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Mathematics 9 Online
OpenStudy (anonymous):

Is sin(α+2π /3) equal to sin(-α+π/3)? HOW?

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

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)

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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