Question

Use an Addition or Subtraction Formula to find the exact value of the expression, as demonstrated in Example 1. cos 41π 12.

Answer 1

Answer: $\frac{\sqrt{6}-\sqrt{2}}{2}$

Step-by-step explanation:

We apply the formula $\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$.

Note that  $\cos(\frac{41}{12}\pi)=\cos((\frac{36}{12}+\frac{7}{12})\pi)=\cos(3\pi + \frac{7}{12})\pi)$. Take  $x=3\pi$ and $y=\frac{7}{12}\pi$ in the formula above to get

$\cos(\frac{41}{12}\pi)=\cos(3\pi)\cos(\frac{7}{12}\pi)-\sin(3\pi)\sin(\frac{7}{12}\pi)=(-1)\cdot \cos(\frac{7}{12}\pi)-0\cdot\sin(\frac{7}{12}\pi)=-\cos(\frac{7}{12}\pi)$

Then the value of this expression is $-\cos(\frac{7}{12}\pi)$

We can use the cosine addition formula again to simplify further. Decompose the fraction in the argument as:

$\cos(\frac{7}{12}\pi)=\cos((\frac{3}{12}+\frac{4}{12})\pi)=\cos((\frac{1}{4}\pi + \frac{1}{3})\pi)$

Applying the formula with $x=\frac{1}{4}\pi$ and $y=\frac{1}{3}\pi$ we obtain

$\cos(\frac{7}{12}\pi)=\cos(\frac{1}{4}\pi)\cos(\frac{1}{3}\pi)-\sin(\frac{1}{4}\pi)\sin(\frac{1}{3}\pi)=\frac{\sqrt{2}}{2}\cdot\frac{1}{2} -\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}=\frac{\sqrt{2}-\sqrt{6}}{2}$

We conclude that this expression has the value $-\frac{\sqrt{2}-\sqrt{6}}{2}=\frac{\sqrt{6}-\sqrt{2}}{2}$