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FP1 Maths question

Discussion in 'Mathematics' started by moontitan, Nov 29, 2018.

  1. moontitan

    moontitan New commenter

    Sinx=(sqrt3-1)/(2sqrt2) for 0<=theta<pi/2
    From the t formulae it is given that t=(sqrt3-1)/(sqrt3+1)
    and sin2x=2t/(1+t^2)
    So sin2x=1/2 which means 0<= theta<pi

    What is the value of x?
    I get 15 and 75 but I can see that 75 does not work when I sub it back into the question.
    So why does this result come about?
  2. Informant

    Informant New commenter

    I'm curious about your problem. For sin(x)= [your value], the principal value of x=15 degrees and in range 0<= x <90 a second value x=75 is indeed a possibility (I assume your reference to theta really means x and that your answers in degrees mean the constraints are not in radians as stated, but in degrees). For t= [your other value], I agree sin(2x) = 1/2 so 2x = 30 degrees (principal value), but also 2x = 150 degrees, so x=75 also as you state. I may be missing something but it appears to me that x=15 and x=75.

    The maths thread on TES Forum isn't heavily used, but if you want informed opinion, typically within 10 mins of posting, try thestudentroom and select Forums followed by Maths Study Help. It's used by students, staff and academics.
  3. adamcreen

    adamcreen Occasional commenter

    You have mixed up x, 2x, theta and 2theta
    Could you give a page reference or a photo of the question? I can post a photo if it's in the FP1 book.

    The Maths thread has plenty of responses when a question comes along, we are happy to join in your mathematical explorations!
  4. moontitan

    moontitan New commenter

    It is on page 119 Q6c of the new FP1 spec.

  5. BG54

    BG54 Occasional commenter

    Sin 75 is the conjugate surd of Sin 15.

    Sin 75 = [sqrt(6) + sqrt(2)]/4
    Sin 15 = [sqrt(6) - sqrt(2)]/4

    If you replace sin 2x by 1/2 in the t^2 equation and solve for t there are two conjugate roots t = 2[1 +/- sqrt(3)], the one given for t in your post is satisfied by sin 15 and the other is satisfied by sin 75.

    That's just an observation, without seeing the original question it's difficult to say if that helps you much.
  6. BG54

    BG54 Occasional commenter

    Edited to correct a typo.
  7. adamcreen

    adamcreen Occasional commenter

    Here it is. Not an x in sight!
    Informant likes this.
  8. gainly

    gainly Star commenter

    As the questions says, you need to find cos (2theta) also. The t formula gives (sqrt3)/2. Therefore only pi/6 satisfies both sin and cos values, giving pi/12 as the only value of theta.
    Informant likes this.
  9. gnulinux

    gnulinux Occasional commenter

    If the question had read 0<=theta<pi/2 then theta would have one value, i.e. theta = pi/12. However, it does not.

    Read again, 0<=theta/2<pi/2 which is equivalent to 0<=theta<pi. That mistake??? invalidates/complicates parts a, b and c and theta would have 2 values even allowing for the mistake in part a.

    Assuming it was intended to read 0<=theta<pi/2 then a and b are ok and c gives tan(2theta) = 1/sqrt(3) where 0<=2theta<=pi. The tan() function on [0,pi] is non-negative and 1-1 only over [0, pi/2) and so 2theta = pi/6 and it follows that theta = pi/12.
    Last edited: Nov 30, 2018
  10. adamcreen

    adamcreen Occasional commenter

    These textbooks are notorious for their errors (I have a blogpost as long as your arm listing all the errors people have found) and it's perfectly believable they meant theta not theta/2 in the inequality. Good spot Gnulinux.
  11. gnulinux

    gnulinux Occasional commenter

  12. gainly

    gainly Star commenter

    I must admit I hadn't noticed the theta/2 and assumed it was theta, as did the OP. I think it must be a mistake, otherwise there would be 2 possible values for tan(theta) but only one value is given in part (a).
  13. moontitan

    moontitan New commenter

    Thanks for all the responses, good to know so many folk still here.

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