Hello Marie,
Recall that in Quadrant IV, cosine is POSITIVE and sine is NEGATIVE.
We'll now use (sin t)^2 + (cos t)^2 = 1 so (sin t)^2 + (1/7)^2= 1 --> (sin t)^2 + (1/49)= 1 --> (sin t)^2 =1- (1/49)
Therefore: (sin t)^2 =48/49.
Again, t is in Quadrant IV, so sin t is NEGATIVE.
That means: sin t = - square root of (48/49) = -square root (48)/square root(49) = -4*(√3)/7
sin t = -4*(√3)/7
Cheers.
Patrick T.
tutor
Hi Kendall. I used the Pythagorean identity that says (sin t)^2 + (cos t)^2 = 1. Since Marie's question said cos t = 1 over 7 aka 1/7, I replaced cos t in the Pythagorean identity by 1/7. So now it is (sin t)^2 + (1/7)^2 = 1. (1/7)^2 = 1^2/7^2 =1/49, so that's how I got the 49
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04/05/24
Kendall O.
How did you get 1 for sin(t)^2+(1/7)^2=1 and where did 49 come from? Just want to know because I'm struggling in a similar topic like this.04/05/24