Respuesta :
sin(theta) = 6/10 and theta is in the second quadrant. Use trigonometric identities to find the following quantities exactly.
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sin = 3/5
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(a) cos(theta)
cos = sqrt(1 - sin^2) = -4/5 (negative in Q2)
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(b) sin(2theta) =
sin(2t) = 2sin(t)*cos(t) = -24/25 --> Q3
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(c) cos(2theta
cos(2t) = sqrt(1 - sin^2(2t)) = -7/25 (negative in Q3)
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(d) tan(2theta) = = sin(2t)/cos(2t) = 24/7 (+ in Q3)
---
sin = 3/5
=============
(a) cos(theta)
cos = sqrt(1 - sin^2) = -4/5 (negative in Q2)
-----
(b) sin(2theta) =
sin(2t) = 2sin(t)*cos(t) = -24/25 --> Q3
-----
(c) cos(2theta
cos(2t) = sqrt(1 - sin^2(2t)) = -7/25 (negative in Q3)
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(d) tan(2theta) = = sin(2t)/cos(2t) = 24/7 (+ in Q3)
[tex]tan \theta = \frac{sin \theta}{cos \theta} = \frac{3/5}{-4/5} = \frac{3}{5}*\frac{-5}{4} = -\frac{3}{4}[/tex]
[tex]csc \theta = \frac{1}{sin \theta} = \frac{5}{3} \\ sec \theta = \frac{1}{cos \theta} = -\frac{5}{4} \\ cot \theta = \frac{1}{tan \theta} = -\frac{4}{3} [/tex]
[tex]csc \theta = \frac{1}{sin \theta} = \frac{5}{3} \\ sec \theta = \frac{1}{cos \theta} = -\frac{5}{4} \\ cot \theta = \frac{1}{tan \theta} = -\frac{4}{3} [/tex]