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Maths C3 - Trigonometry... Help?? Watch

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    (Original post by Philip-flop)
    Thanks for making realise that! I'm still a little stuck on this question though

    If  tan \theta = \frac{3}{4} how do I find  tan \frac{\theta}{4} ?

    Sorry if I'm sounding pretty silly, but I've never come across a question like this
    I would simply find what \theta is (as an exact value) from the given condition since we know it is bounded \pi<\theta<\frac{3}{2}\pi. Then substitute it into the formula.
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    (Original post by Philip-flop)
    Thanks for making realise that! I'm still a little stuck on this question though

    If  tan \theta = \frac{3}{4} how do I find  tan \frac{\theta}{4} ?

    Sorry if I'm sounding pretty silly, but I've never come across a question like this
    Forgot to mention this, but a more straightforward method of doing this is to do the following mapping: \displaystyle 2\theta \mapsto \theta \Rightarrow \tan(\theta)=\frac{2 \tan(\frac{\theta}{2})}{1-\tan^2(\frac{\theta}{2})} and you know the value of \tan(\theta) so just solve for \tan(\frac{\theta}{2}) by forming a quadratic. Your final angle should satisfy the bound of it.
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    (Original post by RDKGames)
    I would simply find what \theta is (as an exact value) from the given condition since we know it is bounded \pi<\theta<\frac{3}{2}\pi. Then substitute it into the formula.
    Ok so solving  tan \theta = \frac{3}{4}

    I get...  \theta = 0.6435011088

    then I have to substitute theta into...  tan \frac{\theta}{2} = \frac{2 tan \frac{\theta}{4}}{1-tan^2 \frac{\theta}{4}} ??

    the only problem is, I don't think I even know how to type something like  tan^2() into my calculator as obviously calculators wont allow you to put 'squared' after tan just before the angle inside the brackets ()
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    (Original post by Philip-flop)
    Ok so solving  tan \theta = \frac{3}{4}

    I get...  \theta = 0.6435011088

    then I have to substitute theta into...  tan \frac{\theta}{2} = \frac{2 tan \frac{\theta}{4}}{1-tan^2 \frac{\theta}{4}} ??

    the only problem is, I don't think I even know how to type something like  tan^2() into my calculator as obviously calculators wont allow you to put 'squared' after tan just before the angle inside the brackets ()
    Think about what \tan^2\theta means : it means "\tan\theta squared" i.e. \left(\tan\theta\right)^2

    So to work out something like \tan^2 1.2 on your calculator you would type in (\tan 1.2)^2.
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    (Original post by notnek)
    Think about what \tan^2\theta means : it means "\tan\theta squared" i.e. \left(\tan\theta\right)^2


    So to work out something like \tan^2 1.2 on your calculator you would type in (\tan 1.2)^2.
    Thank you it turns out I knew how to type it into my calculator but my workings were letting me down!
    But I managed to get there in the end!!

     \theta = 0.6435011088, 3.785093762

    But since...  \pi < \theta < \frac{3 \pi}{2}

    then...
     \theta \not= 0.6435011088 

\theta = 3.785093762

    Then sub theta into... tan \frac{\theta}{2} = \frac{2 tan \frac{\theta}{4}}{1-tan^2 \frac{\theta}{4}}

    So this gives me...  tan \frac{\theta}{2} = \frac{2 tan (\frac{3.785..}{4})}{1-tan^2 (\frac{3.785..}{4})}

     tan \frac{\theta}{2} = -3


    Thanks a lot for your help!

    ...And thanks RDKGames too! You really helped me out on this one!
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    So I've found myself hitting a brick wall (yet again) with another question...
    Name:  C3 EXE7B Q10.png
Views: 24
Size:  2.6 KB

    I've drawn the following...
    Attachment 586832586834

    I realise that I can't use what I've learnt from C2 and apply the 'cosine rule' or 'sine rule'. I can't seem to progress anywhere with this question. All I can work out from the diagram that I've drawn is that angleBAC =  180-3 \theta

    Unless I have to think of the line AC = 5cm as the Hypotenuse and use the fact that H^2 = O^2 + A^2 in order to find the opposite side of the triangle? But this only works with right angled triangles
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    (Original post by Philip-flop)
    So I've found myself hitting a brick wall (yet again) with another question...
    Name:  C3 EXE7B Q10.png
Views: 24
Size:  2.6 KB

    I've drawn the following...
    Attachment 586832586834

    I realise that I can't use what I've learnt from C2 and apply the 'cosine rule' or 'sine rule'. I can't seem to progress anywhere with this question. All I can work out from the diagram that I've drawn is that angleBAC =  180-3 \theta
    Remember \frac{sin(A)}{a} =\frac{sin(B)}{b} and sin2A = 2sin(A)cos(A)
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    (Original post by NotNotBatman)
    Remember \frac{sin(A)}{a} =\frac{sin(B)}{b} and sin2A = 2sin(A)cos(A)
    Oh right. So I was wrong, I do actually have to use the 'sine rule'. I thought I could only do that if I have two side lengths (which there are) and at least one of their opposite angles (in degrees or radians) to find the missing angle, but both angles are missing, so what do I do?
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    (Original post by Philip-flop)
    Oh right. So I was wrong, I do actually have to use the 'sine rule'. I thought I could only do that if I have two side lengths (which there are) and at least one of their opposite angles (in degrees or radians) to find the missing angle, but both angles are missing, so what do I do?
    In this case you can use the sine rule in terms of theta rearrange and use double angle identities, then solve.
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    (Original post by NotNotBatman)
    In this case you can use the sine rule in terms of theta rearrange and use double angle identities, then solve.
    Ok, so using the sine rule I would do...

