Trig Reduction With Pythagoras #2

Question

If $\sin 10^\circ = p$, then determine $$\sin 280^\circ$$ in terms of $p$.

$$\sin 280^\circ=\sin(180^\circ+100^\circ) = -\sin 100^\circ$$ $$-\sin 100^\circ = -\sin(90^\circ+10^\circ) = -\sin 10^\circ$$ $$ =-p $$ Am I correct?

Answer 1

First of all, $$\sin(90^\circ\pm x)=\cos x$$

As $280^\circ$ lies in the third Quadrant where sine ratio is $\le0$

$$\implies\sin(280^\circ)=\sin(3\cdot90^\circ+10^\circ)=-\cos10^\circ$$

As the multiple of $90^\circ$ is $3$ which is odd, the resultant ratio will be cosine

Answer 2

sin(280)= sin(-80) = sin(10-90).