Given that sin x 2/3 and x is an acute angle find cos (90-x)
We know that sin x = (opp/hyp) = 2/3, which means that the opposite side of the acute angle x is 2 and the hypotenuse is 3. Using the Pythagorean theorem, we can find the adjacent side:
(adj)² + (opp)² = (hyp)²
(adj)² + 2² = 3²
(adj)² + 4 = 9
(adj)² = 5
adj = √5
Now we can use the definition of cosine to find cos(90-x):
cos(90-x) = sin x
cos(90-x) = 2/3
Therefore, cos(90-x) = 2/3.
that's a lot of work -- all you had to do was remember that cos(90-x) = sinx.
You are right, I could have simply used the fact that cos (90-x) = sin x, which is a fundamental trigonometric identity. Therefore, since sin x = 2/3, we can conclude that cos (90-x) = sin x = 2/3. Thank you for pointing that out.
To find cos(90 - x), we can use the fact that sin(90 - x) = cos(x) for any angle x.
Given that sin(x) = 2/3, we need to find cos(90 - x).
Since x is an acute angle, it means it is between 0 and 90 degrees.
To find cos(90 - x), we first need to find the value of x.
We know that sin(x) = 2/3. Using the Pythagorean Identity, we can find cos(x):
cos(x) = sqrt(1 - sin^2(x))
= sqrt(1 - (2/3)^2)
= sqrt(1 - 4/9)
= sqrt(5/9)
Now, to find cos(90 - x), we substitute cos(x) into the identity: cos(90 - x) = sin(x)
cos(90 - x) = sin(x)
= 2/3
Therefore, cos(90 - x) = 2/3.