Question

cos^-1(cos(23pi/19))

Answer 1

in the interval? -pi/2 to pi/2

Answer 2

cos is 0 to pi

Answer 3

right, so u need to convert 23pi/19 within 0 to pi, 1st or 2nd quadrant

Answer 4

Why won't 4pi/19 work?

Answer 5

\(\cos x= \cos(2\pi-x)\)

Answer 6

4pi/19 will not work because u get a negative answer then

Answer 7

u just need to figure out by yourself which formula to use each time

Answer 8

aren't we just trying to find the reference angle?

Answer 9

within the given quadrant that is

Answer 10

@watchmath What are the formulas that you are using? could you give me the formula for sine as well?

Answer 11

here the angle was in 3rd quadrant, and u want angle in either 1st or 2nd quadrant, keeping the answer positive

Answer 12

u want list? it can be as long as u want, i would rather suggest u to figure out a formula on you own, rather than seein in a list

Answer 13

for sin^-1(sin(23pi/19)) why would it be a -4pi/19 instead of a 4pi/19?

Answer 14

@hartnn

Answer 15

because sin(-4pi/9) will be negative!, but u want a positive value for sin

Answer 16

sin^-1(sin(23pi/19)) =a
(sin(23pi/19)) =sin a
this indicates that u need a to be in a quadrant where sin is positive

Answer 17

I put down the answer as 4pi/19 but it's wrong. That's why I'm confused as well.

Answer 18

oh, 23pi/19 is in 3rd quadrant where sin is NEGATIVE ! sorry, my bad., i made a mistake.
so u need a angle in which sin is negative.

Answer 19

Ohh, so if the starting x is a negative, the answer has to be a negative too?

Answer 20

not everytime, it depends on what ratio are u evaluating
(sin(23pi/19)) =sin a
here left side was negative(sin negative in 3rd quadrant), so right side needed to be negative.
same will not be true for tan

Answer 21

so u 1st need to know which ratio is positive/negative in which quadrants