Some equations hold only for a special value or two. An identity is bolder: it is true for every angle at once. These few relationships are the grammar of trigonometry — the rules that let you rewrite any tangle of sines and cosines into something simple.
What makes an equation an identity
Most equations you have met are conditional: they are true only for certain values. The equation \( \sin\theta = \tfrac{1}{2} \) is solved by some angles — \( 30^\circ \) is one — but it fails for nearly all the others. You have to find the angles that satisfy it.
An identity is different. It is an equation that holds for every angle where both sides are defined. There is nothing to solve; the two sides are simply two names for the same quantity. A first, almost obvious example:
\[ \sin(\theta + 360^\circ) = \sin\theta \]Spinning a full turn lands you exactly where you started, so the sine cannot change — and this is true no matter which \( \theta \) you pick. That little phrase, "true for every valid angle", is the whole difference. The phrase "valid" matters because some expressions, like \( \tan\theta \), are undefined at certain angles; an identity only has to hold where both sides actually make sense.
Everything below is anchored to the unit circle: the circle of radius \( 1 \) centred at the origin. For an angle \( \theta \) measured from the positive \( x \)-axis, the point where the angle meets the circle has coordinates
\[ x = \cos\theta, \qquad y = \sin\theta. \]Drag the dot around the circle below — watch how the \( x \)-coordinate is the cosine and the \( y \)-coordinate is the sine. Every identity in this lesson is just a fact about this one picture.
The reciprocal identities
Three of the six trig functions are simply the upside-down versions of the other three. They are defined as reciprocals, so the identities are true by definition:
\[ \csc\theta = \frac{1}{\sin\theta}, \qquad \sec\theta = \frac{1}{\cos\theta}, \qquad \cot\theta = \frac{1}{\tan\theta}. \]The names line up in a way that is easy to mix up at first, so notice the pairing carefully: cosecant pairs with sine, and secant pairs with cosine. (The third letter is the giveaway.) Cotangent is the reciprocal of tangent.
A quick check on the unit circle: at \( \theta = 30^\circ \) we have \( \sin 30^\circ = \tfrac{1}{2} \), so
\[ \csc 30^\circ = \frac{1}{\sin 30^\circ} = \frac{1}{1/2} = 2. \]Because they are reciprocals, each pair also multiplies to \( 1 \): \( \sin\theta \cdot \csc\theta = 1 \), and likewise for the other two pairs. We will lean on that fact in a moment.
In words A reciprocal flips a fraction over. So wherever \( \sin\theta \) is small, \( \csc\theta = 1/\sin\theta \) is large, and where \( \sin\theta \) hits \( 0 \), the cosecant is undefined — you cannot divide by zero. The same caution applies to \( \sec\theta \) and \( \cot\theta \).
The quotient identities
Tangent is not a brand-new function — it is the ratio of sine to cosine. Using the unit-circle coordinates \( x = \cos\theta \) and \( y = \sin\theta \), the tangent is the slope \( y/x \) of the line out to the point:
\[ \tan\theta = \frac{\sin\theta}{\cos\theta}. \]Cotangent, its reciprocal, flips the ratio the other way:
\[ \cot\theta = \frac{\cos\theta}{\sin\theta}. \]These are called the quotient identities. They are worth committing to memory, because they let you trade tangents and cotangents for the more fundamental sine and cosine whenever an expression gets messy. A quick check at \( 45^\circ \), where \( \sin 45^\circ = \cos 45^\circ = \tfrac{\sqrt{2}}{2} \):
\[ \tan 45^\circ = \frac{\sin 45^\circ}{\cos 45^\circ} = \frac{\sqrt{2}/2}{\sqrt{2}/2} = 1. \]The Pythagorean identities
This is the keystone of the whole lesson, and it falls straight out of the unit circle. Every point \( (x, y) \) on a circle of radius \( 1 \) satisfies the equation of that circle:
\[ x^2 + y^2 = 1. \]But we already named those coordinates: \( x = \cos\theta \) and \( y = \sin\theta \). Substitute them in and you have the most important identity in trigonometry:
\[ \cos^2\theta + \sin^2\theta = 1. \](The notation \( \sin^2\theta \) is just shorthand for \( (\sin\theta)^2 \).) It says: square the two legs of the little right triangle inside the circle, add them, and you always get \( 1 \) — because the hypotenuse is the radius, and the radius is \( 1 \). It is the Pythagorean theorem in disguise, which is exactly why it carries that name.
Two more identities come for free. Take \( \sin^2\theta + \cos^2\theta = 1 \) and divide every term by \( \cos^2\theta \):
\[ \frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\cos^2\theta} = \frac{1}{\cos^2\theta}. \]Now read each piece through the identities you already know. The first term is \( (\sin\theta/\cos\theta)^2 = \tan^2\theta \); the middle term is just \( 1 \); the right side is \( (1/\cos\theta)^2 = \sec^2\theta \). The result is clean:
\[ \tan^2\theta + 1 = \sec^2\theta. \]Dividing the original identity by \( \sin^2\theta \) instead gives the cotangent–cosecant version by exactly the same reasoning:
\[ 1 + \cot^2\theta = \csc^2\theta. \]Tip You only need to memorise one Pythagorean identity, \( \sin^2\theta + \cos^2\theta = 1 \). The other two are not separate facts to recall — they are what you get by dividing through by \( \cos^2\theta \) or \( \sin^2\theta \). Derive them on the spot and you will never mix up where the \( 1 \) goes.
