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Trigonometry simple problem

  1. Jan 10, 2015 #1
    1. The problem statement, all variables and given/known data
    sinӨ + cosӨ =4/3 ,, then sin2Ө = .... ?

    2. Relevant equations
    sin2Ө = 2cosӨsinӨ
    or any trigonometry functions...

    3. The attempt at a solution
    I tried to write sinӨ + cosӨ =4/3 in the form of cosӨsinӨ but couldn't..
    is there a possible way to solve it ?
     
  2. jcsd
  3. Jan 10, 2015 #2

    Nathanael

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    sin2θ+cos2θ=1
     
  4. Jan 10, 2015 #3
    how is that possible ?
     
  5. Jan 10, 2015 #4

    Nathanael

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    I'm sorry for the unhelpful post, I was assuming you knew that equation and were just forgetting it.

    The equation comes from Pythagorean's theorem and it's always true for all θ (so you can use it in

    Imagine (or draw) the Unit Circle. The x coordinate of a point on the circle is cosθ, and the y coordinate is sinθ (where θ is the angle between the point and the positive x-axis). We know from Pythagorean's theorem that x2+y2=1 (because 1 is the radius of the circle) therefore sin2θ+cos2θ=1

    Edit:
    Or I could have said, the equation of the unit circle is x2+y2=1, therefore sin2θ+cos2θ=1
     
  6. Jan 10, 2015 #5
    yeah i know this equation , what i meant is how can i use it to solve this problem ?
    i tried it but didnt get the answer :(
     
  7. Jan 10, 2015 #6

    Nathanael

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    sin2θ+cos2θ=1
    sinθ+cosθ=4/3

    I haven't done the manipulations, but that is 2 equations with 2 unknowns; it should be enough to uniquely determine cosθ and sinθ
    (In other words, there should be only 1 pair of cosθ and sinθ which add to 4/3 while also having their squares add to one.)
     
  8. Jan 10, 2015 #7

    Nathanael

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    I just realized, it won't uniquely determine cosθ and sinθ, because you could simply interchange cosθ and sinθ and you will have another solution. But this shouldn't change the answer since your formula for sin(2θ) involves the product of the two (which remains unchanged if you interchange them).
     
  9. Jan 10, 2015 #8
    oh i got it know ,, thanks a lot
     
  10. Jan 10, 2015 #9

    Nathanael

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    No problem
     
  11. Jan 10, 2015 #10

    HallsofIvy

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    I assumed you were suggesting he square both sides of his equation: if [itex]sin(x)+ cos(x)= 4/3[/itex] so [itex](sin(x)+ cos(x))^2= sin^2(x)+ 2sin(x)cos(x)+ cos^2(x)= 16/9[/itex]. But [itex]sin^2(x)+ cos^2(x)= 1[/itex] so this is [itex]2sin(x)cos(x)+ 1= 16/9[/itex], [itex]2sin(x)cos(x)= 7/9[/itex].
     
  12. Jan 10, 2015 #11

    Nathanael

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    Nice method :)
     
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