If $$ \sin heta=\frac{-1}{\sqrt{7}} $$ where $$ t0 $$ and $$ heta \in\left[90^{\circ}: 270^{\circ} ight] $$ determine, using a diagram, an expression in terms of $$ t $$ for $$ \sin \left( heta+30^{\circ} ight) $$.

Question

If $$ \sin 	heta=\frac{-1}{\sqrt{7}} $$ where $$ t>0 $$ and $$ 	heta \in\left[90^{\circ}: 270^{\circ}ight] $$ determine, using a diagram, an expression in terms of $$ t $$ for $$ \sin \left(	heta+30^{\circ}ight) $$.

UpStudy AI · Accepted Answer

We begin with the given information:

$$
\sin \theta = \frac{-1}{\sqrt{7}}
$$

and we are told that $$\theta \in [90^\circ, 270^\circ]$$. In this interval the angle lies in quadrants II or III, but since $$\sin \theta$$ is negative the angle must be in quadrant III. (Also, the cosine in quadrant III is negative.) We are also given a positive parameter $$t$$ which will be used to represent the horizontal leg of a right‐triangle (see diagram below).

**Step 1. Draw a Right Triangle**

Construct a right triangle corresponding to the reference angle for $$\theta$$. Let the sides be as follows:

- Opposite side (relative to $$\theta$$): $$1$$ (we take the absolute value of $$\sin \theta$$).
- Hypotenuse: $$\sqrt{1+t^2}$$.
- Adjacent side: $$t$$.

Since the sine is negative (and the angle is in quadrant III) we take

$$
\sin \theta = -\frac{1}{\sqrt{1+t^2}}
$$

We are given that

$$
\sin \theta = -\frac{1}{\sqrt{7}}
$$

Thus, we equate:

$$
\frac{1}{\sqrt{1+t^2}} = \frac{1}{\sqrt{7}} \quad \Longrightarrow \quad \sqrt{1+t^2} = \sqrt{7}
$$

Squaring both sides gives

$$
1+t^2=7 \quad \Longrightarrow \quad t^2=6,
$$

and so

$$
t=\sqrt{6}\quad (\text{since } t>0).
$$

**Step 2. Find $$\cos \theta$$**

Using the triangle, we have

$$
\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}.
$$

Because $$\theta$$ is in quadrant III, $$\cos \theta$$ is negative:

$$
\cos \theta = -\frac{t}{\sqrt{1+t^2}}.
$$

With $$t=\sqrt{6}$$ and $$1+t^2=7$$ we have

$$
\cos \theta = -\frac{\sqrt{6}}{\sqrt{7}}.
$$

But we will keep $$\cos \theta$$ in terms of $$t$$ for our final expression.

**Step 3. Use the Angle Addition Formula**

The formula for the sine of a sum is

$$
\sin(\theta+30^\circ) = \sin\theta\cos30^\circ + \cos\theta\sin30^\circ.
$$

Recall that

$$
\cos30^\circ = \frac{\sqrt{3}}{2} \quad \text{and} \quad \sin30^\circ = \frac{1}{2}.
$$

Substitute $$\sin\theta$$ and $$\cos\theta$$ into the formula:

$$
\sin(\theta+30^\circ)= \left(-\frac{1}{\sqrt{1+t^2}}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(-\frac{t}{\sqrt{1+t^2}}\right)\left(\frac{1}{2}\right).
$$

Factor out the common factor of $$-\frac{1}{2\sqrt{1+t^2}}$$:

$$
\sin(\theta+30^\circ) = -\frac{1}{2\sqrt{1+t^2}} (\sqrt{3} + t).
$$

This is the required expression in terms of $$t$$.

**Step 4. Diagram Summary**

A diagram for the triangle may be sketched as follows:

```
         |\
         | \  hypotenuse = √(1+t²)
   1     |  \
 (Opposite)   \
         |    \
         |_____\
           t
         (Adjacent)
```

In this diagram we take the opposite side as $$1$$ (in magnitude) and the adjacent side as $$t$$ (with a negative sign because the terminal side of $$\theta$$ in quadrant III is both to the left and below the origin). The hypotenuse is $$\sqrt{1+t^2}$$, and from the given $$\sqrt{1+t^2}=\sqrt{7}$$ so that $$t=\sqrt{6}$$.

**Final Answer**

$$
\boxed{\sin(\theta+30^\circ)=-\frac{\sqrt{3}+t}{2\sqrt{1+t^2}}.}
$$
$$ \sin(\theta+30^\circ) = -\frac{\sqrt{3} + t}{2\sqrt{1 + t^2}} $$

If \( \sin \theta=\frac{-1}{\sqrt{7}} \) where \( t>0 \) and \( \theta \in\left[90^{\circ}: 270^{\circ}\right] \) determine, using a diagram, an expression in terms of \( t \) for \( \sin \left(\theta+30^{\circ}\right) \).

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