anonymous
  • anonymous
can someone help me with pre cal
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
what type?
anonymous
  • anonymous
Trig
anonymous
  • anonymous
Let theta be an acute angle such that sin(theta)=(6/7). evaluate the other five trigonometric functions of theta.

Looking for something else?

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

More answers

anonymous
  • anonymous
instead of theta let us use x here
anonymous
  • anonymous
given sinx =6/7 ,cosecx =1/sinx cosx =sqrt( 1-(sinx)^2 ) tanx = sinx / (sqrt( 1-(sinx)^2 )) and cotx = (sqrt( 1-(sinx)^2 ))/sinx
UnkleRhaukus
  • UnkleRhaukus
\[\boxed{\sin\theta=\frac{\text{opposite side}}{\text{hypotenuse}}},\\\boxed{\cos\theta=\frac{\text{adjacent side}}{\text{hypotenuse}}},\\\boxed{\tan\theta=\frac{\text{opposite side}}{\text{adjacent side}}},\] \[\boxed{\csc\theta=\frac1{\sin\theta}=\frac{\text{hypotenuse}}{\text{opposite side}}},\\\boxed{\sec\theta=\frac1{\cos\theta}=\frac{\text{hypotenuse}}{\text{adjacent side}}},\\\boxed{\cot\theta =\frac1{\tan\theta}=\frac{\text{adjacent side}}{\text{opposite side}}}.\]
anonymous
  • anonymous
matricked , i am confused
anonymous
  • anonymous
thank you unkleRhaukus
whpalmer4
  • whpalmer4
Both have told you essentially the same thing, although @UnkleRhaukus' formulation may appear simpler. \[\sin \theta = \frac{6}{7}\]This implies that the opposite side is 6, and the hypotenuse is 7 (or multiples thereof). To find the adjacent side, we'll have to use the Pythagorean theorem: \[a^2 + 6^2 = 7^2\]\[a^2=49-36\]\[a=\sqrt{13}\] Continuing on, \[\cos \theta = \frac{\sqrt{13}}{7}\]\[\sec \theta = \frac{1}{\cos \theta} = \frac{7}{\sqrt{13}} = \frac{7\sqrt{13}}{13}\]\[\tan\theta = \frac{6}{\sqrt{13}} = \frac{6\sqrt{13}}{13}\]\[\csc\theta = \frac{7}{6}\]\[\cot\theta = \frac{\sqrt{13}}{6}\] Now let's do them the @matricked way: \[\sin\theta = \frac{6}{7}\]\[\cos\theta = \sqrt{1-\sin^2\theta} = \sqrt{1-{(6/7)^2}} = \frac{\sqrt{13}}{7}\]\[\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{6}{\sqrt{13}} = \frac{6\sqrt{13}}{13}\]and so on. Exactly the same answers, just looks a bit more frightening because he writes out the taking of a square root to find the unknown side each time he needs it.

Looking for something else?

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