use a sum or difference formula to find the exact value of the trigonometric function. sec(-7π/12)…

Explanation:

Step1: Recall secant - cosine relation

$\sec x=\frac{1}{\cos x}$, so $\sec\left(-\frac{7\pi}{12}\right)=\frac{1}{\cos\left(-\frac{7\pi}{12}\right)}$. Since $\cos(-\alpha)=\cos\alpha$, then $\sec\left(-\frac{7\pi}{12}\right)=\frac{1}{\cos\left(\frac{7\pi}{12}\right)}$.

Step2: Rewrite $\frac{7\pi}{12}$ as a sum

$\frac{7\pi}{12}=\frac{3\pi}{12}+\frac{4\pi}{12}=\frac{\pi}{4}+\frac{\pi}{3}$.

Step3: Use cosine sum formula

The cosine sum formula is $\cos(A + B)=\cos A\cos B-\sin A\sin B$. Here $A=\frac{\pi}{4}$ and $B = \frac{\pi}{3}$. So $\cos\left(\frac{\pi}{4}+\frac{\pi}{3}\right)=\cos\frac{\pi}{4}\cos\frac{\pi}{3}-\sin\frac{\pi}{4}\sin\frac{\pi}{3}$. We know that $\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}$, $\cos\frac{\pi}{3}=\frac{1}{2}$, $\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}$, $\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}$. Then $\cos\left(\frac{\pi}{4}+\frac{\pi}{3}\right)=\frac{\sqrt{2}}{2}\times\frac{1}{2}-\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}=\frac{\sqrt{2}(1 - \sqrt{3})}{4}$.

Step4: Find the value of secant

$\sec\left(-\frac{7\pi}{12}\right)=\frac{1}{\cos\left(\frac{7\pi}{12}\right)}=\frac{4}{\sqrt{2}(1 - \sqrt{3})}$. Rationalize the denominator: [ \begin{align*} \frac{4}{\sqrt{2}(1 - \sqrt{3})}&=\frac{4(1+\sqrt{3})}{\sqrt{2}(1 - \sqrt{3})(1+\sqrt{3})}\ &=\frac{4(1+\sqrt{3})}{\sqrt{2}(1 - 3)}\ &=\frac{4(1+\sqrt{3})}{- 2\sqrt{2}}\ &=-\sqrt{2}(1 + \sqrt{3})\ &=-\sqrt{2}-\sqrt{6} \end{align*} ]

Answer:

$-\sqrt{2}-\sqrt{6}$