Home > Trigonometry > Translating Sine and Cosine Functions

If #sin x = 1/2# and #sec x < 0# then what is #cos x# ?

Answer 1

Oliver Atkinson

#cos x = -sqrt(3)/2#

Note that:

#sin^2 x + cos^2 x = 1#

So:

#cos x = +-sqrt(1-sin^2x) = +-sqrt(1-(1/2)^2) = +-sqrt(3/4) = +-sqrt(3)/2#

We are told #sec x < 0# and hence #cos x = 1/(sec x) < 0#.

So:

#cos x = -sqrt(3)/2#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Violet Anderson

If ( \sin(x) = \frac{1}{2} ) and ( \sec(x) < 0 ), then ( \cos(x) ) can be found using the fact that ( \sec(x) = \frac{1}{\cos(x)} ). Since ( \sec(x) < 0 ), it implies that ( \cos(x) < 0 ) as well.

Knowing that ( \sin(x) = \frac{1}{2} ) and ( \cos(x) < 0 ), we can use the Pythagorean identity for sine and cosine:

[ \sin^2(x) + \cos^2(x) = 1 ]

Substituting the given value for ( \sin(x) ), we have:

[ \left(\frac{1}{2}\right)^2 + \cos^2(x) = 1 ]

[ \frac{1}{4} + \cos^2(x) = 1 ]

[ \cos^2(x) = 1 - \frac{1}{4} ]

[ \cos^2(x) = \frac{3}{4} ]

Since ( \cos(x) < 0 ), ( \cos(x) ) must be the negative square root of ( \frac{3}{4} ). Thus:

[ \cos(x) = -\frac{\sqrt{3}}{2} ]

By signing up, you agree to our Terms of Service and Privacy Policy

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.

If #sin x = 1/2# and #sec x &lt; 0# then what is #cos x# ?

If #sin x = 1/2# and #sec x < 0# then what is #cos x# ?