Every point $(\cos\theta, \sin\theta)$ lies on the unit circle, and the unit circle has the equation $x^2 + y^2 = 1$. Put those two facts together and you get the most used identity in trigonometry.
Key idea
$$\sin^2\theta + \cos^2\theta = 1 \quad \text{for every angle } \theta.$$ To find one value from the other, solve for its square, take the square root, and choose the sign from the quadrant.
Why it works
Substitute $x = \cos\theta$ and $y = \sin\theta$ into $x^2 + y^2 = 1$. It is the Pythagorean theorem for the right triangle with hypotenuse $1$ inside the circle. Dividing the identity by $\cos^2\theta$ gives a second form, $\tan^2\theta + 1 = \sec^2\theta$; dividing by $\sin^2\theta$ gives $1 + \cot^2\theta = \csc^2\theta$.
$\sin^2\theta$ means $(\sin\theta)^2$: find the sine, then square it.
Worked examples
Worked example: A rational answer
$\sin\theta = \frac{3}{5}$ and $\theta$ is in Quadrant II. Find $\cos\theta$.
- $\cos^2\theta = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25}$.
- So $\cos\theta = \pm\frac{4}{5}$.
- In Quadrant II, $x < 0$, so cosine is negative.
Answer: $\cos\theta = -\frac{4}{5}$
Worked example: A radical answer
$\cos\theta = \frac{1}{3}$ with $\frac{3\pi}{2} < \theta < 2\pi$. Find $\sin\theta$.
- $\sin^2\theta = 1 - \frac{1}{9} = \frac{8}{9}$, so $\sin\theta = \pm\frac{\sqrt{8}}{3} = \pm\frac{2\sqrt{2}}{3}$.
- $\theta$ is in Quadrant IV, where $y < 0$.
Answer: $\sin\theta = -\frac{2\sqrt{2}}{3}$
Worked example: Starting from tangent
$\tan\theta = 2$ and $\theta$ is in Quadrant III. Find $\sec\theta$.
- Use $\tan^2\theta + 1 = \sec^2\theta$: $\sec^2\theta = 4 + 1 = 5$, so $\sec\theta = \pm\sqrt{5}$.
- In Quadrant III cosine is negative, and secant has the same sign as cosine.
Answer: $\sec\theta = -\sqrt{5}$
Common mistake
Finding the right size and forgetting the sign: the square root only tells you $\pm$. Always ask where the angle is. A quick sketch of the terminal side shows whether $x$ (cosine) and $y$ (sine) are positive or negative.