T formulae Q
- Thread starter Sonal7
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\(\displaystyle \cos\left(\dfrac \theta 2\right)\) isn't necessarily \(\displaystyle -\dfrac{13}{12}\)
\(\displaystyle \theta = 45.24^\circ \Rightarrow \sin\left(\dfrac \theta 2\right) \approx \dfrac{5}{13} \text{ and }\cos\left(\dfrac \theta 2\right) \approx \dfrac{12}{13}\)
To get the numbers you wrote \(\displaystyle \dfrac \theta 2\) must be in the 2nd quadrant, where sines are positive, and cosines are negative.
\(\displaystyle \dfrac \theta 2 = \pi - \arcsin\left(\dfrac{5}{13}\right) \approx 157.38^\circ \Rightarrow \theta \approx 314.76^\circ\)
with this value of \(\displaystyle \theta\) we obtain the values for sine and cosine you want.
\(\displaystyle \theta = 45.24^\circ \Rightarrow \sin\left(\dfrac \theta 2\right) \approx \dfrac{5}{13} \text{ and }\cos\left(\dfrac \theta 2\right) \approx \dfrac{12}{13}\)
To get the numbers you wrote \(\displaystyle \dfrac \theta 2\) must be in the 2nd quadrant, where sines are positive, and cosines are negative.
\(\displaystyle \dfrac \theta 2 = \pi - \arcsin\left(\dfrac{5}{13}\right) \approx 157.38^\circ \Rightarrow \theta \approx 314.76^\circ\)
with this value of \(\displaystyle \theta\) we obtain the values for sine and cosine you want.
Dr.Peterson
Elite Member
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- Nov 12, 2017
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Please show the whole problem and your work. If you used the half-angle formula, you should know that there is a plus-or-minus in it, which must be filled in based on other information. If you were just given [MATH]\sin\frac{\theta}{2}[/MATH], then you can't know the sign of the cosine.
. Thank you very much.
Dr.Peterson
Elite Member
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Right. The quadrant determines the sign.
I am not familiar with the term "t-formula", which is not very descriptive, but on looking it up I see how you are using it.
I am not familiar with the term "t-formula", which is not very descriptive, but on looking it up I see how you are using it.
