Use the given information to find (a) sin(s+t), (b)...

Use the given information to find (a) $\sin(s+t)$, (b) $\tan(s+t)$, and (c) the quadrant of $s+t$. $\cos s = \frac{3}{5}$ and $\sin t = -\frac{12}{13}$, s and t in quadrant IV (a) $\sin(s+t) =$ (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.) (b) $\tan(s+t) =$ (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.) (c) What is the quadrant of $s+t$? Quadrant IV Quadrant III Quadrant II Quadrant I

Transcript

00:01 Okay, so i'm going to set up both of these triangles in quadrants.

00:06 So both of them are in fourth quadrant, s.

00:14 So cosine is my adjacent over my hypotenuse, and sign is my opposite over my hypotenuse.

00:24 So if i do pythagorean theorem, 3 squared plus b squared equals c squared.

00:30 That's going to give me b squared equals 25 minus 9.

00:35 When you move the 3 squared over, b squared equals 16, you squared the 16, and you're going to get four, but since we're going down, this is a negative 4.

00:47 When i do t, that's a squared plus negative 5 squared, equals 13 squared, a squared plus 25 equals 169.

01:02 So a squared is 169 minus 25 or 144.

01:09 Four again square root it and you get 12.

01:15 From here i'm going to set up all three trig functions for each one.

01:21 So my sign ratio at s is my opposite over my hypotenuse.

01:28 Let me make that for a little meter.

01:32 My cosine ratio is my adjacent over my hypotenuse and my tangent ratio is my opposite over my adjacent.

01:46 I'm going to do this same thing with t.

01:48 My sign of t is my opposite over my hypotenuse.

01:56 Let's do cosine next and not tangent.

01:59 My cosine ratio of t is my adjacent over my hypotenuse and my tangent ratio is my opposite over my adjacent.

02:11 We're going to use all three of these.

02:14 First define sign of s plus t and then we'll find tangent so sign of s plus t is sine cosine plus cosine sign so you're going to multiply those together so my sign of s ratio was negative four -fifths my cosine at t ratio was 12 out of 13 my sign my cosine of s was three -fifths my sign t ratio was negative five out of 13 then i'm going to simplify so it's going to be negative 48 over 65 plus negative 15 over 65 which ends up giving me negative 63 over 65 so note that your sign of s plus t is negative because we'll need that to answer the question where s plus t is in which quadrant.

03:28 Then i want to find, so we're going to move on down to tangent.

03:35 So tangent of s plus t is tangent of s plus t over one minus tangent of s times tangent of t.

03:53 And if we look up above, the numbers that we just did, our tangent at s was negative four -thirds.

04:05 Our tangent of t was negative 5 -12ths.

04:10 1 minus negative 4 -thirds, negative 5 -12s.

04:18 And then we've got to simplify...

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