Use the sum or difference formula for tangent to find the exact value for

Ask by Hilton Lowe. in Virgin Islands

Nov 14,2024

Question

Use the sum or difference formula for tangent to find the exact value for $$ 	an \left(-165^{\circ}ight) $$ $$ 	an \left(-165^{\circ}ight)=\square $$

UpStudy AI · Accepted Answer

To find the exact value of $$ \tan(-165^\circ) $$, we can use the tangent sum formula. The angle $$-165^\circ$$ can be expressed as:

$$
-165^\circ = -180^\circ + 15^\circ
$$

Using the tangent difference formula, which states that:

$$
\tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b}
$$

we can set $$ a = -180^\circ $$ and $$ b = 15^\circ $$. We know that:

$$
\tan(-180^\circ) = 0
$$

Now we need to find $$ \tan(15^\circ) $$. We can use the tangent sum formula for $$ 15^\circ $$:

$$
15^\circ = 45^\circ - 30^\circ
$$

Using the tangent difference formula again:

$$
\tan(15^\circ) = \tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ}
$$

We know:

$$
\tan 45^\circ = 1 \quad \text{and} \quad \tan 30^\circ = \frac{1}{\sqrt{3}}
$$

Substituting these values into the formula:

$$
\tan(15^\circ) = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}}
$$

To simplify this, we can multiply the numerator and denominator by $$\sqrt{3}$$:

$$
\tan(15^\circ) = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}
$$

Now substituting back into our expression for $$\tan(-165^\circ)$$:

$$
\tan(-165^\circ) = \tan(-180^\circ + 15^\circ) = \frac{\tan(-180^\circ) - \tan(15^\circ)}{1 + \tan(-180^\circ) \tan(15^\circ)}
$$

Since $$\tan(-180^\circ) = 0$$:

$$
\tan(-165^\circ) = \frac{0 - \tan(15^\circ)}{1 + 0 \cdot \tan(15^\circ)} = -\tan(15^\circ)
$$

Thus:

$$
\tan(-165^\circ) = -\frac{\sqrt{3} - 1}{\sqrt{3} + 1}
$$

To simplify this expression, we can multiply the numerator and denominator by $$\sqrt{3} - 1$$:

$$
\tan(-165^\circ) = -\frac{(\sqrt{3} - 1)(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = -\frac{3 - 2\sqrt{3} + 1}{3 - 1} = -\frac{4 - 2\sqrt{3}}{2} = -2 + \sqrt{3}
$$

Thus, the exact value of $$ \tan(-165^\circ) $$ is:

$$
\boxed{-2 + \sqrt{3}}
$$
The exact value of $$ \tan(-165^\circ) $$ is $$ -2 + \sqrt{3} $$.

Use the sum or difference formula for tangent to find the exact value for

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