Given #cosalpha=1/3#, how do you find #sin(90^circ-alpha)#?
2 Answers
Apr 2, 2017
Explanation:
From the sketch:
The key is the right-angled triangle.
Apr 4, 2017
Explanation:
Use unit circle and property of complement arcs:
sin (90 - a) = cos a
In this case: