can someone please help me with this problem. i'm doing my review right now and i'm stuck?
Find cos(s + t) given that cos s = 1/3, with s in quadrant 1, and sin t = -1/2, with t in quadrant 4.
s in I
if cos s = 1/3, then sin s = √8/3
t in IV
if sin t = -1/2, then cos t = √3/2
cos(s+t)
= cosscost - sinssint
= (1/3)(√3/2) - (√8/3)(-1/2)
=(√3 + √8)/6
this is the wrong answer..
I am 100% sure my answer is right.
What have they got?
Is is (√3 + 2√2)/6 perhaps.
If it is, I will let you think about it.
yes, (√3 + 2√2)/6 is the correct answer/form. thank you for your help!
Of course, I would be happy to help you with this problem! To find cos(s + t), we can use the cosine addition formula:
cos(s + t) = cos(s)cos(t) - sin(s)sin(t)
First, let's find the value of sin(s). We know that cos(s) = 1/3 and s is in quadrant 1. In quadrant 1, both the sine and cosine values are positive. We can use the Pythagorean identity to find sin(s):
sin^2(s) + cos^2(s) = 1
sin^2(s) + (1/3)^2 = 1
sin^2(s) = 1 - (1/3)^2
sin^2(s) = 1 - 1/9
sin^2(s) = 8/9
Since s is in quadrant 1, sin(s) is positive. Taking the square root of both sides, we get:
sin(s) = √(8/9)
Now, let's find the value of cos(t). We know that sin(t) = -1/2 and t is in quadrant 4. In quadrant 4, cosine values are positive and sine values are negative. We can again use the Pythagorean identity to find cos(t):
cos^2(t) + sin^2(t) = 1
cos^2(t) + (-1/2)^2 = 1
cos^2(t) = 1 - (1/2)^2
cos^2(t) = 1 - 1/4
cos^2(t) = 3/4
Since t is in quadrant 4, cos(t) is positive. Taking the square root of both sides, we get:
cos(t) = √(3/4)
Now that we know the values of sin(s) and cos(t), we can substitute them into the cosine addition formula:
cos(s + t) = cos(s)cos(t) - sin(s)sin(t)
cos(s + t) = (1/3)(√(3/4)) - (√(8/9))(-1/2)
Simplifying this expression will give you the value of cos(s + t).