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C3 Rcosalpha - coordinates where graph crosses the axis? How do I do it? Watch

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    How do I do part B? I don't have a clue how to do it, and I can't find any videos on YouTube either.

    For part a, I got 6.5cos(x-0.395), which I guess would help slightly but I don't know how that links in either.

    Please can someone walk me through it? Thank you
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    differentiate y and set it to 0 and find x coordinates?
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    (Original post by vector12)
    How do I do part B? I don't have a clue how to do it, and I can't find any videos on YouTube either.

    For part a, I got 6.5cos(x-0.395), which I guess would help slightly but I don't know how that links in either.

    Please can someone walk me through it? Thank you
    Just do what you always did at GCSE: set y=0 and then x=0. Use the original form for the exact answers.

    x=0 \Rightarrow y=6 \cos(0)+2.5 \sin(0) and this gives the y-intercept.

    y=0 \Rightarrow 0=6\cos(x)+2.5\sin(x) will give the x-intercepts. [HINT: You can get tan(x) from this form, then easily solve for the two x-values in range, but you can also use your R-alpha form too here (if you prefer) since you'll round anyway and don't need the exact answer]
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    (Original post by sulaimanali)
    differentiate y and set it to 0 and find x coordinates?
    It's not asking for stationary points...
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    (Original post by RDKGames)
    Just do what you always did at GCSE: set y=0 and then x=0. Use the original form for the exact answers.

    x=0 \Rightarrow y=6 \cos(0)+2.5 \sin(0) and this gives the y-intercept.

    y=0 \Rightarrow 0=6\cos(x)+2.5\sin(x) will give the x-intercepts. [HINT: You can get tan(x) from this form, then easily solve for the two x-values in range, but you can also use your R-alpha form too here (if you prefer) since you'll round anyway and don't need the exact answer]
    Okay, so I replace my 6.5cos(x-0.395) with the 6cosx+2.5sinx = 0, so then I have 6.5cos(x-0.395)=0, and I then divide by 6.5 to get cos(x-0.395)=0. How do I then solve that (to get the two coordinates)??
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    (Original post by vector12)
    Okay, so I replace my 6.5cos(x-0.395) with the 6cosx+2.5sinx = 0, so then I have 6.5cos(x-0.395)=0, and I then divide by 6.5 to get cos(x-0.395)=0. How do I then solve that (to get the two coordinates)??
    Should be a straightforward trig equation to solve from C2. Might help if you revisit that briefly.
 
 
 
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