Maths C3 - Trigonometry... Help??

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    (Original post by notnek)
    I was going to explain how to find \sin 300 but that would involve explaining the exact trig values and using CAST. I don't have time now to do that and don't want to give a short crap explanation.

    Instead I'll be lazy and point you to two videos. Together they will show you how to find things like \sin 300:

    https://www.youtube.com/watch?v=MoyVdhHOV_o

    https://www.youtube.com/watch?v=VBhAtctYy8g

    If I have time tomorrow I can help or maybe someone else will be kind enough to explain.

    By the way, I know students who have got an A* in A Level maths without ever learning the exact trig values, since you're allowed a calculator in the exam. But they're a nice tool to be able to attempt certain questions that require them and really understand all the trig concepts.
    Thank you. The only problem is I mentioned that I already know how to solve things like ...  sin(300)...

    I want to know how to find the angle x for example when...  tan (x) = -1... without using a calculator
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    (Original post by Philip-flop)
    Thank you. The only problem is I mentioned that I already know how to solve things like ...  sin(300)...

    I want to know how to find the angle x for example when...  tan (x) = -1... without using a calculator
    \tan(x)=\frac{\sin(x)}{\cos(x)}=-1 \Rightarrow \sin(x)=-\cos(x)

    Sketch the two graphs and see where they intersect, look for whichever solution is closest to 0 (the principal solution, in other words)
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    (Original post by Philip-flop)
    Thank you. The only problem is I mentioned that I already know how to solve things like ...  sin(300)...

    I want to know how to find the angle x for example when...  tan (x) = -1... without using a calculator
    tan 45 = 1.

    This is one of the exact values.

    So to solve tan x = -1 with CAST, draw in an acute angle of 45 then look for quadrants where tan is negative.
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    (Original post by RDKGames)
    \tan(x)=\frac{\sin(x)}{\cos(x)}=-1 \Rightarrow \sin(x)=-\cos(x)

    Sketch the two graphs and see where they intersect, look for whichever solution is closest to 0 (the principal solution, in other words)
    Thank you!! That's a pretty cool way of doing it! Just quite time consuming. But I got there in the end!

    Would like to know how to use the Trig Ratio triangles for these questions at some point though. Although like notnek said, I'll be allowed a calculator in the exam.

    Saying that. Looking at question 1(g) and (h) they look pretty tricky even with a calculator! Not sure how I'm going to rearrange...  arcsin (sin \frac{\pi}{3})

    EDIT: Actually it was pretty easy to solve using a calculator. But still
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    (Original post by RDKGames)
    For \arctan(-1)= -\arctan(1) you need to know that \tan(\frac{\pi}{4})=1. You can easily find this from your ratio triangles by considering the 45 degree one, and you know that tan of an angle is opposite side divided by the adjacent one. Since both sides are equal, you get 1/1 which is just 1, but this is something you just remember. Another way to think about it is that tan is the gradient of the hypotenuse. You can see this clearly from the unit circle and the gradient is 1 at 45 degrees. Sine would be the change in y, and cosine would be the change in x, so one over the other gives the gradient.

    g and h are pretty easy. You are taking an angle, applying a trigonometric function to it, then applying the inverse of that same trigonometric function. Inverse is like going back a step. So the trig function and its inverse cancel eachother out. It's like dividing by n then proceeding to immediately multiply n on the same thing. Though for g, it is slightly different, try and see why it is so. Hint: remember that \sin(x)=\sin(\pi-x)
    Oh right. I'm sort of starting to understand a bit better now. Still a little bit confused as to why things like...
    cos (-x) = cos(x)... but then...  sin(-x) = -sin (x)... Is it because of their positioning on the graph?

    So for 1(g) I would do this...
     arcsin (sin\frac{\pi}{3})

    Use the trig ratio triangles. The triangle with the angle  \frac{\pi}{3} and use trigonometry's SOH to give me...
     =arcsin (\frac{\sqrt 3}{2})

    arcsin can then be re-written as...
     =sin^-^1 (\frac{\sqrt 3}{2})

    But then what from here?
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    (Original post by Philip-flop)
    Thank you!! That's a pretty cool way of doing it! Just quite time consuming. But I got there in the end!