     \frac{sin2 \theta}{5} = \frac{sin \theta}{4}

     sin2 \theta = \frac{5sin \theta}{4}

    Apply double angle formula for sin...
     2sin \theta cos \theta = \frac{5sin \theta}{4}

    not sure I actually know what I'm doing here/next
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    (Original post by Philip-flop)
    Ok, so using the sine rule I would do...

     \frac{sin2 \theta}{5} = \frac{sin \theta}{4}

     sin2 \theta = \frac{5sin \theta}{4}

    Apply double angle formula for sin...
     2sin \theta cos \theta = \frac{5sin \theta}{4}

    not sure I actually know what I'm doing here/next
    rearrange so you have 0 on the RHS, factorise and solve each factor equaling 0. Then reject the obvious wrong one.
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    (Original post by NotNotBatman)
    rearrange so you have 0 on the RHS, factorise and solve each factor equaling 0. Then reject the obvious wrong one.
    Thanks a lot! Managed to work out the answer as \theta = 51.31781255
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    Can someone try to help understand this question...
    Name:  C3 EXE7B Q12.png
Views: 39
Size:  4.6 KB

    I've drawn out a little diagram of what I think this question is describing...
    Attachment 586964586966

    I understand that tan is +ve in the 1st & 3rd quadrant which shows for  y= \frac{3}{4}x ... But how do I find out the value of  tan \theta ? Does it have something to do with the gradient of line L :  y = \frac{3}{4}x ??
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    (Original post by Philip-flop)
    Can someone try to help understand this question...
    Name:  C3 EXE7B Q12.png
Views: 39
Size:  4.6 KB

    I've drawn out a little diagram of what I think this question is describing...


    I understand that tan is +ve in the 1st & 3rd quadrant which shows for  y= \frac{3}{4}x ... But how do I find out the value of  tan \theta ? Does it have something to do with the gradient of line L :  y = \frac{3}{4}x ??
    \tan(\theta) is the gradient. It is easy to see if you refer to your unit circle; as sine is the vertical distance, and cosine is the horizontal distance. Vertical over horizontal gives the gradient, or tan in this case.
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    (Original post by RDKGames)
    \tan(\theta) is the gradient. It is easy to see if you refer to your unit circle; as sine is the vertical distance, and cosine is the horizontal distance. Vertical over horizontal gives the gradient, or tan in this case.
    Oh yeah because of "SOH" and "CAH"...
    \frac{sin \theta}{1} = sin \theta which gives the vertical distance and...
    \frac{cos \theta}{1} = sin \theta which gives the horizontal distance.

    This means coordinates of where y= \frac{3}{4}x meets the unit circle is...
     (cos \theta , sin \theta)
    (4,3)

    because...  tan \theta = \frac{sin \theta}{cos \theta} = \frac{3}{4}


    Thank you soooooo much
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    Came across this question last night (part (i)), and have come home from work to attempt it again but I still find myself stuck
    Name:  C3 EXE7C Q1.png
Views: 36
Size:  8.4 KB

    For part (i) I've managed to get up to...
     \frac{1-tan x}{1+tan x}

    I've used the identity  tan \theta = \frac{sin \theta}{cos\theta} and have done a lot of re-arranging of the equation only to end up with...
     \frac{cos^3x - sinx}{cos^3x + sinx} which I have a feeling isn't right and is stopping me from progressing any further :/
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    (Original post by Philip-flop)
    Came across this question last night (part (i)), and have come home from work to attempt it again but I still find myself stuck
    Name:  C3 EXE7C Q1.png
Views: 36
Size:  8.4 KB

    For part (i) I've managed to get up to...
     \frac{1-tan x}{1+tan x}

    I've used the identity  tan \theta = \frac{sin \theta}{cos\theta} and have done a lot of re-arranging of the equation only to end up with...
     \frac{cos^3x - sinx}{cos^3x + sinx} which I have a feeling isn't right and is stopping me from progressing any further :/
    Your working is correct until you said

     \displaystyle \frac{cos^3x - sinx}{cos^3x + sinx}

    It should be this instead

     \displaystyle \frac{cos x - sin x}{cos x + sin x}

    From here I would recommend multiplying top and bottom by

     \displaystyle cos x - sin x


    Have another go and post your working if you still get

     \displaystyle \frac{cos^3x - sinx}{cos^3x + sinx}
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    (Original post by notnek)
    Your working is correct until you said

     \displaystyle \frac{cos^3x - sinx}{cos^3x + sinx}

    It should be this instead

     \displaystyle \frac{cos x - sin x}{cos x + sin x}

    From here I would recommend multiplying top and bottom by

     \displaystyle cos x - sin x


    Have another go and post your working if you still get

     \displaystyle \frac{cos^3x - sinx}{cos^3x + sinx}
    But how do I get from...
     \frac{1-tan x}{1+tan x}

    to..
    \frac{cos x - sin x}{cos x + sinx}??
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    (Original post by Philip-flop)
    But how do I get from...
     \frac{1-tan x}{1+tan x}

    to..
    \frac{cos x - sin x}{cos x + sinx}??
    Multiply top and bottom by cosine.
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    (Original post by Philip-flop)
    But how do I get from...
     \frac{1-tan x}{1+tan x}

    to..
    \frac{cos x - sin x}{cos x + sinx}??
     \displaystyle \frac{1-tan x}{1+tan x} = \frac{1-\frac{\sin x}{\cos x}}{1+\frac{\sin x}{\cos x}}

    The next step is to multiply top and bottom of the fraction by \cos x.
 
 
 
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