Question

Given cos theta=4sqrt2/7 find sin theta

Answer 1

`\\ \rm\Rrightarrow cos\theta=(4√(2))/(7)`

Now

`\\ \rm\Rrightarrow sin^2\theta=1-cos^2\theta`

`\\ \rm\Rrightarrow sin^2\theta=1-(16(2))/(49)`

`\\ \rm\Rrightarrow sin^2\theta=1-(32)/(49)`

`\\ \rm\Rrightarrow sin^2\theta=(17)/(49)`

`\\ \rm\Rrightarrow sin\theta=(√(17))/(7)`

Answer 2

• 
  `sin \theta=√(17)/7`

Explanation:

Given

• 
  `cos \theta=4√(2)/7`

Find

• 
  `sin \theta`

Solution

Use identity:

• 
  `sin^2x+cos^2x=1`

Substitute and find:

• 
  `cos^2 \theta=(4√(2)/7)^2`
• 
  `cos^2 \theta=16*2/49=32/49`

• 
  `sin \theta=√(1-cos^2 \theta) = √(1-32/49) =√(17/49) =√(17)/7`