Question

Given cos t= 1/2, tan > 0, find sin (t) and sec (t).
A. Sin (t)= - Square root3/2, sec (t) =2
B. Sin (t)= Square root3/2, sec (t) =-2
C. Sin (t)= - Square root3/2, sec (t) =-2
D. Sin (t)= Square root3/2, sec (t) =2

Answer 1

If the cosine is 1/2, then the sin will be...

\[\sin^{2} x + \cos^{2}x = 1 \]

fill in for the cos(x) = 1/2 and find the sin(x)

Answer 2

\[\sin^2(t) +\frac{ 1 }{ 4 } = 1 \]

Answer 3

the sec(t) is also 1 / cos(t)

Answer 4

Or another method, from a triangle

Answer 5

Use pythagorean theorem to get the ? .

2^2 = 1^2 + ?^2

? =

then

sin(t) = ? / 2

Answer 6

\[\sec(t) = \frac{ 1 }{ \cos(t) } = \frac{ 1 }{ 1/2 }\]

Answer 7

This makes no sense to me

Answer 8

just use the relationship identity,

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

cos(t) = 1/2 Given
\[\sin^2(t) + [\frac{ 1 }{ 2 }]^2 = 1\]
\[\sin^2(t) = 1 - \frac{ 1 }{ 4 }\]
\[\sin^2(t) = \frac{ 3 }{ 4 }\]
\[\sin(t) = \sqrt{\frac{ 3 }{ 4 }} = \frac{ \sqrt{3} }{ \sqrt{4} } = \frac{ \sqrt{3} }{ 2 }\]

So the sin(t) is
\[\sin(t) = \frac{ \sqrt{3} }{ 2 }\]

Answer 9

For the secant of t,

\[\sec(t) = \frac{ 1 }{ \cos(t) } = \frac{ 1 }{ 1/2 } = 2\]

sec(t) = 2

Answer 10

Ok know that makes more sense but how did that one become a 3?

Answer 11

That part?
\[1 - \frac{ 1 }{ 4 } ~~=~~ \frac{ 4 }{ 4 } - \frac{ 1 }{ 4 } ~~= ~~\frac{ 3 }{ 4 }\]

Answer 12

u understand it? i am about to log out

Answer 13

Yes. Thank you