How do you use the angle sum or difference identity to find the exact value of #tan((7pi)/12)#?

Question

Emery Adkins · Accepted Answer

#- (2 + sqrt3)# #tan ((7pi)/12) = tan (pi/12 + pi) = - tan (pi/12)# Call #tan (pi/12) = tan t# --># tan 2t = tan (pi/6) = 1/sqrt3# Use trig identity: #tan 2a = (2tan a)/(1 - tan^2 a)# #1/sqrt3 = (2tan t)/(1 - tan^2 t)# Cross multiply --> #1 - tan^2 t = 2sqrt3tan t# #- tan^2 t - 2sqrt3tan t + 1 = 0.# Solve this quadratic equation for tan t by the improved quadratic equation (Socratic Search) #D = d^2 = b^2 - 4ac = 12 + 4 = 16# --> #d = +- 4# There are 2 real roots: #tan t = -b/(2a) +- d/(2a) = (-2sqrt3)/-2 +- d/2 = sqrt3 +- 2#. Since #pi/12# is in Quadrant I, its tan is positive #tan t = tan (pi/12) = sqrt3 + 2# There for: #tan ((7pi)/12) = - tan t = - (sqrt3 + 2)#

Madelyn Adams · Answer

undefined To find the exact value of $\tan\left(\frac{7\pi}{12}\right)$ using the angle sum or difference identity, we can express $\frac{7\pi}{12}$ as the difference between two angles whose tangent values are known.

$\frac{7\pi}{12}$ can be expressed as $\frac{\pi}{3} - \frac{\pi}{4}$.

Since $\tan(\alpha - \beta) = \frac{\tan(\alpha) - \tan(\beta)}{1 + \tan(\alpha)\tan(\beta)}$, we can use this formula to find $\tan\left(\frac{7\pi}{12}\right)$ by substituting $\alpha = \frac{\pi}{3}$ and $\beta = \frac{\pi}{4}$:

$$\tan\left(\frac{7\pi}{12}\right) = \tan\left(\frac{\pi}{3} - \frac{\pi}{4}\right) = \frac{\tan\left(\frac{\pi}{3}\right) - \tan\left(\frac{\pi}{4}\right)}{1 + \tan\left(\frac{\pi}{3}\right)\tan\left(\frac{\pi}{4}\right)}$$

Knowing that $\tan\left(\frac{\pi}{3}\right) = \sqrt{3}$ and $\tan\left(\frac{\pi}{4}\right) = 1$, we can substitute these values into the formula:

$$\tan\left(\frac{7\pi}{12}\right) = \frac{\sqrt{3} - 1}{1 + \sqrt{3} \times 1}$$

Simplify the expression:

$$\tan\left(\frac{7\pi}{12}\right) = \frac{\sqrt{3} - 1}{1 + \sqrt{3}}$$

To rationalize the denominator, multiply both numerator and denominator by the conjugate of the denominator:

$$\tan\left(\frac{7\pi}{12}\right) = \frac{(\sqrt{3} - 1)(1 - \sqrt{3})}{(1 + \sqrt{3})(1 - \sqrt{3})}$$

$$= \frac{-2\sqrt{3} + 1 + \sqrt{3} - 1}{1 - 3}$$

$$= \frac{-\sqrt{3}}{-2}$$

Finally, simplify:

$$\tan\left(\frac{7\pi}{12}\right) = \frac{\sqrt{3}}{2}$$