Even, odd, and putting identities to work
One more pair of facts rounds out the toolkit. Reflecting an angle across the \( x \)-axis — replacing \( \theta \) with \( -\theta \) — keeps the \( x \)-coordinate the same but flips the sign of the \( y \)-coordinate. Since \( x = \cos\theta \) and \( y = \sin\theta \), that picture says immediately:
\[ \cos(-\theta) = \cos\theta, \qquad \sin(-\theta) = -\sin\theta. \]Cosine is an even function (symmetric about the axis); sine is an odd function (it changes sign). These let you strip minus signs out of an angle.
The point of collecting all of these is leverage. Faced with an awkward expression, you replace each piece with a simpler equivalent until the clutter cancels. Watch how the reciprocal identity collapses a product to nothing more than \( 1 \):
\[ \sin\theta \cdot \csc\theta = \sin\theta \cdot \frac{1}{\sin\theta} = 1. \]That is the entire game of "proving an identity": start on one side, swap in known identities, and steer it toward the other side. The two worked examples below show the moves in full.
- Rearrange the Pythagorean identity \( \sin^2\theta + \cos^2\theta = 1 \) to isolate the numerator: \( 1 - \cos^2\theta = \sin^2\theta \).
- Substitute that into the fraction: \( \dfrac{\sin^2\theta}{\sin\theta} \).
- Cancel one factor of \( \sin\theta \) from top and bottom: \( \dfrac{\sin^2\theta}{\sin\theta} = \sin\theta \).
So \( \dfrac{1 - \cos^2\theta}{\sin\theta} = \mathbf{\sin\theta} \).
- Start on the left and replace \( \tan\theta \) using the quotient identity \( \tan\theta = \dfrac{\sin\theta}{\cos\theta} \).
- The left side becomes \( \cos\theta \cdot \dfrac{\sin\theta}{\cos\theta} \).
- The factor of \( \cos\theta \) cancels with the \( \cos\theta \) in the denominator, leaving \( \sin\theta \).
Both sides now read the same, so the identity holds: \( \cos\theta \cdot \tan\theta = \mathbf{\sin\theta} \).
Practice
Try each one yourself, then reveal the full solution.
1. Given \( \sin\theta = \dfrac{3}{5} \) and \( \cos\theta = \dfrac{4}{5} \), find \( \tan\theta \) and verify the Pythagorean identity.
Use the quotient identity \( \tan\theta = \dfrac{\sin\theta}{\cos\theta} \).
Substitute: \( \tan\theta = \dfrac{3/5}{4/5} = \dfrac{3}{5} \cdot \dfrac{5}{4} = \dfrac{3}{4} \).
Check the Pythagorean identity: \( \sin^2\theta + \cos^2\theta = \left(\tfrac{3}{5}\right)^2 + \left(\tfrac{4}{5}\right)^2 = \tfrac{9}{25} + \tfrac{16}{25} = \tfrac{25}{25} = 1 \). ✓
So \( \tan\theta = \mathbf{\dfrac{3}{4}} \).
2. Simplify \( \dfrac{\sin\theta}{\cos\theta} \cdot \cos\theta \).
The first factor is the quotient identity itself: \( \dfrac{\sin\theta}{\cos\theta} = \tan\theta \), but here it is easier to cancel directly.
Multiply: \( \dfrac{\sin\theta}{\cos\theta} \cdot \cos\theta = \dfrac{\sin\theta \cdot \cos\theta}{\cos\theta} \).
Cancel the common factor of \( \cos\theta \) from top and bottom.
So the expression simplifies to \( \mathbf{\sin\theta} \).
3. Use an identity to rewrite \( \tan^2\theta + 1 \) as a single function, then evaluate it at \( \theta = 45^\circ \).
By the Pythagorean identity for tangent, \( \tan^2\theta + 1 = \sec^2\theta \).
At \( 45^\circ \), \( \cos 45^\circ = \dfrac{\sqrt{2}}{2} \), so \( \sec 45^\circ = \dfrac{1}{\cos 45^\circ} = \dfrac{1}{\sqrt{2}/2} = \dfrac{2}{\sqrt{2}} = \sqrt{2} \).
Therefore \( \sec^2 45^\circ = \left(\sqrt{2}\right)^2 = 2 \).
As a check, \( \tan 45^\circ = 1 \), so \( \tan^2 45^\circ + 1 = 1 + 1 = 2 \). ✓
So \( \tan^2\theta + 1 = \sec^2\theta \), which equals \( \mathbf{2} \) at \( 45^\circ \).