Trigonometric Ratios

Triangles are among the most important shapes in geometry. So much so that there’s an entire branch of mathematics, Trigonometry, dedicated to the study of triangles – the relationships between their sides and angles.

Now, in this tutorial, we will look at the concepts that form the foundation of trigonometry – trigonometric ratios.

Sides of a Right Triangle

In addition to the one right angle, there are two acute angles in a right triangle. And we take one of the acute angles as our reference angle (C in the figure below).


Now, there are special names for each of the sides of a right triangle.

  • The side opposite the right angle is the longest and is known as the hypotenuse.
  • The side opposite to the reference angle is named the opposite (or perpendicular).
  • And the third side – the side included between the right angle and our reference angle – is called the adjacent (or base).

So the “opposite” and “adjacent” depend on the choice of the reference angle. If we switch the reference to A –

Having understood what the different terms associated with a right triangle mean, we are now ready to look at the trigonometric ratios.

Trigonometric Ratios

Trigonometric ratios are ratios between the side lengths of a right triangle. And the value of a trigonometric ratio depends on the reference angle alone.

There are a total of six trigonometric ratios. We will study them in two parts (3 + 3), so it’s easier to follow.

Basic Trig Ratios

Here are three basic trigonometric ratios.

Trigonometric Ratios - Sine, Cosine, Tangent

To give you an example, let’s find the three trigonometric ratios for angle C in the figure below.

Alright. If C is our reference angle, AB will be the “opposite” and BC “adjacent”. Let’s redraw the figure and color-label it.

Now, using the formulas we just learned, here are the three ratios.

Memorization Hack

When you have just learned about the ratios, it can be a tad confusing and difficult to remember the formula for each one. So there are several mnemonics to help you recall them easily. Here’s one of them.

Reciprocal Trig Ratios

Now that we are familiar with sine (sin), cosine (cos), and tangent (tan), it’s time for their reciprocal ratios.

Trigonometric Ratios - Cosecant, Secant, Cotangent


Trigonometric Ratios – Examples


Find the values of sin P, sec P, and cot P for using the figure below.


To get the required ratios, we need to know all three sides of the triangle. So, before anything else, we need to find PQ.

Using the Pythagorean theorem, we have

PR2=PQ2+QR2PR^2 = PQ^2 + QR^2

Substituting the values PQ and QR,

PR2=52+122PR=13\begin{align*} PR^2 &= 5^2 + 12^2 \\[1em] PR &= 13 \end{align*}

Now, we just need to use the formulas for the trigonometric ratios and get the values.

sinP=oppositehypotenuse=QRPR=1213\begin{align*} \sin P &= \frac{\text{opposite}}{\text{hypotenuse}} \\[1.3em] &= \frac{QR}{PR} \, = \, \frac{12}{13} \end{align*}
secP=hypotenuseadjacent=PRPQ=135\begin{align*} \sec P &= \frac{\text{hypotenuse}}{\text{adjacent}} \\[1.3em] &= \frac{PR}{PQ} \, = \, \frac{13}{5} \end{align*}
cotP=adjacentopposite=QRPQ=512\begin{align*} \cot P &= \frac{\text{adjacent}}{\text{opposite}} \\[1.3em] &= \frac{QR}{PQ} \, = \, \frac{5}{12} \end{align*}



If sinθ=15\hspace{0.2em} \sin \theta = \frac{1}{\sqrt{5}} \hspace{0.2em}, find cosθ\hspace{0.2em} \cos \theta \hspace{0.2em} and tanθ\hspace{0.2em} \tan \theta \hspace{0.2em}.


The question gives us the value of sin θ.

sinθ=oppositehypotenuse=15\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{1}{\sqrt{5}}

So if we take a triangle in which the length of the “opposite” side is 1 and that of the hypotenuse is √5, we can use it to find the third side (adjacent).# And then we can get the ratios.

Here’s that triangle.