    Would like to know how to use the Trig Ratio triangles for these questions at some point though. Although like notnek said, I'll be allowed a calculator in the exam.

    Saying that. Looking at question 1(g) and (h) they look pretty tricky even with a calculator! Not sure how I'm going to rearrange...  arcsin (sin \frac{\pi}{3})
    For \arctan(-1)= -\arctan(1) you need to know that \tan(\frac{\pi}{4})=1. You can easily find this from your ratio triangles by considering the 45 degree one, and you know that tan of an angle is opposite side divided by the adjacent one. Since both sides are equal, you get 1/1 which is just 1, but this is something you just remember. Another way to think about it is that tan is the gradient of the hypotenuse. You can see this clearly from the unit circle and the gradient is 1 at 45 degrees. Sine would be the change in y, and cosine would be the change in x, so one over the other gives the gradient.

    g and h are pretty easy. You are taking an angle, applying a trigonometric function to it, then applying the inverse of that same trigonometric function. Inverse is like going back a step. So the trig function and its inverse cancel eachother out. It's like dividing by n then proceeding to immediately multiply n on the same thing.

    Though for h, it is slightly different. To understand what is going on, you need to think back to when you did inverse graphs. You know that a function can ONLY have an inverse if it is one-to-one. The sine function is NOT one-to-one therefore we must restrict the domain to \sin(x), -\frac{\pi}{2}\leq x \leq \frac{\pi}{2} and only then we have our inverse. The problem with \frac{2\pi}{3} is that it lies outside our restricted domain. However, thanks to the symmetry of the function, we know that \sin(x)=\sin(\pi-x) and we can use this to get an angle which is inside our domain while keeping the overall value the same.

    Here is a diagram to illustrate this. The black line is x=\frac{2\pi}{3} and you need to figure out what the purple line is in order to get the correct answer, because the purple one has the same value due to symmetry and how the green dotted line shows it. The orange line represents \arcsin(x), the blue dotted line represents \sin(x) while the red line is the blue line with a restricted domain applied to it.

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    (Original post by Philip-flop)
    Oh right. I'm sort of starting to understand a bit better now. Still a little bit confused as to why things like...
    cos (-x) = cos(x)... but then...  sin(-x) = -sin (x)... Is it because of their positioning on the graph?

    So for 1(g) I would do this...
     arcsin (sin\frac{\pi}{3})

    Use the trig ratio triangles. The triangle with the angle  \frac{\pi}{3} and use trigonometry's SOH to give me...
     =arcsin (\frac{\sqrt 3}{2})

    arcsin can then be re-written as...
     =sin^-^1 (\frac{\sqrt 3}{2})

    But then what from here?
    My post above expands on my previous comment.

    As far your question is concerned, it goes back to \frac{\pi}{3}. You can even see it from your ratio triangle with the angle of \frac{\pi}{3}
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    (Original post by RDKGames)
    My post above expands on my previous comment.

    As far your question is concerned, it goes back to \frac{\pi}{3}. You can even see it from your ratio triangle with the angle of \frac{\pi}{3}
    Oh right yeah. I'm reading it now
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    (Original post by RDKGames)

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    Just realised we're talking about Q1(h). Ok. I understand that for sin to be a function it must be one-to-one, therefore a restricted domain is put in place. And I can see how the symmetry is used between the black and purple line because  \frac{2 \pi}{3} doesn't fall within the restricted domain of \frac{\pi}{2}\leq x \leq \frac{\pi}{2}.

    I'm kinda confused with how sin x = sin (\pi - x) because of the symmetry

    Very good graph btw!! It's helping me visualise the problem so much easier!!
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    (Original post by Philip-flop)
    Just realised we're talking about Q1(h).
    This is about 1g now, you've done h.
    Ok. I understand that for sin to be a function it must be one-to-one, therefore a restricted domain is put in place. And I can see how the symmetry is used between the black and purple line because  \frac{2 \pi}{3} doesn't fall within the restricted domain of \frac{\pi}{2}\leq x \leq \frac{\pi}{2}.

