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:
$sin (90 - a) = cos a = 1/3$

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