Answer: [tex]\frac{-3\sqrt{13}+4\sqrt{3}}{20}[/tex]
This is the single fraction of -3*sqrt(13)+4*sqrt(3) up top all over 20.
sqrt = square root
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Explanation:
Angle theta is between pi and 3pi/2. This places the angle in quadrant Q3 where both cosine and sine are negative
Use the pythagorean trig identity to get the following:
[tex]\sin^2 \theta + \cos^2 \theta = 1\\\\\sin^2 \theta + \left(-\frac{\sqrt{3}}{4}\right)^2 = 1\\\\\sin^2 \theta + \frac{3}{16} = 1\\\\\sin^2 \theta = 1 - \frac{3}{16}\\\\\sin^2 \theta = \frac{16}{16} - \frac{3}{16}\\\\\sin^2 \theta = \frac{16-3}{16}\\\\\sin^2 \theta = \frac{13}{16}\\\\\sin \theta = -\sqrt{\frac{13}{16}} \ \text{ ... sine is negative in Q3}\\\\\sin \theta = -\frac{\sqrt{13}}{\sqrt{16}}\\\\\sin \theta = -\frac{\sqrt{13}}{4}\\\\[/tex]
Angle beta is in Q1 where sine and cosine are positive.
Draw a right triangle with legs 3 and 4. The hypotenuse is 5 through the pythagorean theorem. In other words, we have a 3-4-5 right triangle.
Since [tex]\tan \beta = \frac{3}{4}[/tex], this means [tex]\sin \beta = \frac{3}{5} \ \text{ and } \ \cos \beta = \frac{4}{5}[/tex]
Use these ideas:
In this case we have: opposite = 3, adjacent = 4, hypotenuse = 5.
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To recap:
[tex]\cos \theta = -\frac{\sqrt{3}}{4}\\\\\sin \theta = -\frac{\sqrt{13}}{4}\\\\\cos \beta = \frac{3}{5}\\\\\sin \beta = \frac{4}{5}\\\\[/tex]
They lead to this
[tex]\sin\left(\theta + \beta\right) = \sin \theta * \cos \beta - \cos \theta * \sin \beta\\\\\sin\left(\theta + \beta\right) = -\frac{\sqrt{13}}{4} * \frac{3}{5} - \left(-\frac{\sqrt{3}}{4}\right) * \frac{4}{5}\\\\\sin\left(\theta + \beta\right) = -\frac{3\sqrt{13}}{20}+\frac{4\sqrt{3}}{20}\\\\\sin\left(\theta + \beta\right) = \frac{-3\sqrt{13}+4\sqrt{3}}{20}\\\\[/tex]