Sin and Cos

[yangxx2]

---

Members don't see this ad.

Hi everyone,

I have a question about sin and cos.
In a question about calculating tension, the only value they gave was sin(angle) = 0.75, without giving the angle.
When going through the problem, I realized I needed to get cos(angle). TPRH said it was 2/3.

How do you calculate that without knowing the angle? I looked in the math section but there wasn't anything about it 😛 I forgot all my trig/precalc! 😕

Thanks in advance!

 

[bambam02]

sin^2 +cos^2 = 1, call cos(theta) = x

(3/4)^2 +(x)^2=1


1-(9/16) = x^2

X^2 = 7/16

x~~2.6/4 ~~~~2/3


guestimate



or you can do it like this--->

if sin = .75 = 3/4 = O/H, then cos = A/H. just remember pythagarus--> O^2 + A^2 = H^2 ---

then you can just solve for A---> A^2 = 16-9 = 7, so A= again 2.6ish

so then cos is 2.6/4 ---> about 2/3. (more than .5, less than .75 --> cut it about in the middle for the nice good ole round estimate of 2/3)

 

[yangxx2]

Ohhhhh, got it. I forgot that first rule of sin^2 + cos^2 = 1.
Thanks so much!