Using double angles to find value of

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The discussion revolves around using double angle formulas to solve for sin(3x) given sin^2(x) = 8/9 and x in Quadrant II. The user initially misapplies the double angle formula for sin(3x) and is corrected to use sin(3x) = sin(2x + x). Participants suggest using the addition formula along with known values of sin(2x) and cos(2x) to derive sin(3x). The conversation emphasizes substituting the correct double angle formulas and values step-by-step to arrive at the solution. The thread highlights the importance of understanding the relationships between sine and cosine in different quadrants.
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Homework Statement


sin2x.png

Homework Equations


doubleangleformulae.png

The Attempt at a Solution


I know the following:
sin^2x = 8/9
the hyp. is 9, the "y" value is 8.
Therefore, the "x" value has to be \sqrt{17}

What is the next step?
 
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You are given that x is in Quadrant II, so sin(x) >= 0 and cos(x) <= 0.
You are also given that sin2(x) = 8/9, so you can easily find cos2(x).

Now find sin(x) and cos(x) by taking the square root with the appropriate sign, and you can evaluate sin(2x) = 2sin(x)cos(x).
 
Im sorry I should've really thought before I posted. I actually got through A,B, and C. However, d (with sin3(x)) is confusing me.

The double angle formula for it will be:
sin3x = sin0.5xcos0.5x correct?
But then, when I sub in known values, I don't receive the correct answer:

-10*root 2
--------------
27
 
aeromat said:
Im sorry I should've really thought before I posted. I actually got through A,B, and C. However, d (with sin3(x)) is confusing me.

The double angle formula for it will be:
sin3x = sin0.5xcos0.5x correct?
No - why would you think that. sin(3x) = sin(2x + x). Use the addition formula and then the double angle formula to get things in terms of sin(x) and cos(x), which you already know.
aeromat said:
But then, when I sub in known values, I don't receive the correct answer:

-10*root 2
--------------
27
 
I'm sorry but we just started this unit and I am not sure how I would do that:
Use the addition formula and then the double angle formula to get things in terms of sin(x) and cos(x), which you already know.
 
sin(3x) = sin(2x + x) = sin(2x)cos(x) + cos(2x)sin(x)

Then you use the double-angle formulas for sin(2x) and cos(2x).
 
Bohrok said:
sin(3x) = sin(2x + x) = sin(2x)cos(x) + cos(2x)sin(x)

Then you use the double-angle formulas for sin(2x) and cos(2x).

I'm sorry but I have no idea how you would do that. Would you mind explaining to me step-by-step? :confused:
 
Well you have sin(2x)cos(x) + cos(2x)sin(x). In other words. Use the double angle formulas for the bold areas. Like the first part would be 2sin(x)cos(x)cos(x) or 2sin(x)cos²(x). The second part is to substitute one of the double angle formulas for cos(2x), look at the second version as I think that one will help a lot, and then substitute your x values for each. Hope I make sense! :D
 
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