Question

Question on inverse trig function. Will post pic and what I tried.

Answer 1

for a, what I did:
I got triangles for each of the trig function inside.
sin^-1(-1/2) = -pi/6

for the sec^-1(2)...i don't know. I got pi/3 but when I tried it on calculator it was wrong.

and...I'm stuck...


for b
I got triangles for each of trig inside.
tan^-1(2) = a
tan a = 2
but how do you get a? I tried the calculator but I dont know what it is in terms of pi.

tan^-1(3) = b
tan b = 3
I dont know how to continue.

Thanks for helping.

Answer 2

I think letter a doesn't exist

Answer 3

\[\sec^{-1} x = \cos^{-1} \frac{1}{x}\]

Answer 4

there's no inverse secant function in most calculators, just inverse functions for sine, cosine and tangent

Answer 5

oh you have a point but
isnt sec^ 1(x) = (1/ cos ^-1(x))?

Answer 6

nope. that identity is good for sec x, not sec^-1 x

Answer 7

haha. didn't know

Answer 8

let me show why. suppose \[\alpha = \sec^{-1} x\\\sec \alpha = x\\\frac{1}{\cos \alpha}=x\\\cos\alpha = \frac{1}{x}\\\alpha = \cos^{-1} \frac{1}{x}\]

Answer 9

for questions in (b), the angles are not special. you just have you use your calculator to get the approximate values

Answer 10

check out dr Pan http://www.youtube.com/watch?v=aVVZOwo18k8&list=FLxFDZsq7_sDYcfK7w9lp-WA&index=6

Answer 11

i see.. so in a.. will have cot( -pi/6 - pi/3)
do i just add them up then solve or does this have an identity? or will i use the identity of tan?

Answer 12

you can add them first

Answer 13

then solve? ok ill try
so.. it is
1/tan(-pi/2)..i used the calculator and its undefined.

Answer 14

yes. the answer is undefined.

Answer 15

ok...so why..if i used wolframalpha..its 0..

Answer 16

so for 6b....I can do the same, right? I mean...cos( someDecimal + someDecimal )..
I can just add them up and not use identities?

Answer 17

oh, sorry. cot(-pi/2)=cos(-pi/2)/sin(-pi/2) = 0/(-1) = 0

Answer 18

Its ok... so...that's solved...

For b...i just use the decimal form
add them
then solve for cos or do I use cos Identity?

Answer 19

use your calculator to evalulate tan^-1 3

Answer 20

ok...for tan ^-1 2...how?

Answer 21

If I did tan ^-1 (2)..
tan y = 2

so then the value of y is a decimal.

Answer 22

yep. it's not a special angle

Answer 23

ok..so that would be cos(Decimal + Decimal) so... theres no exact values (in terms of pi) for tan ^-1 2 and tan ^-1 3.

Answer 24

that's right

Answer 25

ok Thank you! :D

Answer 26

I found out you can solve this without calculator.

build 2 triangles...for tan -1 (2) and tan-1(3)

then
use cos( x + y ) identity

then subsitute in terms of cos and sin.