Question

HELP ME PLEASE
sec theta= 9/8, theta in the quadrant IV
Find tan theta

Answer 1

@satellite73 could you maybe explain?

Answer 2

draw a triangle

Answer 3

secant is "hypotenuse over adjacent" so we can label it like this

Answer 4

okidoki
im following

Answer 5

all that is missing is the opposite side, which you can find via pythagoras
you good with that?

Answer 6

so sqrt of 17

Answer 7

if you have done a couple of these before,you can go right to the answer
it is
\[\sqrt{9^2-8^2}=\sqrt{81-64}=\sqrt{17}\] yeah you got it
not quite done yet though

Answer 8

okay whats next

Answer 9

Ok, the first thing to consider is the fact that secant theta is the same as 1/cosine theta. So if secant theta = 9/8, then 1/cosine theta = 9/8 and cos theta = 8/9, flipping both fractions. See that part first? That's important! In the 4th Q, it would look like thisYou need to solve for x because x is the side opposite the angle, and you need that for tangent (opposite over adjacent). To find x, use Pythagorean's theorem. x^2 + 8^2 = 9^2. Or\[x ^{2}=(9^{2})-(8^{2})=81-64=\sqrt{17}\]'So the length of the missing side is square root of 17. That's the side opposite the angle, so opposite over adjacent would be \[\frac{ \sqrt{17} }{ 8 }\]

Answer 10

now you have the opposite side is \(\sqrt{17}\) and the adjacent side is \(8\) and you are in quadrant 4 which makes tangent negative

Answer 11

the ratio is therefore
\[\tan(\theta)=-\frac{\sqrt{17}}{8}\]
don't forget the minus sign!

Answer 12

okay I have another question if that is okay?

Answer 13

sure

Answer 14

y=4sin(3x) which part is the amplitude, I totally forgot

Answer 15

the absolute value of the number out front, which in this case is \(4\)

Answer 16

okidoki thank you thank you

Answer 17

Oops sorry that is a negative value!

Answer 18

yw