$\sin^2 \theta + \cos^2 \theta = 1$ is known as the *Pythagorean Trigonometric Identity* for reasons that will become obvious in the proof below. Having proved the first identity, we can divide this by $sin^2 \theta$ and $cos^2 \theta$ respectively to derive two further identities. The three identities are: $ \begin{align*} \sin^2 \theta + \cos^2 \theta &= 1 \tag{1} \\ \csc^2 \theta - \cot^2 \theta &= 1 \tag{2} \\ \sec^2 \theta - \tan^2 \theta &= 1 \tag{3} \\ \end{align*} $ ## Use of these identities These identities are typically used to simplify expressions, or in some cases to derive $\sin \theta$ when we have $\cos \theta$ or vice versa. *Note*: When using these identities to solve a problem, one of the final steps is often going to involve taking a square root. It is very easy to make the mistake of just using the principal root and not considering the negative value of the square root. Problems will often express some bounds on the quadrant in which the solution falls eg "Find $\cos \theta$ of the angle $\theta$ in Quadrant III for which $\sin⁡ \theta = -\frac{24}{25}quot; which allow you to select which one is appropriate. To understand the reference to "quadrant" above, if we consider points on the radius of a unit circle starting with a ray in the positive direction of the $x$ axis and measuring the angle $\theta$ counterclockwise [^1], under the [[Unit Circle Definitions of Trigonometric Functions|unit circle definition]], we establish that the coordinates of these points on the circumference are $(x,y) = (\cos \theta, \sin \theta)$. Since the question states we are considering Quadrant III, it follows that $x$ and $y$ will both be negative in this case. Therefore, having solved for $\cos^2 \theta$, when taking the final square root we should probably use the negative value so that the solution falls in Quadrant III. [^1]: Angles are measured counterclockwise, and so are Quadrants numbered. Our starting point is a ray along the $x$ axis in a positive direction from the origin. This is all purely a matter of convention, but is critical for everyone to share a common frame of reference when talking about quadrants and angle measurements. ## Proof that $\sin^2 \theta + \cos^2 \theta = 1$ Consider a right triangle $\Delta ABC$ with sides $a$, $b$ and $c$ and a right angle at vertex $B$ as follows: ![[Basic Triangle Diagram.png]] [^2] Let $m \angle CAB = \theta$, then $ \begin{align*} \text{Given} \sin \theta &= \frac{a}{b} \text{ and } \cos \theta = \frac{c}{b} \\ \\ \sin^2 \theta + \cos^2 \theta &= \frac{a^2}{b^2} + \frac{c^2}{b^2} \\ &= \frac{a^2 + c^2}{b^2} \\ \text{But} \; b^2 &= a^2 + c^2 \text{ from Pythagoras}\\ \text{so} \; \sin^2 \theta + \cos^2 \theta &= \frac{b^2}{b^2} \\ &= 1 \\ \\ \text{therefore } \sin^2 \theta + \cos^2 \theta &= 1 \tag{1}\\ \end{align*} $ ## Derivation of $\csc^2 \theta - \cot^2 \theta = 1$ $ \begin{align*} \sin^2 \theta + \cos^2 \theta &= 1 & \text{...from(1)} \\ \frac{\sin^2 \theta}{\sin^2 \theta} + \frac{\cos^2 \theta}{\sin^2 \theta} &= \frac{1}{\sin^2 \theta} & \text{...dividing by} \; \sin^2 \theta\\ 1 + \cot^2 \theta &= \csc^2 \theta & \text{...simplifying}\\ \\ \text{therefore } \csc^2 \theta - \cot^2 \theta &= 1 \tag{2} \end{align*} $ ## Derivation of $\sec^2 \theta - \tan^2 \theta = 1$ $ \begin{align*} \sin^2 \theta + \cos^2 \theta &= 1 & \text{...from(1)} \\ \frac{\sin^2 \theta}{\cos^2 \theta} + \frac{\cos^2 \theta}{\cos^2 \theta} &= \frac{1}{\cos^2 \theta} & \text{...dividing by} \; \cos^2 \theta\\ \tan^2 \theta + 1 &= \sec^2 \theta & \text{...simplifying} \\ \\ \text{therefore } \sec^2 \theta - \tan^2 \theta &= 1 \tag{3} \end{align*} $ ## References Check out the [Khan Academy Review](https://www.khanacademy.org/math/trigonometry/unit-circle-trig-func/pythagorean-identity/a/pythagorean-identity-review) on this topic. [^2]: Diagram available on [geogebra](https://www.geogebra.org/calculator/npw4rmcw)