Now again, by the Pythagorean theorem –

AB2=AC2+BC2AB^2 = AC^2 + BC^2

Substituting the known values and solving for BC, we have

(5)2=12+BC2BC=2\begin{align*} \left( \sqrt{5} \right )^2 &= 1^2 + BC^2 \\[1em] BC &= 2 \end{align*}

And finally, the required ratios would be,

cosθ=adjacenthypotenuse=BCAB=25\begin{align*} \cos \theta &= \frac{\text{adjacent}}{\text{hypotenuse}} \\[1.3em] &= \frac{BC}{AB} \, = \, \frac{2}{\sqrt{5}} \end{align*}
tanθ=oppositeadjacent=ACBC=12\begin{align*} \tan \theta &= \frac{\text{opposite}}{\text{adjacent}} \\[1.3em] &= \frac{AC}{BC} \, = \, \frac{1}{2} \end{align*}

Note - The two sides do not necessarily have lengths 1 and √5. They just need to be in the form k and √5·k (k being some constant) so their ratio remains 1:√5.

But since here we are interested in the ratios only and not in the actual lengths, it’s perfectly fine to leave out k (it would have canceled out anyway).

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Triangle Size Doesn't Matter

It’s crucial to understand that the value of each trigonometric ratio depends only on the (reference) angle and not on the size of the triangle.

Trigonometric ratios don't depend on the size of the triangle, but only on the reference angle

So the ratios for angle A will remain constant, regardless of which of the three triangles – ABC, AB’C’, or AB”C” – we use for our calculations.

For example,

sinA=BCAB=BCAB=BCAB\sin A \,=\, \frac{ {\color{Orchid} BC} }{AB} \,=\, \frac{ {\color{Orchid} B'C'} }{AB'} \,=\, \frac{ {\color{Orchid} B''C''} }{AB''}
But Why is That?

For a given acute angle (A in our example above), all the right-angled triangles we can have will be similar (angle-angle similarity) – they will have the same shape. In other words, they will all be scaled up (or down) versions of the same triangle.

So the ratios between their side lengths will be the same. Hence, the trigonometric ratios will be the same.

Trigonometric Ratios Table

In most cases, you will be using a calculator or trigonometric tables to get the values of trigonometric ratios for different angles.

However, depending on where you study, you might be expected to memorize the values for certain standard angles. And even if you don’t need to, it is good to have some idea of what those values are and how they vary as the angle changes.

So here’s a table for your reference. Rows of related ratios (reciprocals) are in the same color.

Points to Note

  • All trigonometric ratios have non-negative values for acute angles (θ\hspace{0.2em} \theta \hspace{0.2em} < 90°\hspace{0.2em} 90 \degree \hspace{0.2em}).
  • sinθ\hspace{0.2em} \sin \theta \hspace{0.2em} increases from 0\hspace{0.2em} 0 \hspace{0.2em} to 1\hspace{0.2em} 1 \hspace{0.2em} as θ\hspace{0.2em} \theta \hspace{0.2em} increases from 0\hspace{0.2em} 0 \hspace{0.2em} to 90°\hspace{0.2em} 90 \degree \hspace{0.2em}.
  • cosθ\hspace{0.2em} \cos \theta \hspace{0.2em} decreases from 1\hspace{0.2em} 1 \hspace{0.2em} to 0\hspace{0.2em} 0 \hspace{0.2em} as θ\hspace{0.2em} \theta \hspace{0.2em} increases from 0\hspace{0.2em} 0 \hspace{0.2em} to 90°\hspace{0.2em} 90 \degree \hspace{0.2em}.
  • tanθ\hspace{0.2em} \tan \theta \hspace{0.2em} increases indefinitely from 0\hspace{0.2em} 0 \hspace{0.2em} as θ\hspace{0.2em} \theta \hspace{0.2em} increases from 0\hspace{0.2em} 0 \hspace{0.2em} to 90°\hspace{0.2em} 90 \degree \hspace{0.2em}. At 90°\hspace{0.2em} 90 \degree \hspace{0.2em}, tanθ\hspace{0.2em} \tan \theta \hspace{0.2em} is undefined.
  • tanθ\hspace{0.2em} \tan \theta \hspace{0.2em} is the ratio of sinθ\hspace{0.2em} \sin \theta \hspace{0.2em} and cosθ\hspace{0.2em} \cos \theta \hspace{0.2em}.
tanθ=sinθcosθ\tan \theta = \frac{\sin \theta}{\cos \theta}

Basic Trigonometric Identities

In this last section, let’s look at some of the basic, yet important trigonometric identities.

