Finding Trigonometric Value: tan(7π/8) Explained

This explanation details how to calculate the exact value of the trigonometric function tangent for the angle $\dfrac{7\pi}{8}$. We will use fundamental trigonometric identities and the half-angle formula.

Understanding the Angle 7π/8

The angle $\dfrac{7\pi}{8}$ radians is located in the second quadrant of the unit circle. We can relate this angle to a reference angle in the first quadrant.

Using the property $\tan(x) = -\tan(\pi - x)$, we can rewrite the expression:

$$ \tan \left( \dfrac{7\pi}{8} \right) = -\tan \left( \pi - \dfrac{7\pi}{8} \right) $$

$$ \tan \left( \dfrac{7\pi}{8} \right) = -\tan \left( \dfrac{8\pi - 7\pi}{8} \right) $$

$$ \tan \left( \dfrac{7\pi}{8} \right) = -\tan \left( \dfrac{\pi}{8} \right) $$

Now, the problem reduces to finding the value of $\tan \left( \dfrac{\pi}{8} \right)$.

Calculation Using Half-Angle Formula

We can find the value of $\tan \left( \dfrac{\pi}{8} \right)$ using the half-angle formula for tangent. The formula is:

$$ \tan \left( \dfrac{\theta}{2} \right) = \dfrac{1 - \cos(\theta)}{\sin(\theta)} $$

Let's choose $\theta = \dfrac{\pi}{4}$. Then $\dfrac{\theta}{2} = \dfrac{\pi}{8}$.

We know the values for $\cos(\dfrac{\pi}{4})$ and $\sin(\dfrac{\pi}{4})$:

• $\cos \left( \dfrac{\pi}{4} \right) = \dfrac{\sqrt{2}}{2}$
• $\sin \left( \dfrac{\pi}{4} \right) = \dfrac{\sqrt{2}}{2}$

Substitute these values into the half-angle formula:

$$ \tan \left( \dfrac{\pi}{8} \right) = \dfrac{1 - \cos \left( \dfrac{\pi}{4} \right)}{\sin \left( \dfrac{\pi}{4} \right)} $$

$$ \tan \left( \dfrac{\pi}{8} \right) = \dfrac{1 - \dfrac{\sqrt{2}}{2}}{\dfrac{\sqrt{2}}{2}} $$

To simplify, multiply the numerator and denominator by 2:

$$ \tan \left( \dfrac{\pi}{8} \right) = \dfrac{2 \times \left( 1 - \dfrac{\sqrt{2}}{2} \right)}{2 \times \left( \dfrac{\sqrt{2}}{2} \right)} $$

$$ \tan \left( \dfrac{\pi}{8} \right) = \dfrac{2 - \sqrt{2}}{\sqrt{2}} $$

Rationalize the denominator by multiplying the numerator and denominator by $\sqrt{2}$:

$$ \tan \left( \dfrac{\pi}{8} \right) = \dfrac{(2 - \sqrt{2}) \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} $$

$$ \tan \left( \dfrac{\pi}{8} \right) = \dfrac{2\sqrt{2} - (\sqrt{2})^2}{2} $$

$$ \tan \left( \dfrac{\pi}{8} \right) = \dfrac{2\sqrt{2} - 2}{2} $$

$$ \tan \left( \dfrac{\pi}{8} \right) = \sqrt{2} - 1 $$

Final Calculation for tan(7π/8)

Now, substitute the value of $\tan \left( \dfrac{\pi}{8} \right)$ back into our earlier equation:

$$ \tan \left( \dfrac{7\pi}{8} \right) = -\tan \left( \dfrac{\pi}{8} \right) $$

$$ \tan \left( \dfrac{7\pi}{8} \right) = -(\sqrt{2} - 1) $$

$$ \tan \left( \dfrac{7\pi}{8} \right) = 1 - \sqrt{2} $$

Conclusion

The calculated value of $\tan \left( \dfrac{7\pi}{8} \right)$ is $1 - \sqrt{2}$.