    I'm kinda confused with how sin x = sin (\pi - x) because of the symmetry

    Very good graph btw!! It's helping me visualise the problem so much easier!!
    One way to explain why \sin(x)=\sin(\pi-x) is to use the graph above. With \sin(x) you are going along the x-axis to your chosen x value and then hitting the curve to see what y-value you get. With \sin(\pi-x) you are doing the similar thing, except you are first going forward by \pi radians, and then going BACK by the amount you need. Look on the diagram, you can see that the distance between x=0 and the purple line is the SAME as the distance between x=\pi and the black line as two parts of the curve are symmetrical ABOUT x=\frac{\pi}{2}

    Another way to show it is to use graph transformations:

    1. Start with \sin(x)



    2. Translate by vector [-\pi,0] which means \sin(x) \Rightarrow \sin(x+\pi)



    3. Reflect in the y-axis (x=0): \sin(x+\pi) \Rightarrow \sin(-x+\pi)




    And as you can see, the green graphs is exactly the same as the red one, therefore \sin(x)=\sin(\pi-x)

    Also note, you can use this same transformation technique (with graphs sketches if you wish) to show that \cos(-x)=\cos(x) and \sin(-x)=-\sin(x)
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    (Original post by Philip-flop)
    Just realised we're talking about Q1(h). Ok. I understand that for sin to be a function it must be one-to-one, therefore a restricted domain is put in place. And I can see how the symmetry is used between the black and purple line because  \frac{2 \pi}{3} doesn't fall within the restricted domain of \frac{\pi}{2}\leq x \leq \frac{\pi}{2}.

    I'm kinda confused with how sin x = sin (\pi - x) because of the symmetry

    Very good graph btw!! It's helping me visualise the problem so much easier!!
     \sin x is a function, we restrict the domain so that it has an inverse.
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    (Original post by RDKGames)
    This is about 1g now, you've done h.
    So we're working through Q1(g)...
     arcsin(sin \frac{\pi}{3})

    Oh dear. Now I'm confused as to where the \frac{2 \pi}{3} has come from Likewise, I understand the process of your transformations from the diagrams you showed me, but I'm not sure why it has been transformed in that way? (sorry if that doesn't make sense).

    But overall I think I'm beginning to get a grasp on this question. It's almost like sin is cancelled out by the inverse of sin (arcsin). So that...

     arcsin(sin \frac{\pi}{3})...the angle from sin \frac{\pi}{3} can be used to give a ratio using the trig ratio triangle. This gives...

     arcsin (\frac {\sqrt 3}{2})

    Which is the same as, and can be re-written...
     sin^-^1 (\frac {\sqrt 3}{2})

    to which can sort of be written as...
     sin \theta = \frac {\sqrt 3}{2}  ... Where \theta is the angle we are looking for which can be found using the trig ratio triangles.

    so...
     \theta = \frac{\pi}{3}
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    (Original post by Philip-flop)
    So we're working through Q1(g)...
     arcsin(sin \frac{\pi}{3})

    Oh dear. Now I'm confused as to where the \frac{2 \pi}{3} has come from Likewise, I understand the process of your transformations from the diagrams you showed me, but I'm not sure why it has been transformed in that way? (sorry if that doesn't make sense).

    But overall I think I'm beginning to get a grasp on this question. It's almost like sin is cancelled out by the inverse of sin (arcsin). So that...

     arcsin(sin \frac{\pi}{3})...the angle from sin \frac{\pi}{3} can be used to give a ratio using the trig ratio triangle. This gives...

     arcsin (\frac {\sqrt 3}{2})

    Which is the same as, and can be re-written...
     sin^-^1 (\frac {\sqrt 3}{2})

    to which can sort of be written as...
     sin \theta = \frac {\sqrt 3}{2}  ... Where \theta is the angle we are looking for which can be found using the trig ratio triangles.

    so...
     \theta = \frac{\pi}{3}
    Oh sorry, having looked back at the questions, the the \frac{2\pi}{3} one is indeed 1h. Not sure what I was thinking saying that at 3am, aha, ignore that part.

    You got 1g correct with a solid method. Hopefully 1h now makes sense as well after the explanations.

    The graph has been transformed that way because it has loads of symmetry, so you can do it differently if you wish, but you'd end up with a slightly different form which still gives the right answer.
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    (Original post by RDKGames)
    Oh sorry, having looked back at the questions, the the \frac{2\pi}{3} is indeed 1h. Not sure what I was thinking saying that at 3am, aha, ignore that part.

