Find cos0 if sin0 =sqrt 2/2 and 0 terminates in quadrant I
In quadrant I
cos θ = √2 / 2
for
θ = π / 4 rad (45°)
sin θ = sin π / 4
sin θ = √2 / 2
tan θ = sin θ / cos θ = ( √2 / 2 ) / ( √2 / 2 ) = 1
sin^2θ + cos^2θ = 1
1/2 + cos^2θ = 1
...
sin A = y/r = sqrt2/2
x^2 + y^2 = r^2
x^2 + 2 = 4
X = sqrt2.
Cos A = x/r = sqrt2/2.
To find the value of cos(0), we can use the Pythagorean identity, which states that the square of the sine of an angle plus the square of the cosine of the same angle is always equal to 1.
Given:
sin(0) = √2/2 (quadrant I)
We know that sin(0) is the y-coordinate of a point on the unit circle in the first quadrant. Since sin(0) = √2/2, we can conclude that the y-coordinate of the point on the unit circle is √2/2.
To find cos(0), we need to find the x-coordinate of the point on the unit circle in the first quadrant. Since the point lies on the unit circle, the x-coordinate is equal to the square root of 1 minus the square of the y-coordinate.
So, cos(0) = √(1 - (√2/2)^2)
= √(1 - 2/4)
= √(1 - 1/2)
= √(1/2)
= 1/√2
However, we should rationalize the denominator (multiply the numerator and denominator by √2) to simplify the expression:
cos(0) = (1/√2) * (√2/√2)
= √2/2
Therefore, cos(0) = √2/2.