Answer:
![(-3√(13)+4√(3))/(20)](https://img.qammunity.org/2023/formulas/mathematics/high-school/tndps8sfbfw0vgdln913abs1lq0zr7x7jd.png)
This is the single fraction of -3*sqrt(13)+4*sqrt(3) up top all over 20.
sqrt = square root
=======================================================
Step-by-step 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:
![\sin^2 \theta + \cos^2 \theta = 1\\\\\sin^2 \theta + \left(-(√(3))/(4)\right)^2 = 1\\\\\sin^2 \theta + (3)/(16) = 1\\\\\sin^2 \theta = 1 - (3)/(16)\\\\\sin^2 \theta = (16)/(16) - (3)/(16)\\\\\sin^2 \theta = (16-3)/(16)\\\\\sin^2 \theta = (13)/(16)\\\\\sin \theta = -\sqrt{(13)/(16)} \ \text{ ... sine is negative in Q3}\\\\\sin \theta = -(√(13))/(√(16))\\\\\sin \theta = -(√(13))/(4)\\\\](https://img.qammunity.org/2023/formulas/mathematics/high-school/1kq5g3hr5fobc24geb0afdcd3p3cgh2k1s.png)
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
, this means
![\sin \beta = (3)/(5) \ \text{ and } \ \cos \beta = (4)/(5)](https://img.qammunity.org/2023/formulas/mathematics/high-school/x55lwia566r89ribxckuaee6wy5sir9fzi.png)
Use these ideas:
- sin = opposite/hypotenuse
- cos = adjacent/hypotenuse
- tan = opposite/adjacent
In this case we have: opposite = 3, adjacent = 4, hypotenuse = 5.
-------------------------------------
To recap:
![\cos \theta = -(√(3))/(4)\\\\\sin \theta = -(√(13))/(4)\\\\\cos \beta = (3)/(5)\\\\\sin \beta = (4)/(5)\\\\](https://img.qammunity.org/2023/formulas/mathematics/high-school/9fxmp9dmrkoswx4cqt2psgst1ye1pgpz9k.png)
They lead to this
![\sin\left(\theta + \beta\right) = \sin \theta * \cos \beta - \cos \theta * \sin \beta\\\\\sin\left(\theta + \beta\right) = -(√(13))/(4) * (3)/(5) - \left(-(√(3))/(4)\right) * (4)/(5)\\\\\sin\left(\theta + \beta\right) = -(3√(13))/(20)+(4√(3))/(20)\\\\\sin\left(\theta + \beta\right) = (-3√(13)+4√(3))/(20)\\\\](https://img.qammunity.org/2023/formulas/mathematics/high-school/wapk90og9wgrfd5fll9s998taoojoqyagq.png)