Trigonometric Identities
Two relationships, both provable from the unit circle, let you convert freely between sin, cos and tan — and let you prove that two very different-looking expressions are secretly the same thing.
The two identities that tie everything together
Recall that on the unit circle, a point at angle $\theta$ has coordinates $(\cos\theta, \sin\theta)$. Two relationships fall straight out of that single fact, and let you convert between the ratios:
1. $\tan\theta \equiv \dfrac{\sin\theta}{\cos\theta}$ — directly from the coordinates (height over width).
2. $\sin^2\theta + \cos^2\theta \equiv 1$ — the Pythagorean identity.
On the unit circle, a point at angle $\theta$ has coordinates $(\cos\theta, \sin\theta)$, and its distance from the centre is the radius, $1$. Pythagoras on the right triangle formed by that point gives $(\cos\theta)^2 + (\sin\theta)^2 = 1^2$. That's the identity — nothing more than Pythagoras applied to a radius-1 circle. The $\equiv$ sign means it holds for every angle, not just some, which is what makes it an identity rather than an equation to solve.
An identity (signalled by $\equiv$) is true for every value of $\theta$ — it's a statement about the ratios themselves, not about one particular angle. An equation like $\sin\theta = 0.5$ is only true for specific angles, which is what the next lesson is about. Confusing the two is a common source of wasted working: you don't "solve" an identity, you prove it, usually by manipulating one side until it matches the other.
Worked example 1
Step 3 needs the sign chosen deliberately: square-rooting gives $\pm\tfrac45$, and you use the fact that $\theta$ is acute to pick the positive one. Always read the given range — it's there precisely to fix the sign, and dropping the $\pm$ decision loses marks.
Worked example 2
Proving an identity means starting from one side and manipulating it — using algebra and other identities — until it exactly matches the other side.
Two apparently unrelated-looking expressions — a product of brackets, and a single squared ratio — turn out to be identical for every angle $\theta$. That's the whole point of an identity: it's not coincidence for a few angles, it's algebraically forced. Spotting a difference-of-two-squares pattern is one of the most common ways the Pythagorean identity gets disguised in exam questions.
Worked example 3
The trick is to spot a rearrangement of the Pythagorean identity hiding in the numerator.
The whole question turned on recognising $1 - \cos^2\theta$ as $\sin^2\theta$. Examiners love hiding the Pythagorean identity in a rearranged form — $1-\cos^2\theta$, $1-\sin^2\theta$, or $\sin^2\theta - 1$. Whenever you see "1 minus a square" of a trig ratio, suspect the identity is in play, and rewriting it usually cracks the problem open.
$\sin^2\theta$ means $(\sin\theta)^2$ — square the whole ratio — not $\sin(\theta^2)$. And $\sin^2\theta + \cos^2\theta = 1$ does not mean $\sin\theta + \cos\theta = 1$; you cannot "take the square root" term by term. The identity only holds with the squares in place.
Check yourself
Given that $\cos\theta = \dfrac{5}{13}$ and $\theta$ is acute, find the exact values of $\sin\theta$ and $\tan\theta$.
(Hint: Pythagorean identity for $\sin\theta$, then $\tan\theta = \sin\theta / \cos\theta$.)
The numbers $5, 12, 13$ form a right-triangle triple, which is why the surds came out clean — the identity is doing Pythagoras on a $5$–$12$–$13$ triangle behind the scenes.