Given that cos0 = sqrt 2/2 of a triangle on the unit circle what is the sine function of the same triangle

1 answer

In a right triangle on the unit circle, the values of sine and cosine can be derived from the angles. The cosine of an angle is defined as the x-coordinate of the point on the unit circle corresponding to that angle, while the sine is defined as the y-coordinate.

Given that \(\cos(0) = \frac{\sqrt{2}}{2}\), this actually corresponds to an angle of \(45^\circ\) (or \(\frac{\pi}{4}\) radians), as the cosine of \(45^\circ\) is indeed \(\frac{\sqrt{2}}{2}\).

For \(45^\circ\), the sine function is also:

\[ \sin(45^\circ) = \frac{\sqrt{2}}{2} \]

So, in a right triangle where \(\cos(0) = \frac{\sqrt{2}}{2}\) suggests an angle of \(45^\circ\), the sine function for the same angle is:

\[ \sin(45^\circ) = \frac{\sqrt{2}}{2} \]