Given #cosalpha=1/3#, how do you find #sin(90^circ-alpha)#?

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Answer 1 · 2017-04-02T17:04:58

#1/3#

From the sketch: #cos alpha = sin (90 - alpha)#

The key is the right-angled triangle.

Wassily Kandinsky

Answer 2 · 2017-04-04T01:18:27

#1/3#

Use unit circle and property of complement arcs:
sin (90 - a) = cos a
In this case:
#sin (90 - a) = cos a = 1/3#