Trigonometry simple problem

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  • #1
Samurai44
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Homework Statement


sinӨ + cosӨ =4/3 ,, then sin2Ө = .... ?

Homework Equations


sin2Ө = 2cosӨsinӨ
or any trigonometry functions...

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 ?
 

Answers and Replies

  • #2
Nathanael
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sin2θ+cos2θ=1
 
  • #3
Samurai44
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sin2θ+cos2θ=1
how is that possible ?
 
  • #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
 
  • #5
Samurai44
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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
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 :(
 
  • #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.)
 
  • #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).
 
  • #8
Samurai44
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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.)
oh i got it know ,, thanks a lot
 
  • #9
Nathanael
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No problem
 
  • #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].
 
  • #11
Nathanael
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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].
Nice method :)
 

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