Math  /  Trigonometry

QuestionUse an addition or subtraction formula to write the expression as a trigonometric function of one number: cos3π7cos2π21+sin3π7sin2π21=cosπA=B2\cos \frac{3 \pi}{7} \cos \frac{2 \pi}{21}+\sin \frac{3 \pi}{7} \sin \frac{2 \pi}{21}=\cos \frac{\pi}{A}=\frac{B}{2}. B=B=

Studdy Solution

STEP 1

What is this asking? We're asked to rewrite a trigonometric expression using an addition or subtraction formula and then find the values of AA and BB. Watch out! Remember those angle addition and subtraction formulas!
Also, be careful with those fractions – common denominators are key!

STEP 2

1. Rewrite using a formula
2. Simplify the expression
3. Find A
4. Find B

STEP 3

The given expression looks a lot like the cosine subtraction formula: cos(xy)=cosxcosy+sinxsiny\cos(x - y) = \cos x \cos y + \sin x \sin y.
Let's **use this**!

STEP 4

In our case, x=3π7x = \frac{3\pi}{7} and y=2π21y = \frac{2\pi}{21}, so we can rewrite the expression as: cos(3π72π21). \cos\left(\frac{3\pi}{7} - \frac{2\pi}{21}\right).

STEP 5

To subtract the fractions inside the cosine, we need a common denominator.
The **common denominator** for 7 and 21 is 21.

STEP 6

We can rewrite 3π7\frac{3\pi}{7} as 9π21\frac{9\pi}{21} by multiplying the top and bottom by 3.
So, our expression becomes: cos(9π212π21). \cos\left(\frac{9\pi}{21} - \frac{2\pi}{21}\right).

STEP 7

Now, we can subtract the fractions inside: cos(9π212π21)=cos(7π21). \cos\left(\frac{9\pi}{21} - \frac{2\pi}{21}\right) = \cos\left(\frac{7\pi}{21}\right).

STEP 8

We can simplify 7π21\frac{7\pi}{21} to π3\frac{\pi}{3} by dividing the top and bottom by 7.
So, our **simplified expression** is: cos(π3). \cos\left(\frac{\pi}{3}\right).

STEP 9

The problem states that our simplified expression equals cosπA\cos\frac{\pi}{A}.
We found that our simplified expression is cosπ3\cos\frac{\pi}{3}.

STEP 10

By comparing cosπ3\cos\frac{\pi}{3} with cosπA\cos\frac{\pi}{A}, we can see that A=3A = \mathbf{3}.

STEP 11

We know that cosπ3=12\cos\frac{\pi}{3} = \frac{1}{2}.

STEP 12

The problem also states that cosπA=B2\cos\frac{\pi}{A} = \frac{B}{2}.
We found that cosπA=cosπ3=12\cos\frac{\pi}{A} = \cos\frac{\pi}{3} = \frac{1}{2}.

STEP 13

By comparing 12\frac{1}{2} with B2\frac{B}{2}, we can see that B=1B = \mathbf{1}.

STEP 14

A=3A = 3 and B=1B = 1.

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