Using Fundamental Identities

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Nov 10, 2013
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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
 
Last edited:
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 do not see any 11sin - but I do see 11sin2(Θ) - and that comes from 7sin2(Θ) + 4sin2(Θ)

Use a pencil and paper - instead of staring at the solution.

77 is created by multiplying the numerator and the denominator by 11 - so that the denominator is rationalized (a common practice in pre-calculator days)
 
Last edited by a moderator:
I do not see any 11sin - but I do see 11sin2(Θ) - and that comes from 7sin2(Θ) + 4sin2(Θ)
So, it looks like:

\(\displaystyle 7(1 - \sin^2 Θ)\)

became:

\(\displaystyle 7 - 7 \sin^2 Θ\)

and then that left us with:

\(\displaystyle 7 = 11 \sin^2 Θ\)

Is that correct?

Use a pencil and paper - instead of staring at the solution.
I actually am using a pen and paper and was attempting to figure it out.

77 is created by multiplying the numerator and the denominator by 11 - so that the denominator is rationalized (a common practice in pre-calculator days)
I think I understand now:

\(\displaystyle 11 \sin^2 Θ = 7\)

\(\displaystyle \sqrt{11 \sin^2 Θ} = \sqrt{7}\)

\(\displaystyle \sqrt{11} \sin Θ = \sqrt{7}\)

\(\displaystyle \sin Θ = \frac{\sqrt{7}}{\sqrt{11}}\)

\(\displaystyle \sin Θ = \frac{\sqrt{77}}{11}\)

Is that right?
 
So, it looks like:

\(\displaystyle 7(1 - \sin^2 Θ)\) = 4 sin2(Θ)

became:

\(\displaystyle 7 - 7 \sin^2 Θ\) = 4 sin2(Θ)

and then that left us with:

\(\displaystyle 7 = 11 \sin^2 Θ\)

Is that correct? ............ Now it is correct .. you were leaving part of the equation out

I actually am using a pen and paper and was attempting to figure it out.

I think I understand now:

\(\displaystyle 11 \sin^2 Θ = 7\)

\(\displaystyle \sqrt{11 \sin^2 Θ} = \sqrt{7}\)

\(\displaystyle \sqrt{11} \sin Θ = \sqrt{7}\)

\(\displaystyle \sin Θ = \frac{\sqrt{7}}{\sqrt{11}}\)

\(\displaystyle \sin Θ = \frac{\sqrt{77}}{11}\)

Is that right? ............ Yes
.
 
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