original post by ghostwalker)
    :facepalm2:

    \dfrac{\pi}{2} is 90 degrees.
    oops
    haha
    (I knew I was going to do something stupid - but I couldn't be ****** to check it)
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    (Original post by Bio_Science)
    Hi,
    Thanks for your reply. Further to b), it says on my assignment sheet that I should find a value of x between -pi and pi. I don't quite understand this interval... On the unit circle, pi is 180°, and minus pi would be minus 180°, which is the same as plus 180°. Do you know what this interval means? Does it mean the same as the interval between 0° and 360°?

    Thanks
    So when I said 0<3x<1080, all I did was multiply everything by 3 (because that's what they did with x from cosx to cos3x).

    If the interval was -180<x<180, then for 3x the interval will be -540<3x<540

    If you are looking for -pi<x<pi, then for 3x the interval will be -3pi<3x<3pi

    Once you have worked out all your values of 3x, divide them all by 3 to give you solutions for x There should be quite a few. Remember that pi radians = 180 degrees; this particularly helps me (it might not help you - depends how you work) because I always like to sketch the graph if the interval is not too big.

    Here's a tip by the way: if the interval is in pi, then your answers should probably also be in pi - though if you hate working in pi then you can always work out your answers in degrees and then convert back to pi.
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    (Original post by Toneh)
    \sin(\frac{\pi}{2}-\alpha) = \sin(\frac{\pi}{2}) \cos(\alpha) - \cos(\frac{\pi}{2}) \sin(\alpha)
    now you should be able to work out the values of \sin(\frac{\pi}{2}) and \cos(\frac{\pi}{2})
    thus:
    \sin(\frac{\pi}{2}-\alpha) =  \cos(\alpha)

    I think that should be sufficient

    Can you do the rest?
    Thanks. How did you get the formula \sin(\frac{\pi}{2}-\alpha) = \sin(\frac{\pi}{2}) \cos(\alpha) - \cos(\frac{\pi}{2}) \sin(\alpha)?
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    (Original post by Bio_Science)
    Thanks. How did you get the formula \sin(\frac{\pi}{2}-\alpha) = \sin(\frac{\pi}{2}) \cos(\alpha) - \cos(\frac{\pi}{2}) \sin(\alpha)?
    using the compound angle formula!
    let pi/2 = A, alpha=B
    sin(A-B)=sinAcosB-cosAsinB
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    (Original post by Toneh)
    using the compound angle formula!
    let pi/2 = A, alpha=B
    sin(A-B)=sinAcosB-cosAsinB
    Ohhh, I see. Now I fully understand. I thought there must be more to it than this. But it really is very easy. Thanks so much
 
 
 
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