Question

Let sin x = .22. Which sec x?
A. 1.03
B. 1.13
C. 1.33
D. 1.30

Answer 1

You should really give it a try yourself first.....

Sove for x arcsin0.22

Do I/cosx...

you will get the answer.

Answer 2

I think arcsine(0.22) = angles 12.709 and 167.29. So if the value of x = 12.709. So
sec x= 1/cosx ... could you explain your train of thought? @Jhannybean

Answer 3

I see where you were going with that.

Answer 4

I thought the answer was 1.03..... am I wrong?

Answer 5

Nevermind :) The question looked a little confusing to me.

So we'll do \[\large \sin^{-1}(0.22)=x \] We get our x-value. Using this x-value, we plug it into 1/cos(x) found by the identity \(\large \sec(x) = \cfrac{1}{\cos(x)}\)plugging in the x-value we will have \[\large \cfrac{1}{\cos(\sin^{-1}(0.22))} =?\]

Answer 6

Few I though I was the one that was confused ;) my skills haven't left me yet haha

Answer 7

\[\large \sin^2(x) +\cos^2(x) = 1\]\[\large \sin(x) = 0.22\]\[\large (0.22)^2 +\cos^2(x) =1\]\[\large \cos^2(x) = 1-(0.22)^2\]\[\large \cos(x) = \pm \sqrt{1-(0.22)^2}\]And since 0.22 is in quad 1,we're only taking positive values, so, \[\large \cos(x) = +\sqrt{1-(0.22)^2}\]We know that \(\large \sec(x) = \cfrac{1}{\cos(x)}\) So apply this to the problem.\[\large \sec(x) = \cfrac{1}{\sqrt{1-(0.22)^2}}\]

Answer 8

how do you make such nice bold print writing.. Its looks very professional @Jhannybean

Answer 9

equation editor down below :)

Answer 10

but look when I type it looks like this...\[\sin ^{2} +\cos ^{2} = 1\]

Answer 11

o shoot nvm....