Trigonometric Ratios of Complementary Angles

Two angles whose sum is 90°\hspace{0.2em} 90 \degree \hspace{0.2em} are known as complementary angles. For example, 30°\hspace{0.2em} 30 \degree \hspace{0.2em} and 60°\hspace{0.2em} 60 \degree \hspace{0.2em} or 15°\hspace{0.2em} 15 \degree \hspace{0.2em} and 75°\hspace{0.2em} 75 \degree \hspace{0.2em}.

So for an angle θ\hspace{0.2em} \theta \hspace{0.2em}, the complementary angle would be (90°\hspace{0.2em} 90 \degree \hspace{0.2em}θ\hspace{0.2em} \theta \hspace{0.2em}).

That said, here’s our first set of trigonometric identities – those involving ratios of complementary angles.

sin(90°θ)=cosθcsc(90°θ)=secθtan(90°θ)=cotθ\begin{align*} \sin (90 \degree - \theta) = \cos \theta \\[1em] \csc (90 \degree - \theta) = \sec \theta \\[1em] \tan (90 \degree - \theta) = \cot \theta \end{align*}
cos(90°θ)=sinθsec(90°θ)=cscθcot(90°θ)=tanθ\begin{align*} \cos (90 \degree - \theta) = \sin \theta \\[1em] \sec (90 \degree - \theta) = \csc \theta \\[1em] \cot (90 \degree - \theta) = \tan \theta \end{align*}
Here’s Why?

In any right triangle, the two acute angles are always complementary. So θ and 90°\hspace{0.2em} 90 \degree \hspace{0.2em}θ\hspace{0.2em} \theta \hspace{0.2em} are both present in the same right triangle.

Consequently, the trigonometric ratios for both θ and 90°\hspace{0.2em} 90 \degree \hspace{0.2em}θ\hspace{0.2em} \theta \hspace{0.2em} come from the same triangle. And hence involve the same three sides. It’s just that the “adjacent” for θ is “opposite” for 90°\hspace{0.2em} 90 \degree \hspace{0.2em}θ\hspace{0.2em} \theta \hspace{0.2em} and “opposite” for θ is “adjacent” for 90°\hspace{0.2em} 90 \degree \hspace{0.2em}θ\hspace{0.2em} \theta \hspace{0.2em}. Hypotenuse remains the same for both angles.

So that explains the relation between the trigonometric ratios of complementary angles.


If sin30°=12\hspace{0.2em} \sin 30 \degree = \frac{1}{2} \hspace{0.2em}, find the value of sec60°\hspace{0.2em} \sec 60 \degree \hspace{0.2em}.


Perhaps the first thing that pops out in this question is that we have 30°\hspace{0.2em} 30 \degree \hspace{0.2em} and 60°\hspace{0.2em} 60 \degree \hspace{0.2em} – complementary angles.

So we can use the identity below to get cos60°\hspace{0.2em} \cos 60 \degree \hspace{0.2em} from sin30°\hspace{0.2em} \sin 30 \degree \hspace{0.2em}.

cosθ=sin(90°θ)\cos \theta = \sin (90 \degree - \theta)

Substituting θ=60°\hspace{0.2em} \theta = 60 \degree \hspace{0.2em}, we have

cos60°=sin(90°60°)=sin30°=12\begin{align*} \cos 60 \degree &= \sin (90 \degree - 60 \degree) \\[1.3em] &= \sin 30 \degree \\[1.3em] &= \frac{1}{2} \end{align*}

But the question requires us to find sec60°\hspace{0.2em} \sec 60 \degree \hspace{0.2em}. So, we’ll need to take the reciprocal of cos60°\hspace{0.2em} \cos 60 \degree \hspace{0.2em}.

secθ=1cosθsec60°=1cos60°=112=2\begin{align*} \sec \theta &= \frac{1}{\cos \theta} \\[1.3em] \sec 60 \degree &= \frac{1}{\cos 60 \degree} \\[1.3em] &= \frac{1}{\frac{1}{2}} \, = \, 2 \end{align*}

There’s our answer.

