Question

The sine of an angle whose terminal side falls in the first quadrant is .4067. Find the cosine of this angle. (Hint: Use the Pythagorean identity.)
a. .165
b. .9136
c. .4067
d. not enough information is given

Answer 1

Use the Pythagorean Theorem. Or maybe just remember the Pythagorean Identity:
\[\sin ^x+\cos ^2x = 1\]

Answer 2

\[\sin ^2x+\cos ^2x=1\]

Answer 3

Sorry for the previous typo

Answer 4

using the Pythagorean identity:

\[\sin ^{2}\theta + \cos ^{\theta} = 1\]

\[\cos ^{2}\theta = 1-(0.4067)^{2} \]

now just solve for cos

Answer 5

I'm still not coming up with one of those answers.

Answer 6

what do you get when you take 1-(0.4067)^2 ?

Answer 7

Did you remember to hit the square root button after you did 1 - .4067 squared?

Answer 8

that expression will \[=\cos ^{2}\theta \]

Answer 9

@jonnymiller, I got .83459511

Answer 10

then just take the square root of that

Answer 11

I got θ=2 π ℕ_1 + arccos(((3 sqrt(9273279))/(10000)))

Answer 12

i think you are making it more complicated.......

set it up again,

\[\cos ^{2}\theta + \sin ^{2}\theta = 1\]

\[\cos ^{2}\theta + (0.4067)^{2} = 1 \]

\[\cos ^{2}\theta = 1-0.16540489\]
\[\cos ^{2}\theta =0.83459511\]
\[\cos \theta = 0.9135617713\]

Answer 13

Oh, okay. That makes sense. I was making it way too complicated, haha. Thank you for explaining!

Answer 14

no problem, i hope you understand

Answer 15

I do, thank you.

Answer 16

you're welcome

Answer 17

:)