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Given sin 60degrees √ 3/2 find cos 60°

cos 60°

Given sin 60degrees √ 3/2 find cos 60° cos 60°-example-1

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Explanation:

if I understand you correctly, then you got

sin(60°) = sqrt(3/2)

and now we have to find cos(60°) based on that.

but : sin(60°) = sqrt(3/4) = sqrt(3)/2

I will continue with that.

one major observation with the trigonometric functions in a norm circle (radius = 1) is that sine and cosine are the legs of a right-angled triangle.

the radius at the angle is the Hypotenuse.

and so, Pythagoras applies

c² = a² + b²

c is the Hypotenuse, a and b are the legs.

so,

radius² = sin²(x) + cos²(x)

and because radius = 1

1² = 1 = sin²(x) + cos²(x)

in our case

1 = sin²(60) + cos²(60)

1 = (sqrt(3)/2)² + cos²(60) = 3/4 + cos²(60)

cos²(60) = 1 - 3/4 = 1/4

cos(60) = sqrt(1/4) = 1/2

User Anil  Panwar
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