Relations Between Squares of Ratios

For our second set of identities, we have those involving the squares of trigonometric ratios.

  • sin2θ+cos2θ=1\hspace{0.2em} \sin ^2 \theta + \cos ^2 \theta = 1 \hspace{0.2em}
  • sec2θtan2θ=1\hspace{0.2em} \sec ^2 \theta - \tan ^2 \theta = 1 \hspace{0.2em}
  • csc2θ+cot2θ=1\hspace{0.2em} \csc ^2 \theta + \cot ^2 \theta = 1 \hspace{0.2em}

If even one of the trigonometric ratios is known for any angle, these identities can help us find any of the remaining five ratios.


Consider the triangle below.

The Pythagorean theorem tells us –

a2+b2=c2a^2 + b^2 = c^2

Dividing throughout by c, we have,

a2c2+b2c2=c2c2a2c2+b2c2=1(1)\begin{align*} \frac{a^2}{c^2} + \frac{b^2}{c^2} &= \frac{c^2}{c^2} \\[1.3em] \frac{a^2}{c^2} + \frac{b^2}{c^2} &= 1 \hspace{0.25cm} \rule[0.1cm]{1cm}{0.1em} \hspace{0.15cm} (1) \end{align*}

Also, from the triangle, it’s clear that,

ac=sinθ(2)bc=cosθ(3)\begin{align*} \frac{a}{c} &= \sin \theta \hspace{0.25cm} \rule[0.1cm]{1cm}{0.1em} \hspace{0.15cm} (2) \\[1.3em] \frac{b}{c} &= \cos \theta \hspace{0.25cm} \rule[0.1cm]{1cm}{0.1em} \hspace{0.15cm} (3) \end{align*}

Combining (1)\hspace{0.2em} (1) \hspace{0.2em}, (2)\hspace{0.2em} (2) \hspace{0.2em}, and (3)\hspace{0.2em} (3) \hspace{0.2em}, we get

sin2+cos2=1\sin ^2 + \cos ^2 = 1

And we have proved the first identity on our list.

If we divide this identity, throughout, by sin2θ\hspace{0.2em} \sin ^2 \theta \hspace{0.2em} (or cos2θ\hspace{0.2em} \cos ^2 \theta \hspace{0.2em}) and rearrange, we’ll get the second (or third) identity.

And we have proved the first identity on our list.


If sinθ=13\hspace{0.2em} \sin \theta = \frac{1}{\sqrt{3}} \hspace{0.2em} and θ<90°\hspace{0.2em} \theta < 90 \degree \hspace{0.2em}, find cosθ\hspace{0.2em} \cos \theta \hspace{0.2em} and tanθ\hspace{0.2em} \tan \theta \hspace{0.2em}.


If we want to find the cosine of an angle from its sine ratio, we can use the following trigonometric identity.

sin2θ+cos2θ=1\sin ^2 \theta + \cos ^2 \theta = 1

Substituting the value of sin θ and solving for cos θ, we get

(13)2θ+cos2θ=1cosθ=±23\begin{align*} &\left ( \frac{1}{\sqrt{3}} \right )^2 \theta + \cos ^2 \theta = 1 \\[1.3em] & \cos \theta = \pm \sqrt{\frac{2}{3}} \end{align*}

. That means cos θ (and all other trigonometric ratios) would be non-negative. So,

cosθ=23\cos \theta = \sqrt{\frac{2}{3}}

Alright. Now that we know the values of both sin θ and cos θ, we can find tan θusing the following relation

tanθ=sinθcosθ\tan \theta = \frac{\sin \theta}{\cos \theta}

Plugging in the values for sine and cosine –

tanθ=1323=12\begin{align*} \tan \theta &= \frac{\frac{1}{\sqrt{3}}}{\sqrt{\frac{2}{3}}} \\[1.75em] &= \frac{1}{\sqrt{2}} \end{align*}

That’s it.

And with that, we come to the end of this tutorial on trigonometric ratios. Until next time.