If sin=4/5 then that is cos =

Hi Alan,

$$\\\cos{(\alpha)}=\textcolor[rgb]{1,0,0}{\pm}\frac{3}{5}\\ \text{if }{+\frac{3}{5}} \Rightarrow \alpha = \cos^{-1}(0.6) = 53.1301023542\ensuremath{^\circ}\\ \text{if }{-\frac{3}{5}} \Rightarrow \alpha = \cos^{-1}(-0.6) = 126.869897646\ensuremath{^\circ}\\ \sin{(53.1301023542\ensuremath{^\circ})}=\sin{(126.869897646\ensuremath{^\circ})}=0.8=\frac{4}{5}$$

Latex code:

\text{if }{+\frac{3}{5}} \Rightarrow \alpha = \cos^{-1}(0.6) = 53.1301023542\ensuremath{^\circ}\\
\text{if }{-\frac{3}{5}} \Rightarrow \alpha = \cos^{-1}(-0.6) = 126.869897646\ensuremath{^\circ}\\
\sin{(53.1301023542\ensuremath{^\circ})}=\sin{(126.869897646\ensuremath{^\circ})}=0.8=\frac{4}{5}

Using the arcsin function

asin(4/5) = about 53.13 degrees

This angle could lie in the 2nd quadrant, too=  (180 - 53.13) = 126.87 degrees

If sin=4/5 then we can picture the well known 3, 4, 5 right-angled triangle; so cos = 3/5.

Thanks...alan....I've answered so many questions that I'm going snowblind!!!

What, even with those shades!!!

You are right heureka;  sin is also positive in the second quadrant, where cos is negative.