Question

Use Pythagorean identity cos^2(theta) + sine^2(thetaof theta cosine

Answer 1

Well, you can certainly isolate \(\cos\theta\), but you need more than just that identity to find it.

Answer 2

sin (theta) = sqrt(7)/7

Answer 3

\[
\sin^2\theta + \cos^2\theta = 1\implies |\cos\theta| = \sqrt{1-\sin^2\theta}
\]Since we know that \(\theta\) is acute, we know that \(\cos\theta > 0\), and so \(|\cos\theta| = \cos\theta\).

Answer 4

Given the equation you just gave, we have: \[
\cos\theta = \sqrt{1- \left(\frac{\sqrt 7}{7}\right)^2}
\]

Answer 5

You just need to simplify.

Answer 6

I'm having trouble with the simplification, sorry.

Answer 7

Start simple: \[
\left(\frac{\sqrt 7}{7}\right)^2=\left(\frac{\sqrt 7}{7}\right)\left(\frac{\sqrt 7}{7}\right)=\frac{\sqrt 7\cdot \sqrt 7}{7\cdot 7} =\ldots
\]Can you simplify this?

Answer 8

I'm getting sqrt(6)/7

Answer 9

Not quite it

Answer 10

the answer to your last question is 1/7

Answer 11

What is \(1 - 1/7\)?

Answer 12

0

Answer 13

No, \[
1-\frac17
\]

Answer 14

\[
1-\frac 17 = \frac77-\frac17 = \ldots
\]

Answer 15

I'm getting 6/7

Answer 16

Okay, so: \[
\sqrt{\frac 67} = \frac{\sqrt 6}{\sqrt7}
\]To rationalize the denominator, you multiply top and bottom by it: \[
\frac{\sqrt 6}{\sqrt7}= \frac{\sqrt 6\cdot\sqrt7}{\sqrt7\cdot\sqrt7} = \ldots
\]

Answer 17

sqrt(42)/7

Answer 18

That looks right.

Answer 19

\[
\left(\frac{\sqrt{7}}7\right)^2+\left(\frac{\sqrt{42}}7\right)^2 = 1
\]

Answer 20

Also, \(\sqrt{42}/2 < 1 < \pi /2\), so it is acute.

Answer 21

thank you