Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Find the exact value of sin (arccos (3/5)). For full credit, explain your reasoning.

See more answers at
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


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

@dk02266 Can you do it?
@wio one cannot assume if cosx=3/5 the sides are of 3 units and 5 units respectively as sine,cosine etc are ratios..

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

aliter let arccos (3/5)= x then cos x=3/5 now we sinx =sart(1-cos^2x)=sqrt(1- 9/25)=sqrt (16/25) =4/5 thus sinx=4/5 or sin( arccos (3/5))=4/5 as x= arccos (3/5)
@matricked |dw:1358194140456:dw| \[\begin{split} 1 &= \sqrt{(a/c)^2+(b/c)^2} \\ &= \sqrt{a^2/c^2+b^2/c^2} \\ &= \sqrt{(a^2+b^2)/c^2} \\ &= \sqrt{(a^2+b^2)}/c \\ c&= \sqrt{(a^2+b^2)} \\ \end{split} \]You can proportion the triangle however you want. Why wouldn't you keep it as integers? My method is mathematically sound.

Not the answer you are looking for?

Search for more explanations.

Ask your own question