Trigonometry is a fundamental branch in mathematics that deals with the right-angled triangle; The relationship between it’s sides and angles. Trigonometry has many applications in the real world for solving problems that involve angles and distances. Fields like engineering and physics make use of it daily. Check out Trigonometry: Basic calculations for Right-Angled Triangles where the basics of trigonometry was covered. This article will be explaining about trigonometric identities or trig identities for short along with four worked examples on proving them.
Trigonometric identities are equalities involving trigonometric functions that hold true for all values of the variables. Learning and understanding these identities allow you to simplify numeros complex equations you will come across with ease.
First trigonometric identities you should know
The first basic trig identities that you should already be familiar with are the six trig functions. Trigonometric functions are side relationships in a right angle triangle based on the acute angle \(\theta\). These are: sine, cosine, tangent, cosecant, secant, and cotangent.
Where sine or \(\sin\) is \(\frac{\text{opposite}}{\text{hypotenuse}}\)
Cosine \(\cos\) is \(\frac{\text{adjacent}}{\text{hypotenuse}}\)
Tangent \(\tan\) is \(\frac{\text{opposite}}{\text{adjacent}}\)
The next three are actually part of the reciprocal identities
Like the name suggests, it’s simply the reciprocal of the three trig functions shown above:
Cosecant \(\csc\) is \(\frac{1}{\sin}\)
Secant \(\sec\) is \(\frac{1}{\cos}\)
Cotangent \(\cot\) is \(\frac{1}{\tan}\)
Next, you should also be familiar with the quotient identities, which thankfully is only two:
\(\tan \theta = \frac{\sin \theta}{\cos \theta}\)
\(\cot \theta = \frac{\cos \theta}{\sin \theta}\) (Like the reciprocal identity)
Example one
With the trig identities you’ve just been taught you should be able to find six trig ratios for the triangle shown above.
Solution
\(\sin\theta = \frac{4}{5}, \quad \csc\theta = \frac{5}{4}\)
\(\cos\theta = \frac{3}{5}, \quad \sec\theta = \frac{5}{3}\)
\(\tan\theta = \frac{4}{3}, \quad \cot\theta = \frac{3}{4}\)
Pythagorean identities
Based on the pythagorean theorem \(a^2 + b^2 = c^2\) you will end up with three pythagorean identities.
First one is: \(\sin^2\theta + \cos^2\theta = 1\)
By dividing the LHS and RHS (Left hand side and right hand side) of the first equation with \(\sin^2\theta\), you will end up with the second identity:
\(\frac{\sin^2\theta}{\sin^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \frac{1}{\sin^2\theta}\)
Simplify:
\(1 + \cot^2\theta = \csc^2\theta\)
The last one is gotten by dividing the LHS and RHS with \(\cos^2\theta\):
\( \frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\cos^2\theta} = \frac{1}{\cos^2\theta}\)
Simplify:
\(\tan^2\theta + 1 = \sec^2\theta\)
Odd and even identities
The even and odd identities describe how trig functions behave when you plug in negative angles; even functions stay the same when you use a negative input, while odd functions become negative when you use a negative input.
Even functions \(f(-x) = f(x)\):
\(\sin(-x) = -\sin(x)\)
\(\cos(-x) = \cos(x)\)
Odd functions \(f(-x) = -f(x)\):
\(\tan(-x) = -\tan(x)\)
\(\cot(-x) = -\cot(x)\)
\(\sec(-x) = \sec(x)\)
\(\csc(-x) = -\csc(x)\)
Co-function identities
The Cofunction identities show how trigonometric functions relate to each other when the angles add up to \(90^\circ\) or in \(\pi/2\) radians.
