assignment 9.2: sum and difference ide
score: 3/8 answered: 3/8
question 4
calculate an exact answer using a formula learned in class.
tan \frac{\pi}{12}

Question

Accepted Answer

# Explanation:
## Step1: Use the tangent difference formula
The formula for $\tan(A - B)=\frac{\tan A-\tan B}{1 + \tan A\tan B}$. We can write $\frac{\pi}{12}=\frac{\pi}{3}-\frac{\pi}{4}$.
So, $\tan\frac{\pi}{12}=\tan(\frac{\pi}{3}-\frac{\pi}{4})$.
Substitute $A = \frac{\pi}{3}$ and $B=\frac{\pi}{4}$ into the formula:
$$
\begin{align*}
\tan(\frac{\pi}{3}-\frac{\pi}{4})&=\frac{\tan\frac{\pi}{3}-\tan\frac{\pi}{4}}{1+\tan\frac{\pi}{3}\tan\frac{\pi}{4}}\
\end{align*}
$$
## Step2: Substitute the values of $\tan\frac{\pi}{3}$ and $\tan\frac{\pi}{4}$
We know that $\tan\frac{\pi}{3}=\sqrt{3}$ and $\tan\frac{\pi}{4} = 1$.
$$
\begin{align*}
\frac{\tan\frac{\pi}{3}-\tan\frac{\pi}{4}}{1+\tan\frac{\pi}{3}\tan\frac{\pi}{4}}&=\frac{\sqrt{3}- 1}{1+\sqrt{3}\times1}\
&=\frac{\sqrt{3}-1}{1 + \sqrt{3}}\
\end{align*}
$$
## Step3: Rationalize the denominator
Multiply the numerator and denominator by $\sqrt{3}-1$:
$$
\begin{align*}
\frac{(\sqrt{3}-1)(\sqrt{3}-1)}{(1+\sqrt{3})(\sqrt{3}-1)}&=\frac{3-2\sqrt{3}+1}{3 - 1}\
&=\frac{4-2\sqrt{3}}{2}\
&=2-\sqrt{3}
\end{align*}
$$

# Answer:
$2-\sqrt{3}$

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QUESTION IMAGE

Question

Explanation:

Step1: Use the tangent difference formula

Step2: Substitute the values of \(\tan\frac{\pi}{3}\) and \(\tan\frac{\pi}{4}\)

Step3: Rationalize the denominator

Answer: