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    (Original post by Rager6amer)
    Edexcel make you do it but aqa never has so i would hope not but im gonna prepare for it anyway cos you never know
    yeah i hope it doesnt because i just cant remember the proofs for them
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    (Original post by TercioOfParma)
    Hey. I have done a bunch of questions, and while I did get them all correct, I used a different method to what is specified in the mark scheme (I used the binomial formula method and made it (1+ax)^n rather than nCr like is taught at C2). Would I still get full marks for this or would I lose the working marks?
    Please expand upon what you mean because im not fully understanding how you have written it :P
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    Okay, another question, sorry folks

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    Why do they add 90 degrees to get the second answer, I drew out the tan graph and from my drawing it doesn't look like there's an intersection there

    Cheers
    Blake
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    (Original post by Rager6amer)
    Please expand upon what you mean because im not fully understanding how you have written it :P
    So you are aware of binomial expansions? There are two methods of finding the coefficients, nCr and the expansion formula ( 1 + nax + (n(n-1)(ax)^2)/2! ........). The mark schemes I have read relating to questions like (2 - 3x)^5 or whatever only specify solutions based off of nCr, whereas I have used the expansion formula and have removed the 2 by going 2^5(1 - 3/2x)^5 as learned in C4. Would I get full marks using the unspecified formulae or would I have to use nCr in order to get full marks?
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    (Original post by Blake Jones)
    Okay, another question, sorry folks

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    Why do they add 90 degrees to get the second answer, I drew out the tan graph and from my drawing it doesn't look like there's an intersection there

    Cheers
    Blake
    They add 180 degrees to 2x.
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    (Original post by TercioOfParma)
    They add 180 degrees to 2x.
    ah yeah I see, cheers
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    Okay, so I was wondering whether someone could summarise for me how to find the different angles when there are multiple on sin, cos and tan graphs.

    If you get the first value on a tan graph you add or take 180 to get the others right?
    Do you add and take 360 for sin and cos? How do you find their multiple angles?

    Cheers
    Blake
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    (Original post by Rager6amer)
    Yeah sure what is the problem you experience with them when doing these types of questions?
    It's just when they start to throw in summations and things. That's when i get confused.
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    (Original post by TercioOfParma)
    So you are aware of binomial expansions? There are two methods of finding the coefficients, nCr and the expansion formula ( 1 + nax + (n(n-1)(ax)^2)/2! ........). The mark schemes I have read relating to questions like (2 - 3x)^5 or whatever only specify solutions based off of nCr, whereas I have used the expansion formula and have removed the 2 by going 2^5(1 - 3/2x)^5 as learned in C4. Would I get full marks using the unspecified formulae or would I have to use nCr in order to get full marks?
    defo, my teacher taught me that way for C4 as well and taught the C2 year 12's to use that one so im sure its valid!
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    (Original post by Blake Jones)
    Okay, so I was wondering whether someone could summarise for me how to find the different angles when there are multiple on sin, cos and tan graphs.

    If you get the first value on a tan graph you add or take 180 to get the others right?
    Do you add and take 360 for sin and cos? How do you find their multiple angles?

    Cheers
    Blake
    Draw out each graph and use the symmetry, to work out the angles, e.g. you get a value of Sin as 1/2 inverse sin that angle and you will get 30 degrees, using symmetry, 30 degrees will be a solution, and 180 - 30,= 150 will be, also 360 - 30 = 330 will be your solution, if you draw the sine graph and then a line that represents 1/2 you can clearly see this
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    just gone over optimisation still hate it..
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    (Original post by SunDun111)
    defo, my teacher taught me that way for C4 as well and taught the C2 year 12's to use that one so im sure its valid!
    I hope so, thanks a lot
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    (Original post by SunDun111)
    just gone over optimisation still hate it..
    What's that? Does it sometimes have a different name?
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    (Original post by SunDun111)
    Draw out each graph and use the symmetry, to work out the angles, e.g. you get a value of Sin as 1/2 inverse sin that angle and you will get 30 degrees, using symmetry, 30 degrees will be a solution, and 180 - 30,= 150 will be, also 360 - 30 = 330 will be your solution, if you draw the sine graph and then a line that represents 1/2 you can clearly see this
    I do that for the basic trig graphs but do you do the same for 4cos0 for example where 0 is theta?

