Double Angle Identites (Trig Ratios)

Dhkajsd

New member
Q. from textbook :

Working:

Thanks a lot for any assistance

MarkFL

Super Moderator
Staff member
Hello, 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?

pka

Elite Member
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 $$\displaystyle -1\le\sin(2A)\le 0.$$

Dhkajsd

New member
Than
Hello, 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?
Thank you, didn’t realise it was so simple it must be negative