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Given #cosalpha=1/3#, how do you find #sinalpha#?

Answer 1

Oliver Atkinson

#sin(a) = +-(2sqrt2)/3#

You can use a trig identity that states that #cos^2(a) = 1 - sin^2(a)#

so we get that

#1/9 = 1 - sin^2(a) => sin^2(a) = 8/9 => sin(a) = +-(2sqrt2)/3#

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Answer 2

Daniel Acosta

Use the identity #cos^2(alpha) + sin^2(alpha) = 1# and solve for #sin(alpha)#:

#sin(alpha) = +-sqrt(1 - cos^2(alpha))#

Substitute #(1/3)^2# for #cos^2(alpha)#

#sin(alpha) = +-sqrt(1 - (1/3)^2)#

#sin(alpha) = +-sqrt(1 - 1/9)#

#sin(alpha) = +-sqrt(8/9)#

#sin(alpha) = +-sqrt(8)/3#

#sin(alpha) = +-(2sqrt(2))/3#

Because we are not given any clue whether #alpha# is in the first or the fourth quadrant, then we cannot determine whether the sine function is positive or negative.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.