Given that sinx = 7 /25 pi/2 <or equal to x <or equal to pi Find the value of cos 2x
sinx = 7/25 , π/2 ≤ x ≤ π
so x is in quadrant II , making 2x fall in quadrant IV
So the opposite side is 7, and the hypotenuse is 25, so the adjacent side would be 24 by Pythagoras ,
and cosx = -24/25
cos 2x = cos^2 x - sin^2 x
= 576/625 - 49/625 = 527/625
Just a note:
If π/2 ≤ x ≤ π, π ≤ 2x ≤ 2π which means 2x could be in QIII or QIV.
In this case, since x > 3π/4, 2x>3π/2, and is indeed in QIV, but it need not have been so.
To find the value of cos 2x, we'll need to use trigonometric identities and the given information about sin(x).
First, let's recall the formula for cos 2x:
cos 2x = cos^2(x) - sin^2(x)
From the given information, we know that sin(x) = 7/25 and π/2 ≤ x ≤ π.
To find cos(x), we can use the Pythagorean Identity:
sin^2(x) + cos^2(x) = 1
Since we know sin(x), we can substitute it into the equation:
(7/25)^2 + cos^2(x) = 1
49/625 + cos^2(x) = 1
Rearranging the equation, we find:
cos^2(x) = 1 - 49/625
cos^2(x) = 576/625
From here, we can take the square root of both sides to find cos(x):
cos(x) = ±√(576/625)
cos(x) = ±(24/25)
Now, let's use the double-angle identity to find cos 2x:
cos 2x = cos^2(x) - sin^2(x)
Substituting the values we found earlier:
cos 2x = (±24/25)^2 - (7/25)^2
cos 2x = (576/625) - (49/625)
cos 2x = 527/625
Therefore, the value of cos 2x is 527/625.