Question

Write the first expression in terms of the second if the terminal point determined by t is in the given quadrant.

sin(t), sec(t);
Quadrant IV

Answer 1

To write sin(t) in terms of sec(t) in Quadrant IV, use the identity sin^2(t) + cos^2(t) = 1 and the fact that sec(t) is 1/cos(t); then take the negative square root since sine is negative in Quadrant IV to get sin(t) = -sqrt(1 - (1/sec(t))^2).

Step-by-step explanation:

We are asked to write the function sin(t) in terms of sec(t) for an angle t in Quadrant IV. In Quadrant IV, sine is negative and secant (which is 1/cos(t)) is positive as cosine is positive. To express sin(t) in terms of sec(t), use the identity sin^2(t) + cos^2(t) = 1 and the fact that sec(t) = 1/cos(t). We can solve for sin(t) as follows:

Firstly, solve the Pythagorean identity for sin(t):
sin(t) = ±sqrt(1 - cos^2(t)).

Then, express cos(t) in terms of sec(t):
cos(t) = 1/sec(t).

Substitute cos(t) into the equation for sin(t):
sin(t) = ±sqrt(1 - (1/sec(t))^2).

As we are in Quadrant IV, our sin(t) value should be negative, so we take the negative solution:
sin(t) = -sqrt(1 - (1/sec(t))^2) or sin(t) = -sqrt(1 - cos^2(t)) = -sqrt(1 - (1/sec(t))^2).

Answer 2

`\bf sin^2(\theta)+cos^2(\theta)=1\implies sin^2(\theta)=1-cos^2(\theta)\\\\ sin(\theta)=\pm√(1-cos^2(\theta))\\\\ cos(\theta)=\cfrac{1}{sec(\theta)}\\\\ -----------------------------\\\\ sin(t)\implies \pm √(1-cos^2(t))\implies \pm \sqrt{1-\cfrac{1^2}{sec^2(t)}}\\\\\\ \pm \sqrt{\cfrac{sec^2(t)-1}{sec^2(t)}}`

so.. hmm which is it? the positive or the negative one?

well, we know angle "t" is in the IV quadrant, sine is negative on the IV quadrant, thus, is the negative one
`\bf -\sqrt{\cfrac{sec^2(t)-1}{sec^2(t)}}`