- Thread starter Dhkajsd
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- Jan 29, 2005

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Given that \(\displaystyle 90o\le A\le 180^o\) means that \(\displaystyle 180^o\le 2A \le 360^o\) is in either the third or fourth quadrant.

So then \(\displaystyle -1\le\sin(2A)\le 0.\)

Thank you, didn’t realise it was so simple it must be negativeHello, and welcome to FMH!

Your working would be correct if the angle \(A\) were a first quadrant angle. But, it's a second quadrant angle. So, what must the sign of the cosine function there be?