Maths C3 - Trigonometry... Help??

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    Ok. How do I even do part (c) and (d)??

    Name:  C3 Exe 7A Q12(a).png
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    For part (c)... Shall I start off by using the fact that...
     tan \theta = \frac{sin \theta}{cos \theta} ?
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    ^ Anyone??
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    (Original post by Philip-flop)
    Ok. How do I even do part (c) and (d)??

    Name:  C3 Exe 7A Q12(a).png
Views: 13
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    For part (c)... Shall I start off by using the fact that...
     tan \theta = \frac{sin \theta}{cos \theta} ?
    Look back at question 5H that you did successfully a few days back.

    You need to use \tan 45 = 1 again to write the fraction in the form

    \displaystyle \frac{\tan A + \tan B}{1-\tan A \tan B}

    For d), use the fact that \displaystyle \sin 45 = \cos 45 = \frac{1}{\sqrt{2}}

    These are all quite tricky questions that you don't really see in exams. Don't worry too much if you find them hard.
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    (Original post by notnek)
    Look back at question 5H that you did successfully a few days back.

    You need to use \tan 45 = 1 again to write the fraction in the form

    \displaystyle \frac{\tan A + \tan B}{1-\tan A \tan B}

    For d), use the fact that \displaystyle \sin 45 = \cos 45 = \frac{1}{\sqrt{2}}

    These are all quite tricky questions that you don't really see in exams. Don't worry too much if you find them hard.
    Yeah I had a feeling that part (c) was similar to Question 5(h) that I did not long ago. Managed to figure that one out now. But am still stuck on part(d).

    Thanks a lot for your help
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    (Original post by Philip-flop)
    Yeah I had a feeling that part (c) was similar to Question 5(h) that I did not long ago. Managed to figure that one out now. But am still stuck on part(d).

    Thanks a lot for your help
    \displaystyle \frac{1}{\sqrt{2}}\left( \sin \theta + \cos \theta \right) = \frac{1}{\sqrt{2}} \sin \theta + \frac{1}{\sqrt{2}} \cos \theta

    \displaystyle sin 45 = cos 45 =  \frac{1}{\sqrt{2}}

    Try substituting these to get something of the form \sin A \cos B + \sin B \cos A
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    (Original post by notnek)
    \displaystyle \frac{1}{\sqrt{2}}\left( \sin \theta + \cos \theta \right) = \frac{1}{\sqrt{2}} \sin \theta + \frac{1}{\sqrt{2}} \cos \theta

    \displaystyle sin 45 = cos 45 =  \frac{1}{\sqrt{2}}

    Try substituting these to get something of the form \sin A \cos B + \sin B \cos A
    Oh yeah of course! I actually started doing that then realised. So I've done...

     \frac{1}{\sqrt 2} (sin \theta + cos \theta)

     = \frac{1}{\sqrt 2} sin \theta + \frac{1}{\sqrt 2} cos \theta

     = sin \theta cos \frac{\pi}{4} + cos \theta sin \frac{\pi}{4}

     = sin( \theta + \frac{\pi}{4})

    Alternatively I could have done....
     \frac{1}{\sqrt 2} (sin \theta + cos \theta)

     = \frac{1}{\sqrt 2} sin \theta + \frac{1}{\sqrt 2} cos \theta

    = sin \frac{\pi}{4} sin \theta + cos \frac{\pi}{4} cos \theta

     = cos (\frac{\pi}{4} - \theta)

    Am I right??
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    (Original post by Philip-flop)
    Oh yeah of course! I actually started doing that then realised. So I've done...

