Give the value without a calculator: sin[arc sec(-3)]?

Answer 1

This is one of those math jokes, right?

Isn't the answer P-nes? The negative log of Nes or something like that...I don't remember, but I DO know the answer is supposed to make you say "Penis."

Math rules the local vicinity.

Answer 2

The sin of anything is between -1 and 1, so do ignore the previous answers...

Let x=arcsec(-3). This means sec x = -3 or cos x = -1/3.

Note that the arcsec function has range [0,pi] so x having a negative cos will be in the 2nd quadrant.

We are looking for
sin x = sqrt(1-cos²x) ---> positive root because sin is positive in quadrant 2

= sqrt(1 - 1/9)
= sqrt(8/9)
= (2/3)sqrt2

Answer 3

sec = secant = a million/cosine pi/3 is an perspective in radians: one hundred eighty/3 tiers = 60 tiers so it relatively is seeking a million/cos(60) you need to understand by making use of coronary heart the "a million-2-?3" triangle. it particularly is a suitable-angled triangle with angles 30-60-ninety tiers, and corresponding aspects a million-?3-2. From this you get cos(60) = adjoining/hypotenuse = a million/2 so sec(60) = a million/(a million/2) = 2

Answer 4

Let
sec(x)=-3

x=arc sec(-3)

1/cosx=-3

then cosx=-1/3

sinx= - 1/3 * sqrt(8)
sin[arc sec(-3)] =sinx =-2/3*sqrt(2)

Answer 5

Sin[ArcSec[z]] = Sqrt[1 - 1/z^2] ==> 2 Sqrt[2]/3

Answer 6

the answer is -3...

because arc sec is hypotenuse/ opposite (if using the sohcahtoa method i learned in HS)

and sin is opposite/ hypotenuse.

so the two cancel eachother out and you are left with -3.

Answer 7

sec is 1/cos

-3 = 1/cos ang
cos ang = -1/3

(sin ang)^2 = 1-(cos ang)^2 from Pythag.

So, sin ang = sdqrt[1-(1-/3)^2] = sqrt[1-1/9] =
= sqrt(8/9) = 2sqrt(2)/3