Question

sin(sec^-1(x+3/4)) help, I don't even know where to start with this one.

Answer 1

ugliness for sure
draw a triangle first

Answer 2

ok did the triangle... but bec its sec, wasnt sure how to label it

Answer 3

find the third side via pythagoras

Answer 4

thinking real basic as
secant equals hypotenuse over adjacent" that is how i labelled it

Answer 5

i'm sure you meant (x+3)/4 yes?

Answer 6

but before we go further, is it
\[x+\frac{3}{4}\] or is it
\[\frac{x+3}{4}\]

Answer 7

the second one

Answer 8

what @xxferrocixx
said
if it is the second then label differently

Answer 9

put the hypotenuse as \(x+3\) and the adjacent as \(4\)

Answer 10

draw tool is no longer working for me, but i assume you get the idea

Answer 11

did you ever get help with that last problem btw? the one where you had to rotate about x = -1?

Answer 12

ok
so for the other side I got: \[\sqrt{x ^{2}+6x+7}\]

Answer 13

I did I struggled with it... but I guess with time I will become more confident with this stuff

Answer 14

if that is
\[\sqrt{(x+3)^2-4^2}\]then yes

Answer 15

hmm but i think it is not

Answer 16

oops let me do it over

Answer 17

should be
\[\sqrt{x^2+6x-7}\] i think you subtracted wrong
and that is not the final answer in any case, you have to put that over the hypotenuse

Answer 18

thats the answer? that simple?

Answer 19

wow i guess it is that simple

Answer 20

It is a little more complicated than that (just a little)

Answer 21

that is not the answer, that is the "opposite" side
the answer for
\[\sin\left(\sec^{-1}(\frac{x+3}{4})\right)\]would be
\[\frac{\sqrt{x^2+6x-7}}{x+3}\] as in "opposite over hypotenuse"

Answer 22

the most complicated part is writing it!

Answer 23

what do you get if you plug in -10 for x

Answer 24

i give up what?

Answer 25

not the answer if you use the function

Answer 26

you need


\[\frac{\sqrt{x^2+6x-7}}{|x+3|}\]

Answer 27

the range of the arcsec is between 0 and pi and the sin on 0 to pi is positive

Answer 28

ooh jeez it is always something

Answer 29

OK ... well thank you satellite, for your help, according to the software, the answer you helped me with is right

Answer 30

if you want, repost the last problem we can do it

Answer 31

@Zarkon yeah i see my mistake too
if i did it the real way the denominator would be
\[\sqrt{(x+3)^2}\]

Answer 32

@itsmecaro tell the software people you found a mistake (even though it was @Zarkon that found it)
maybe they will offer you a reward

Answer 33

Like one of those Christmas tree Debbie cakes?

Answer 34

hahaha a reward??? right they will just blow me off

Answer 35

i was thinking more along the lines of unused halloween kit kats

Answer 36

ewwww

Answer 37

What? They are unused?

Answer 38

saw someone returning a whole bag to target earlier today
they said they have to throw them out

Answer 39

i dont know. ive heard some creepy things where they inject stuff in the chocolates

Answer 40

it's called fat`

Answer 41

You can solve this using trig identities.

sin (arcsec((x+3)/4))

Let y = arcsec (x+3)/4

sec y = (x+3)/4

cosy = 4/(x+3)

Trig identity:
sin^2(y) + cos^2(y) = 1

sin^2(y) = 1 - cos^2(y)

sin (y) = + - sqrt( 1 - cos^2(y))

sin(y) = + - sqrt[1 - (4/(x+3))^2]

Answer 42

you don't get a \(\pm\) because \(\sin(y)\ge0\)

(because \(0\le y\le \pi\))

Answer 43

was that given in the problem , or you derived that

Answer 44

i guess because the range of arcsec is positive

Answer 45

the range of the arcsec(x) is \([0,\pi/2)\cup(\pi/2,\pi]\)

Answer 46

ok and sine of that is positive.

Answer 47

yes

Answer 48

that makes sense. how did you know that satellite needed an absolute value. its not obvious with the 'triangle' method

Answer 49

if you draw a triangle, i mean

Answer 50

because of the domains and ranges of arcsec(x) and sin(x)

Answer 51

you can also get it from the triangle...which satellite73 mentions

Answer 52

it seems to work algebraically though. also the hypotenuse is x+3, so that should always be positive, so i guess | x + 3 | makes sense.


sin (arcsec((x+3)/4))

Let y = arcsec (x+3)/4

sec y = (x+3)/4

cosy = 4/(x+3)

Trig identity:
sin^2(y) + cos^2(y) = 1

sin^2(y) = 1 - cos^2(y)

sin (y) = + sqrt( 1 - cos^2(y))

sin(y) = + sqrt[1 - (4/(x+3))^2]

sin(y) = sqrt[ (x+3)^2 - 4^2 ) / (x+3)^2]

sin(y) = sqrt( x^2 + 6x -7 ) / sqrt((x+3)^2)

sin(y) = sqrt(x^2 + 6x -7) / |x+3|

Answer 53

satellite says
"yeah i see my mistake too if i did it the real way the denominator would be
sqrt((x+3)^2) "

I can only imagine satellite meant by the 'real' way, using the trig identities.

Answer 54

since you guys are math genius' can u help me out with this next one?