Question

gsdagasdf

Answer 1

@sourwing

Answer 2

@radar

Answer 3

Not familiar with this problem, it appears that "t" is Quadrant IV.

Answer 4

Draw a diagram.

Answer 5

draw a pic of what?

Answer 6

he reference angle for t.

Answer 7

so we know that angle "t" is between \(\bf \cfrac{3\pi}{2}<t<2\pi\) so as radar said, that puts it in the IV Quadrant

so... we know that in the IV Quadrant, cosine is positive, and sine is negative

Answer 8

ok so what do we do with this info

Answer 9

so \(\bf sec(t)=\cfrac{13}{5}\iff \cfrac{1}{cos(t)}\implies \cfrac{1}{cos(t)}=\cfrac{13}{5}\\ \quad \\
5=13cos(t)\implies \cfrac{5}{13}=cos(t)\\ \quad \\ \quad \\
cos(t)=\cfrac{5}{13} \implies \cfrac{adjacent}{hypotenuse} \implies \cfrac{a=5}{c=13} \\ \quad \\
\textit{using the pythagorean theorem }c^2=a^2+b^2\implies \sqrt{c^2-a^2}={\color{blue}{ b}}\\ \quad \\ \quad \\
\textit{keeping in mind that }sin(t)=\cfrac{{\color{blue}{ b}}}{c}\)

Answer 10

the pythagorean theorem root, doesn't tell us whether is negative or positive
but we know is in the IV Quadrant, thus you'd use the negative value for "b"

once you have "b", you'd have the sin(t)
the cos(t) is given, and then just plug them in the given expression

Answer 11

thanks so much

Answer 12

yw