    I struggle to draw it when it's not the basic graph
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    (Original post by TercioOfParma)
    So you are aware of binomial expansions? There are two methods of finding the coefficients, nCr and the expansion formula ( 1 + nax + (n(n-1)(ax)^2)/2! ........). The mark schemes I have read relating to questions like (2 - 3x)^5 or whatever only specify solutions based off of nCr, whereas I have used the expansion formula and have removed the 2 by going 2^5(1 - 3/2x)^5 as learned in C4. Would I get full marks using the unspecified formulae or would I have to use nCr in order to get full marks?
    Oh i see well in A2 maths i think you need both so you might as well get comfortable with them now but my teacher taught us both and said this: if you are given (1+x)^n use the 1 + nx + n(n-1)/2! (x)^2 formula because its the quickest but if you are given (a+b)^n where a > 1 then you should use the nCr method but both should be absolutely acceptable on the marks scheme yep
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    (Original post by SunDun111)
    just gone over optimisation still hate it..
    Oh hey so basically its obviously just like differentiation but in the form of physical shapes e.g cubes or cylinders. Now i think its pretty easy. You just have to ask yourself what the formula of volume or surface area is e.g. for a cube lets say. Break it down! What is the surface area? Well its the area of 6 different faces; top/bottom, left/right and front/back.

    A random variable e.g. 2x is assgined to each of the length, width and depth so just take a step back and think about it. Work out the area of each of the 3 types of face and multiply the area by 2 because there are 2 for each type, e.g if the height is h and the length of the cube is 2x then the area of that one face is 2xh but you have one on the other side so it becomes 4xh. When you have done this for all faces they normally tell you the numerical value of the surface area or whatever it is so just add together all the algebraic terms of the surface area and set that equal to the numerical value.

    Then they may ask you to find V (volume) or whatever but DONT let it put you off by asking you to put it in an unfamiliar form. For now just concentrate on manipulating the area to get the formula for V. Sometimes they ask you to put V in a form that doesn't involve one of the variables, e.g. without h (only numbers and x's) so just rearrange the above equation to make h the subject and then sub this rearranged formula into the V formula.

    After this they may ask you to calculate dV/dx or whatever but dont be put off by this, think of v as y like in dy/dx and then it becomes easy differentiation. They then might ask to verify a minimum/max point at x = whatever but this is easy once you know that dv/dx at a certain point = 0. I hope this is clear and that this has at least marginally helped you, let me know how you get on with it and dont give up
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    (Original post by Blake Jones)
    I do that for the basic trig graphs but do you do the same for 4cos0 for example where 0 is theta?

    I struggle to draw it when it's not the basic graph
    when you have 4 cos x = a then you would divide a by 4 to get cos x = a/4. Is this what you meant?
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    (Original post by melinalouise)
    thankyou
    no problem mate
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    (Original post by Rager6amer)
    Oh hey so basically its obviously just like differentiation but in the form of physical shapes e.g cubes or cylinders. Now i think its pretty easy. You just have to ask yourself what the formula of volume or surface area is e.g. for a cube lets say. Break it down! What is the surface area? Well its the area of 6 different faces; top/bottom, left/right and front/back.

    A random variable e.g. 2x is assgined to each of the length, width and depth so just take a step back and think about it. Work out the area of each of the 3 types of face and multiply the area by 2 because there are 2 for each type, e.g if the height is h and the length of the cube is 2x then the area of that one face is 2xh but you have one on the other side so it becomes 4xh. When you have done this for all faces they normally tell you the numerical value of the surface area or whatever it is so just add together all the algebraic terms of the surface area and set that equal to the numerical value.

    Then they may ask you to find V (volume) or whatever but DONT let it put you off by asking you to put it in an unfamiliar form. For now just concentrate on manipulating the area to get the formula for V. Sometimes they ask you to put V in a form that doesn't involve one of the variables, e.g. without h (only numbers and x's) so just rearrange the above equation to make h the subject and then sub this rearranged formula into the V formula.

    After this they may ask you to calculate dV/dx or whatever but dont be put off by this, think of v as y like in dy/dx and then it becomes easy differentiation. They then might ask to verify a minimum/max point at x = whatever but this is easy once you know that dv/dx at a certain point = 0. I hope this is clear and that this has at least marginally helped you, let me know how you get on with it and dont give up
    thanks man, kidna nervous, getting 70's in past papers but always feel like i mess up these exams. In core 1 before the exam i was getting high 60's and low 70's, now i can only count 61 marks, and thats due to absolute mistakes would of easily got 65+ If i didnt mess up that integration question.
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    (Original post by RueXO)
    I get confused with cos graphs 😒
    If you remember that the y axis is between -1 and 1 then its fairly easy.

    You don't need to remember them. Simply type Cos(0), Cos(90), Cos(180), Cos(270) and Cos(360) and it will tell you whereabouts on the y axis those points are (always either 0, 1 or -1. You can then plot them on and if you know the shape draw the graph!
 
 
 
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