Question

Find the exact value, given that $\sin A = -\frac{4}{5}$, with A in quadrant IV, $\tan B = \frac{7}{24}$ with B in quadrant III, and $\cos C = -\frac{5}{13}$ with C in quadrant II.\newline$\cos 2A$

          Find the exact value, given that $\sin A = -\frac{4}{5}$, with A in quadrant IV, $\tan B = \frac{7}{24}$ with B in quadrant III, and $\cos C = -\frac{5}{13}$ with C in quadrant II.\newline$\cos 2A$
        
Find the exact value, given that sin A = -(4)/(5), with A in quadrant IV, tan B = (7)/(24) with B in quadrant III, and cos C = -(5)/(13) with C in quadrant II.cos 2A

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Precalculus with Limits
Precalculus with Limits
Ron Larson 2nd Edition
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Find the exact value, given that sinA=-(4)/(5), with A in quadrant IV, tanB=(7)/(24) with B in quadrant III, and cosC=-(5)/(13) with C in quadrant II. cos2A with B in quadrant III,and cos C--5/13 with C in quadrant II Cos2A
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Transcript

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00:01 Ok, let's begin off then with this one here.
00:03 Sine a is minus 3 fifths in quadrant number 3.
00:07 Let's draw the angle.
00:08 Here are the four quadrants.
00:11 The third quadrant is over here.
00:14 So this is the angle a.
00:18 And reference angle, we'll call it alpha, is down here.
00:25 So we have the sine of a is negative sine of alpha.
00:32 And sine a, we're told, is minus 3 fifths.
00:35 So minus 3 over 5 is minus sine alpha.
00:41 Well, this triangle here is right angled, of course.
00:47 And sine of alpha, here is 3, here is 5, over h for sine.
00:55 Missing side is, of course, 4, 3, 4, 5 triangle.
00:59 And we have here sine alpha 3 fifths.
01:06 Now, i want to work out the value of cosine a.
01:15 Cosine of a will have the same value as negative cosine of alpha.
01:23 In that quadrant, cosine is negative.
01:28 And cosine alpha, i can see quite clearly, is 4 over 5 a over h.
01:34 So cosine a, then, is negative 4 fifths.
01:37 That one we'll do in a moment.
01:40 Next, i'll deal with this one.
01:42 Cosine b is minus 3 quarters...
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