I know an alternative way of showing Sin C is cos 'theta'.
What is angle C? I can't make it out in the picture... Also could you show your attempt of the problems?
Angle C is pi/2 - theta (radians) "pi over two minus theta" ie. 90 deegrees - theta.
I did show my attempt at it, i couldn't get further.
Certainly, since the angles in a triangle add to [itex]2\pi[/itex] radians, [itex]B= 2\pi- \lambda- (\pi/2- \theta)= 3\pi/2- (\theta- \lambda)[/itex].
Okay, using sin(a+ b)= sin(a)cos(b)- cos(a)sin(b)[/itex], what is
[itex] sin(3\pi/2+ (-(\theta- \lambda)))[/itex]?
The angles of a triangle add upto [itex]\pi[/itex] radians
The expression you require is therefore [itex]\sin(\pi/2+(\theta-\lambda))[/itex] = ?
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