     \frac{1}{\sqrt 2} (sin \theta + cos \theta)

     = \frac{1}{\sqrt 2} sin \theta + \frac{1}{\sqrt 2} cos \theta

     = sin \theta cos \frac{\pi}{4} + cos \theta sin \frac{\pi}{4}

     = sin( \theta + \frac{\pi}{4})

    Alternatively I could have done....
     \frac{1}{\sqrt 2} (sin \theta + cos \theta)

     = \frac{1}{\sqrt 2} sin \theta + \frac{1}{\sqrt 2} cos \theta

    = sin \frac{\pi}{4} sin \theta + cos \frac{\pi}{4} cos \theta

     = cos (\frac{\pi}{4} - \theta)

    Am I right??
    Yes that's all correct.

    And another way you can see that those two forms are equivalent is by using

    \displaystyle \sin A = \cos \left(\frac{\pi}{2}-A \right) :

    \displaystyle \sin \left( \theta + \frac{\pi}{4} \right) = \cos \left[ \frac{\pi}{2} - \left(\theta + \frac{\pi}{4}\right) \right] = \cos \left(\frac{\pi}{4} - \theta \right)
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    (Original post by notnek)
    Yes that's all correct.

    And another way you can see that those two forms are equivalent is by using

    \displaystyle \sin A = \cos \left(\frac{\pi}{2}-A \right) :

    \displaystyle \sin \left( \theta + \frac{\pi}{4} \right) = \cos \left[ \frac{\pi}{2} - \left(\theta + \frac{\pi}{4}\right) \right] = \cos \left(\frac{\pi}{4} - \theta \right)
    Oh yeah! Because sin is a translation of cos by 90 degrees (to the right) parallel to the x-axis, correct?

    Thanks again! Things are definitely making sense now
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    (Original post by Philip-flop)
    Oh yeah! Because sin is a translation of cos by 90 degrees (to the right) parallel to the x-axis, correct?

    Thanks again! Things are definitely making sense now
    That's basically correct although a shift of \cos x by \frac{\pi}{2} to the right gives

    \cos \left(x-\frac{\pi}{2} \right)

    But as you discovered in an older post, cos is an even function so you have

    \displaystyle \cos \left(x-\frac{\pi}{2}\right) = \cos \left[-\left(x-\frac{\pi}{2}\right) \right] = \cos \left(\frac{\pi}{2}-x\right)
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    (Original post by notnek)
    That's basically correct although a shift of \cos x by \frac{\pi}{2} to the right gives

    \cos \left(x-\frac{\pi}{2} \right)

    But as you discovered in an older post, cos is an even function so you have

    \displaystyle \cos \left(x-\frac{\pi}{2}\right) = \cos \left[-\left(x-\frac{\pi}{2}\right) \right] = \cos \left(\frac{\pi}{2}-x\right)
    Yes, cos is an "even function" due to it's symmetry on either side of the y-axis. Whereas sin is an odd function.
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    Ok so I'm stuck again already

    Name:  C3 EXE7A Q13.png
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    Have I even been doing this one right??
    Attachment 584984584986

    Then I don't really know where to go from there.
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    (Original post by Philip-flop)
    Ok so I'm stuck again already

    Name:  C3 EXE7A Q13.png
Views: 17
Size:  6.6 KB

    Have I even been doing this one right??
    Attachment 584984584986

    Then I don't really know where to go from there.
    You have 3\cos \theta on one side of the equation and \sqrt{3}\cos \theta on the other.

    You can move them to the same side and combine them.
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    (Original post by notnek)
    You have 3\cos \theta on one side of the equation and \sqrt{3}\cos \theta on the other.

    You can move them to the same side and combine them.
    Yeah I thought about doing that, so...

     (3- \sqrt 3) cos\theta = sin \theta

    But then I had no idea where to go from there so I just assumed I was wrong
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    (Original post by Philip-flop)
    Yeah I thought about doing that, so...

