Calculate cos(θ/2) Given sin(θ)=-5/13

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To find cos(θ/2) given sin(θ) = -5/13 and θ in the range π < θ < 3π/2, first calculate cos(θ) using the Pythagorean identity x² + y² = r², where y = -5 and r = 13. This results in x = ±12, leading to cos(θ) = -12/13 since θ is in the third quadrant. Then, apply the cosine half-angle identity, cos(θ/2) = ±√((1 + cos(θ))/2). Given the range for θ/2, determine the sign of the square root accordingly. The final result provides the value of cos(θ/2).
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Homework Statement



If sin(θ) = -5/13, π < θ < 3π/2, find cos(θ/2)
 
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MathRaven said:

Homework Statement



If sin(θ) = -5/13, π < θ < 3π/2, find cos(θ/2)

Given that sin(θ) = y/r, that we're in Quadrant III, and that x2 + y2 = r2, you should be able to find x, and from that, cos(θ).

Then use the cosine half-angle identity
\cos \left( \frac{\theta}{2} \right) = \pm \sqrt{\frac{1 + \cos \theta}{2}}
to find cos(θ/2). Since π < θ < 3π/2, you know the range of values that θ/2 lie in, so that will tell you whether the square root should be positive or negative.
 
MathRaven said:

Homework Statement



If sin(θ) = -5/13, π < θ < 3π/2, find cos(θ/2)

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