    You got 1g correct. Hopefully 1h now makes sense as well.
    No worries. I will be moving onto question 1(h) now so part of it already makes sense to me with the help of your graphs
    I seriously owe you one!!!
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    (Original post by Philip-flop)
    So we're working through Q1(g)...
     arcsin(sin \frac{\pi}{3})

    Oh dear. Now I'm confused as to where the \frac{2 \pi}{3} has come from Likewise, I understand the process of your transformations from the diagrams you showed me, but I'm not sure why it has been transformed in that way? (sorry if that doesn't make sense).

    But overall I think I'm beginning to get a grasp on this question. It's almost like sin is cancelled out by the inverse of sin (arcsin). So that...

     arcsin(sin \frac{\pi}{3})...the angle from sin \frac{\pi}{3} can be used to give a ratio using the trig ratio triangle. This gives...

     arcsin (\frac {\sqrt 3}{2})

    Which is the same as, and can be re-written...
     sin^-^1 (\frac {\sqrt 3}{2})

    to which can sort of be written as...
     sin \theta = \frac {\sqrt 3}{2}  ... Where \theta is the angle we are looking for which can be found using the trig ratio triangles.

    so...
     \theta = \frac{\pi}{3}
    These are standard trig angles. But  \arcsin (\sin x )\equiv \sin (\arcsin x)\equiv x .
    This just comes from the fact that when you compose a function with its inverse  f \circ f^{-1}(x)\equiv f^{-1} \circ f(x)\equiv x .
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    (Original post by B_9710)
    These are standard trig angles. But  \arcsin (\sin x )\equiv \sin (\arcsin x)\equiv x .
    This just comes from the fact that when you compose a function with its inverse  f \circ f^{-1}(x)\equiv f^{-1} \circ f(x)\equiv x .
    Of course things get a bit tricky with trig functions, since for example \arcsin \sin x is defined over \mathbb{R} whilst \sin \arcsin x is defined over [-1,1].
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    (Original post by B_9710)
    These are standard trig angles. But  \arcsin (\sin x )\equiv \sin (\arcsin x)\equiv x .
    This just comes from the fact that when you compose a function with its inverse  f \circ f^{-1}(x)\equiv f^{-1} \circ f(x)\equiv x .
     \arcsin (\sin x )\equiv \sin (\arcsin x)\equiv x is not correct. The latter identity is, but the former is not true and is the probably the most misunderstood concept by students when learning trig. \arcsin is not strictly the inverse of \sin.
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    (Original post by IrrationalRoot)
     \arcsin (\sin x )\equiv \sin (\arcsin x)\equiv x is not correct. The latter identity is, but the former is not true and is the probably the most misunderstood concept by students when learning trig. \arcsin is not strictly the inverse of \sin.
    ^ My head hurts
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    (Original post by Philip-flop)
    ^ My head hurts
    I'm assuming your 'head hurts' (lol) because you always thought \sin had an inverse. If so, answer me this question: if \sin x=0, what is x? Is your answer x=0? Wrong, x=2\pi. In fact x could be any even multiple of \pi.

    Hopefully you now see why \sin does not have an inverse. If you're given the value of \sin x, you cannot determine x. \sin is not one-to-one.

    Now that begs the question, what is \arcsin x? I would explain it but it can be quite hard to understand (and will make this post even harder to digest), so it's best to look it up and if you have any questions ask here . The basic idea though is that \arcsin is the inverse of \sin for a certain portion of the domain of \sin. This portion is chosen so that \sin is one-to-one in this portion.

    Anyway, it might make you feel better if I told you that none of my teachers understood that \sin does not actually have an inverse lol. They all assumed that \arcsin(\sin x) \equiv x, which as I've explained isn't true.
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    (Original post by IrrationalRoot)
    ...If so, answer me this question: if \sin x=0, what is x? Is your answer x=0? Wrong, x=2\pi. In fact x could be any even multiple of \pi.
    But 0 is a multiply of \pi so it's not wrong, it's just a principal solution, and it doesn't have to be an even multiple. It would be a mess to lead OP to general solutions of sine at this point.
 
 
 
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