Prove and apply the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$ and its rearrangements, and the ratio identity $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$, to prove further trigonometric identities by transforming one side to the other and to simplify trigonometric expressions

What this dot point is asking

There is one equation that ties the three trigonometric ratios together, and once you have it you can rewrite almost any trigonometric expression in a simpler form. It is the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$, true for every angle $\theta$, alongside the ratio identity $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$. This dot point asks you to know why both are true, to prove further identities by transforming one side until it matches the other, and to simplify trigonometric expressions down to a single ratio. Everything on this page is in degrees, and none of it needs calculus or the compound-angle formulae, which come later.

The skill divides cleanly. Knowing the two identities is the easy half. The exam-worthy half is fluency with the rearrangements ($\sin^2\theta = 1 - \cos^2\theta$ and $\cos^2\theta = 1 - \sin^2\theta$ are used far more often than the identity in its tidy form) and the discipline of a proof: you work on one side only, you never shuffle terms across the equals sign, and you write the reason beside each line. Get those habits and identities stop being puzzles and become routine.

A word on notation. The standard shorthand $\sin^2\theta$ means $(\sin\theta)^2$, the whole ratio squared, not $\sin(\theta^2)$. The square sits on the function, so $\sin^2 30\degree = (\sin 30\degree)^2 = \left(\tfrac12\right)^2 = \tfrac14$.

The answer

The Pythagorean identity and why it is always true

The identity $\sin^2\theta + \cos^2\theta = 1$ is not a coincidence to memorise blindly; it is the theorem of Pythagoras read off the unit circle. Place a point $P$ on a circle of radius $1$ centred at the origin, with the radius $OP$ at an angle $\theta$ measured from the positive $x$-axis. By the unit-circle definition of the ratios, the coordinates of $P$ are exactly $(\cos\theta, \sin\theta)$: the horizontal coordinate is $\cos\theta$ and the vertical coordinate is $\sin\theta$.

Drop a perpendicular from $P$ straight down to the $x$-axis. This makes a right-angled triangle whose horizontal leg has length $\cos\theta$, whose vertical leg has length $\sin\theta$, and whose hypotenuse is the radius, length $1$. Pythagoras' theorem on that triangle says

which is precisely $\sin^2\theta + \cos^2\theta = 1$.

The same picture works for any angle, not just the acute one drawn. For an obtuse or reflex angle, $P$ sits in another quadrant and $\cos\theta$ or $\sin\theta$ turns negative, but squaring removes the sign, so the sum of the squares is still $1$. That is why the identity holds for every angle, with no restriction.

The two rearrangements you will actually use

In its tidy form $\sin^2\theta + \cos^2\theta = 1$ is memorable, but inside a proof you almost always need it rearranged so it replaces one squared ratio with an expression in the other:

Read each as a substitution licence. The first lets you swap $\sin^2\theta$ for $1 - \cos^2\theta$ (or, going the other way, collapse $1 - \cos^2\theta$ down to $\sin^2\theta$); the second does the same for $\cos^2\theta$. The single most common simplifying move in the whole topic is spotting a $1 - \cos^2\theta$ and rewriting it as $\sin^2\theta$, or a $1 - \sin^2\theta$ as $\cos^2\theta$. Train your eye to see those two patterns and most identities fall open.

The ratio identity and what it does

The second identity is

valid whenever $\cos\theta \ne 0$. On the unit circle this is immediate: $\tan\theta$ is defined as the vertical coordinate over the horizontal coordinate of $P$, which is $\dfrac{\sin\theta}{\cos\theta}$. Its job in algebra is to turn a $\tan$ into $\sin$ and $\cos$ so that everything is written in two ratios instead of three, after which the Pythagorean identity can finish the work. Read the other way, it also lets you collapse a stray $\dfrac{\sin\theta}{\cos\theta}$ back into a single $\tan\theta$. Whenever an expression mixes $\tan$ with $\sin$ and $\cos$, rewriting the $\tan$ first is almost always the right opening move.

Proving an identity: work one side to the other

An identity is an equation that is true for every value of $\theta$ for which both sides are defined. That is different from an equation you solve, where you are hunting for the particular angles that make it true. Because an identity is already claimed to be true everywhere, you do not solve it. You demonstrate it, by transforming one side until it becomes the other, with a justified reason at each step.

The method that earns full marks every time is this. Choose the more complicated side to work on, usually the side with a fraction, a product, or more terms, because there is more you can do to it. Write it down labelled as $\text{LHS}$ (or $\text{RHS}$). Transform it line by line, using the Pythagorean identity, the ratio identity, and ordinary algebra (common denominators, factorising, cancelling). Keep going until it is identical to the other side, then write $= \text{RHS}$. The golden rule is that you never move terms across the equals sign: you are not allowed to treat an unproven identity like a solved equation, because that would assume the very thing you are trying to show.

Simplifying a trigonometric expression

Simplifying is proving without a target: you are handed one expression and asked to reduce it to its shortest equivalent, typically a single ratio such as $\sin\theta$ or $\cos\theta$, or the constant $1$. The toolkit is identical. Rewrite any $\tan$ as $\dfrac{\sin\theta}{\cos\theta}$, put fractions over a common denominator, replace $1 - \cos^2\theta$ by $\sin^2\theta$ (or $1 - \sin^2\theta$ by $\cos^2\theta$), then cancel common factors. The expression usually collapses to one ratio. A good final answer is as short as it can be and is written in a single trigonometric function where possible.

How exam questions ask about trigonometric identities

The command word tells you exactly what is wanted:

• "Prove that ... = ..." or "Show that ... = ..." is a one-sided proof. Pick the messier side, transform it, justify each line, and finish on the other side. Both sides are printed, so the marks are entirely for the valid chain between them.
• "Simplify ..." has no target on the page; reduce the expression to a single ratio or a constant. Rewrite $\tan$, use a Pythagorean rearrangement, and cancel.
• "Given that $\cos\theta = \dots$, find the exact value of $\sin\theta$ (or $\tan\theta$)" means use $\sin^2\theta = 1 - \cos^2\theta$ to get the other squared ratio, take the square root, and choose the sign from the quadrant stated. Then the ratio identity gives $\tan\theta$.
• "Hence ..." tells you to reuse the identity you just proved or the value you just found; do not start the next part from scratch.
• "Find the exact value" forbids a rounded decimal, so leave a surd such as $\dfrac{\sqrt5}{3}$ and rationalise where natural.
• A printed right-hand side in a "show that" is a gift: it tells you the form to aim for, so engineer the working to land on it exactly.