     (3- \sqrt 3) cos\theta = sin \theta

    But then I had no idea where to go from there so I just assumed I was wrong
    Now divide by cosine to get tan and solve it that way.
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    (Original post by RDKGames)
    Now divide by cosine to get tan and solve it that way.
    Omg, I'm actually going to kick myself!! Managed to get there in the end.

     tan \theta = 3- \sqrt 3

    and then solved it from there to get...  \theta = 51.7,  231.7







    Thank you RDKGames and notnek I'm forever grateful for your help!!
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    Name:  C3 EXE7A Q13.png
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    For part (c) should I start by doing?...

     cos (\theta + 25) + sin (\theta +65) = 1

     (cos \theta cos 25 - sin \theta sin 25) + (sin \theta cos 65 +cos \theta sin65) = tan 45

     (cos \theta cos 25 - sin \theta sin 25) + (sin \theta cos 65 +cos \theta sin65) = \frac{sin45}{cos45}

    and then should I times by cos(45)?? Or am I being completely retarded here? actually have no idea what I'm doing.
    Why do I suck at trigonometry?
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    (Original post by Philip-flop)
    Name:  C3 EXE7A Q13.png
Views: 17
Size:  6.6 KB

    For part (c) should I start by doing?...

     cos (\theta + 25) + sin (\theta +65) = 1

     (cos \theta cos 25 - sin \theta sin 25) + (sin \theta cos 65 +cos \theta sin65) = tan 45

     (cos \theta cos 25 - sin \theta sin 25) + (sin \theta cos 65 +cos \theta sin65) = \frac{sin45}{cos45}

    and then should I times by cos(45)?? Or am I being completely retarded here? actually have no idea what I'm doing.
    Why do I suck at trigonometry?
    Since 25+65 = 90 we have

    \cos 65 = \sin 25 and \cos 25 = \sin 65

    Try using this here and look out for angle sum of 90 in future questions - you need it in different types of trig questions.

    Also, don't immediately change 1 into \tan 45 unless you think it will be useful. It was useful for a couple of questions that you've done but not here.
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    (Original post by notnek)
    Since 25+65 = 90 we have

    \cos 65 = \sin 25 and \cos 25 = \sin 65

    Try using this here and look out for angle sum of 90 in future questions - you need it in different types of trig questions.

    Also, don't immediately change 1 into \tan 45 unless you think it will be useful. It was useful for a couple of questions that you've done but not here.
    Ohhhh I see.

    I never knew that if...  cos(a) and sin(b) where a and b sum up to 90 degrees that...  cos(a) = sin(b)
    Again, the book fails to explain any of that!

    Ok, so for this question I would do?...

     cos (\theta + 25) + sin (\theta +65) = 1

     (cos \theta cos 25 - sin \theta sin 25) + (sin \theta cos 65 +cos \theta sin65) = 1

     (cos \theta sin 65 - sin \theta cos 65) + (sin \theta cos 65 +cos \theta sin65) = 1

     2(cos \theta sin65) = 1

    But then where from there? :/
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    (Original post by Philip-flop)
    Ohhhh I see.

    I never knew that if...  cos(a) and sin(b) where a and b sum up to 90 degrees that...  cos(a) = sin(b)
    Again, the book fails to explain any of that!

    Ok, so for this question I would do?...

     cos (\theta + 25) + sin (\theta +65) = 1

     (cos \theta cos 25 - sin \theta sin 25) + (sin \theta cos 65 +cos \theta sin65) = 1

     (cos \theta sin 65 - sin \theta cos 65) + (sin \theta cos 65 +cos \theta sin65) = 1

     2(cos \theta sin65) = 1

    But then where from there? :/
    From there \sin 65 is just a number you can work out on your calculator. So you can move it to the other side along with the 2.

    The 'adding to 90' rule is just an application of the identity \sin x = \cos \left(90-x\right) e.g. where x is 25.
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    (Original post by notnek)
    From there \sin 65 is just a number you can work out on your calculator. So you can move it to the other side along with the 2.

    The 'adding to 90' rule is just an application of the identity \sin x = \cos \left(90-x\right) e.g. where x is 25.
    Thank you! Not sure why I panicked when I saw...
     2(cos \theta sin65) = 1

 
 
 
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