anonymous
  • anonymous
Use Pythagorean identity cos^2(theta) + sine^2(thetaof theta cosine
Mathematics
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
Well, you can certainly isolate \(\cos\theta\), but you need more than just that identity to find it.
anonymous
  • anonymous
sin (theta) = sqrt(7)/7
anonymous
  • anonymous
\[ \sin^2\theta + \cos^2\theta = 1\implies |\cos\theta| = \sqrt{1-\sin^2\theta} \]Since we know that \(\theta\) is acute, we know that \(\cos\theta > 0\), and so \(|\cos\theta| = \cos\theta\).

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Given the equation you just gave, we have: \[ \cos\theta = \sqrt{1- \left(\frac{\sqrt 7}{7}\right)^2} \]
anonymous
  • anonymous
You just need to simplify.
anonymous
  • anonymous
I'm having trouble with the simplification, sorry.
anonymous
  • anonymous
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?
anonymous
  • anonymous
I'm getting sqrt(6)/7
anonymous
  • anonymous
Not quite it
anonymous
  • anonymous
the answer to your last question is 1/7
anonymous
  • anonymous
What is \(1 - 1/7\)?
anonymous
  • anonymous
0
anonymous
  • anonymous
No, \[ 1-\frac17 \]
anonymous
  • anonymous
\[ 1-\frac 17 = \frac77-\frac17 = \ldots \]
anonymous
  • anonymous
I'm getting 6/7
anonymous
  • anonymous
Okay, so: \[ \sqrt{\frac 67} = \frac{\sqrt 6}{\sqrt7} \]To rationalize the denominator, you multiply top and bottom by it: \[ \frac{\sqrt 6}{\sqrt7}= \frac{\sqrt 6\cdot\sqrt7}{\sqrt7\cdot\sqrt7} = \ldots \]
anonymous
  • anonymous
sqrt(42)/7
anonymous
  • anonymous
That looks right.
anonymous
  • anonymous
\[ \left(\frac{\sqrt{7}}7\right)^2+\left(\frac{\sqrt{42}}7\right)^2 = 1 \]
anonymous
  • anonymous
Also, \(\sqrt{42}/2 < 1 < \pi /2\), so it is acute.
anonymous
  • anonymous
thank you

Looking for something else?

Not the answer you are looking for? Search for more explanations.