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OpenStudy (anonymous):
Is sin(α+2π /3) equal to sin(-α+π/3)?
HOW?
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OpenStudy (anonymous):
Try expanding using:
sin (A+B)=sinA cosB = cosA sinB
OpenStudy (anonymous):
sorry, the second = sign should be read as +
OpenStudy (anonymous):
I am getting confused with the signs while expanding........ But the terms are equal in both......
OpenStudy (anonymous):
what did you get?
OpenStudy (anonymous):
SinAcos120+cosAsin120 and - sinAcos120-cosAsin120
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OpenStudy (anonymous):
LHS is correct. But the RHS should be sin(-A)cos60+cos(-A)sin60
OpenStudy (anonymous):
You don't need to expand it. Actually
\[\sin \theta = \sin (180 - \theta)\]
Right ? @bajji
OpenStudy (anonymous):
Ya? @ShailKumar
OpenStudy (anonymous):
So, put \[ \theta = \alpha + \frac{2\pi}{3}\]
OpenStudy (anonymous):
How dos that help the problem.... I did put but the result is sin((540-alpha+2)/3)
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OpenStudy (anonymous):
Sorry 3alpha.... And if i do the same for the next... It is not equal right?
OpenStudy (anonymous):
Actually \[ 180^o = \pi \ rad\]
So \[180^o - (\alpha+2\pi/3) = \pi - (\alpha+2\pi/3) = (-\alpha + \pi /3 )\]
OpenStudy (anonymous):
Does this help now ?
OpenStudy (anonymous):
Ya..... Great! Both are equal......
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