Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

dk02266

  • 3 years ago

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

  • This Question is Open
  1. wio
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1358142890651:dw|

  2. wio
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    @dk02266 Can you do it?

  3. matricked
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  4. matricked
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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)

  5. wio
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    @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.

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

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy