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    Hi, I have a proving identity question in Core 3 trigonometry.

    cos(x + pi/3) + root3sinx = sin(x + pi/6)

    I have got as far as:

    cosxcospi/3 - sinxsinpi/3 + root3sinx,

    1/2cosx - root3/2sinx + root3sinx,

    1/2cosx + root3/2sinx,

    From here though, I am at a loss of where to go next. I know I have to make the sin addition formula, but I do not know how.

    Can someone help please?
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    (Original post by Tygra)
    Hi, I have a proving identity question in Core 3 trigonometry.

    cos(x + pi/3) + root3sinx = sin(x + pi/6)

    I have got as far as:

    cosxcospi/3 - sinxsinpi/3 + root3sinx,

    1/2cosx - root3/2sinx + root3sinx,

    1/2cosx + root3/2sinx,

    From here though, I am at a loss of where to go next. I know I have to make the sin addition formula, but I do not know how.

    Can someone help please?
    What you have is correct.

    Now you need to show that 1/2cosx + root3/2sinx=sin(x + pi/6). You can do this by using the addition rules on the right hand side. So expand out the sin(x + pi/6).
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    (Original post by Tygra)
    Hi, I have a proving identity question in Core 3 trigonometry.

    cos(x + pi/3) + root3sinx = sin(x + pi/6)

    I have got as far as:

    cosxcospi/3 - sinxsinpi/3 + root3sinx,

    1/2cosx - root3/2sinx + root3sinx,

    1/2cosx + root3/2sinx,
    \frac{1}{2}\cos x + \frac{\sqrt3}{2}\sin x = \sin (\frac{\pi}{6})\cos x + \cos (\frac{\pi}{6})\sin x
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    What I mean is guys is how do I get the pi/6 from the left-hand side?

    Well actually the inverse of sin(1/2) and the inverse of cos(root3/2) is pi/6, but when dealing with proving trig identities I've never had to inverse any of the trig functions before.
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    Set it equal to Rsin(x + a) and you'll find the value of R is 1 and the value of a is pi/6


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    (Original post by Tygra)
    What I mean is guys is how do I get the pi/6 from the left-hand side?

    Well actually the inverse of sin(1/2) and the inverse of cos(root3/2) is pi/6, but when dealing with proving trig identities I've never had to inverse any of the trig functions before.
    It is expected that you know and recognise the trig ratios for 0,30,45,60,90
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    Yes I do know them, I wasn't sure that was the root I had to take.

    Thanks for your help.
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    Hi again guys, thanks for all your help.

    I'm working through the C3 Edexcel trig section. There's a set of questions that are giving me problems. The first four I can do quite comfortably. It is all based around double-angled formulae.

    The first four are very easy to see which double angle formulae you use. Two of the questions are simplify to one trig function 6sin2xcos2x and 2-4sin^2x/2. I have no problems here. However, a couple of the harder ones are:

    squareroot of 1+ cos2x and

    4sinxcosxcos2x.

    The first harder example I do not know even where to start and the second I get as far as:

    (2sin2x)(1-2sin^2x)

    2sin2x - 4sin^3 2x

    Please can you help?
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    (Original post by Tygra)
    Hi again guys, thanks for all your help.

    I'm working through the C3 Edexcel trig section. There's a set of questions that are giving me problems. The first four I can do quite comfortably. It is all based around double-angled formulae.

    The first four are very easy to see which double angle formulae you use. Two of the questions are simplify to one trig function 6sin2xcos2x and 2-4sin^2x/2. I have no problems here. However, a couple of the harder ones are:

    squareroot of 1+ cos2x and

    4sinxcosxcos2x.

    The first harder example I do not know even where to start and the second I get as far as:

    (2sin2x)(1-2sin^2x)

    2sin2x - 4sin^3 2x

    Please can you help?
    For the first one, if you're taking the square root of something, it would be nice if that "something" were already a square. Do you know any identity involving cos2x that contains squares in it (hint: you're going to want to get rid of that 1 at the start, so you're looking for something that says "cos2x = A - 1" where hopefully the A has a square in it!)

    For the second one I think you've gone down the wrong route. Can you convert the sinxcosx into something else, and then can you combine the "something else" with the cos2x using a repeat of the double angle formula?
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    Hi Davros, thanks for your help.

    With the first example the squareroot is over the whole expression. Did you realise that? The way I wrote it was a bit ambiguous.
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    (Original post by Tygra)
    Hi Davros, thanks for your help.

    With the first example the squareroot is over the whole expression. Did you realise that? The way I wrote it was a bit ambiguous.
    Yes I did!

    That's why I said you want to "get rid of the 1" at some point i.e. it's helpful if you can find an identity which helps you write cos2x as "something - 1"
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    Yes Davros, the identity your hinting towards is the 2cos^2x -1. I'm still a bit lost though.
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    (Original post by Tygra)
    Yes Davros, the identity your hinting towards is the 2cos^2x -1. I'm still a bit lost though.
    So what's the problem? Write down what you've got so far
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    I have so far:

    squareroot of 1 + cos2x,

    squareroot of 1 + 2cos^2x -1,

    squareroot of 2cos^2x
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    (Original post by Tygra)
    I have so far:

    squareroot of 1 + cos2x,

    squareroot of 1 + 2cos^2x -1,

    squareroot of 2cos^2x
    You need to quote me if you want me to see that you've replied

    OK, so what you've written can be simplified to

    \sqrt{2}|\cos 2x|

    (N.B. if your book isn't very precise they may have left out the modulus signs! What is the answer given?)
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    (Original post by davros)
    You need to quote me if you want me to see that you've replied

    OK, so what you've written can be simplified to

    \sqrt{2}|\cos 2x|

    (N.B. if your book isn't very precise they may have left out the modulus signs! What is the answer given?)
    Hi Davros,

    The answer is:

    \sqrt{2}cosx

    There are no modulus signs and cosx is not a double angle.
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    (Original post by Tygra)
    Hi Davros,

    The answer is:

    \sqrt{2}cosx

    There are no modulus signs and cosx is not a double angle.
    Yes, sorry - I misread your cos^2x as a double angle.

    The technically correct answer is \sqrt{2}|\cos x| unless you know for certain that cos x is positive in which case you can omit the modulus signs.
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    (Original post by davros)
    Yes, sorry - I misread your cos^2x as a double angle.

    The technically correct answer is \sqrt{2}|\cos x| unless you know for certain that cos x is positive in which case you can omit the modulus signs.
    Davros, could you show me the steps you take to get the answer please? I'm sure I could learn from that.
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    (Original post by Tygra)
    Davros, could you show me the steps you take to get the answer please? I'm sure I could learn from that.
    You've already got to \sqrt{2\cos^2x} = \sqrt{2} \times \sqrt{\cos^2x}

    So the only issue is what to do with the square root of a square.

    If the domain of x is restricted so that you know that cos x is always positive, then you can write down \sqrt{\cos^2 x} = \cos x but because you don't know whether cos x is positive or negative you technically need a modulus sign around the cos x. I suspect the author of the book has been a bit lazy or hasn't considered the possibility that cos x could be negative!
 
 
 
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