getoutofmylaboratory
New member
- Joined
- Nov 10, 2013
- Messages
- 23
Hello again,
I have the following problem with the answer, but have a few questions about some of the steps.
\(\displaystyle \tan θ = -\frac{\sqrt{7}}{2}, \sec θ > 0\)
\(\displaystyle \tan θ = \frac{\sin θ}{\cos θ}\)
\(\displaystyle -\frac{\sqrt{7}}{2} = \frac{\sin θ}{\cos θ}\)
\(\displaystyle -\sqrt{7} = 2 \frac{\sin θ}{\cos θ}\)
\(\displaystyle -\sqrt{7} \cos θ = 2 \sin θ\)
\(\displaystyle 7 \cos^2 θ = 4 \sin^2 θ\)
\(\displaystyle 7(1 - \sin^2 θ) = 4 \sin^2 θ\)
\(\displaystyle 11 \sin^2 θ = 7\)
\(\displaystyle \sin θ = \pm \frac{-\sqrt{77}}{11}\)
\(\displaystyle \sin θ = - \frac{\sqrt{77}}{11}\)
OK, so my first question is, where did the 11 sin come from? My second question is, where did the 77 come from in the answer?
Thanks
I have the following problem with the answer, but have a few questions about some of the steps.
\(\displaystyle \tan θ = -\frac{\sqrt{7}}{2}, \sec θ > 0\)
\(\displaystyle \tan θ = \frac{\sin θ}{\cos θ}\)
\(\displaystyle -\frac{\sqrt{7}}{2} = \frac{\sin θ}{\cos θ}\)
\(\displaystyle -\sqrt{7} = 2 \frac{\sin θ}{\cos θ}\)
\(\displaystyle -\sqrt{7} \cos θ = 2 \sin θ\)
\(\displaystyle 7 \cos^2 θ = 4 \sin^2 θ\)
\(\displaystyle 7(1 - \sin^2 θ) = 4 \sin^2 θ\)
\(\displaystyle 11 \sin^2 θ = 7\)
\(\displaystyle \sin θ = \pm \frac{-\sqrt{77}}{11}\)
\(\displaystyle \sin θ = - \frac{\sqrt{77}}{11}\)
OK, so my first question is, where did the 11 sin come from? My second question is, where did the 77 come from in the answer?
Thanks
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