In radians:
\(\sin\left(\frac{\pi}{2} – x\right) = \cos(x)\)
\(\cos\left(\frac{\pi}{2} – x\right) = \sin(x)\)
\(\tan\left(\frac{\pi}{2} – x\right) = \cot(x)\)
\(\cot\left(\frac{\pi}{2} – x\right) = \tan(x)\)
\(\sec\left(\frac{\pi}{2} – x\right) = \csc(x)\)
\(\csc\left(\frac{\pi}{2} – x\right) = \sec(x)\)
In degrees:
\(\sin(90^\circ – x) = \cos(x)\)
\(\cos(90^\circ – x) = \sin(x)\)
\(\tan(90^\circ – x) = \cot(x)\)
\(\cot(90^\circ – x) = \tan(x)\)
\(\sec(90^\circ – x) = \csc(x)\)
\(\csc(90^\circ – x) = \sec(x)\)
Double angle identities
We have three double angle identities
\(\sin(2x) = 2\sin(x)\cos(x)\)
\(\cos(2x) = \cos^2(x) – \sin^2(x)\)
Alternate form 1: \(\cos(2x) = 2\cos^2(x) – 1\)
Alternate form 2: \(\cos(2x) = 1 – 2\sin^2(x)\)
\(\tan(2x) = \frac{2\tan(x)}{1 – \tan^2(x)}\)
To derive the second form for the double angle cos identity, we must go back to the first Pythagorean identity;
\( \sin^2\theta + \cos^2\theta = 1 \)
Subtract both sides of the equation by \(\cos^2\theta\) to get
\(\sin^2\theta = 1 – \cos^2\theta\)
Replace \(\cos^2\theta\) in \(\cos^2\theta – \sin^2\theta\) to end up with
\(1 – 2\sin^2\theta\)
The steps are identical for the third form but you will subtract LHS and RHS with
\(\sin^2\theta\)
to get
\(\cos^2\theta = 1 – \sin^2\theta\)
Replace \(\sin^2\theta\) in \(\cos^2\theta – \sin^2\theta\):
\(2\cos^2\theta – 1\)
Half angle identity
\(\sin\left(\frac{x}{2}\right) = \pm\sqrt{\frac{1 – \cos(x)}{2}}\)
\(\cos\left(\frac{x}{2}\right) = \pm\sqrt{\frac{1 + \cos(x)}{2}}\)
\(\tan\left(\frac{x}{2}\right) = \pm\sqrt{\frac{1 – \cos(x)}{1 + \cos(x)}}\)
Moreover, there are two other forms of the half angle tan identity:
Alternate form 1: \(\tan\left(\frac{x}{2}\right) = \frac{\sin(x)}{1 + \cos(x)}\)
Alternate form 2: \(\tan\left(\frac{x}{2}\right) = \frac{1 – \cos(x)}{\sin(x)}\)
Instead of memorizing each one here is how you can derive the second and third form from the first one:
Multiply the numerator and denominator under the square root by \(1 – \cos x\):
\(\tan\left(\frac{x}{2}\right) = \frac{1 – \cos x}{\sqrt{(1 + \cos x)(1 – \cos x)}} \)
Recall the identity:
\(1 + \cos x)(1 – \cos x) = 1 – \cos^2 x = \sin^2 x \)
So the expression becomes:
\(\tan\left(\frac{x}{2}\right) = \frac{1 – \cos x}{\sin x} \)
Now multiply numerator and denominator by \(1 + \cos x\):
\(\tan\left(\frac{x}{2}\right) = \frac{1 – \cos x}{\sin x} \cdot \frac{1 + \cos x}{1 + \cos x} = \frac{(1 – \cos x)(1 + \cos x)}{\sin x (1 + \cos x)}\)
Using the identity again:
\(1 – \cos x)(1 + \cos x) = \sin^2 x \)
Simplify it:
\(\tan\left(\frac{x}{2}\right) = \frac{\sin^2 x}{\sin x(1 + \cos x)} = \frac{\sin x}{1 + \cos x}\)
Therefore
\(\tan\left(\frac{x}{2}\right) = \frac{\sin x}{1 + \cos x}\)
The second form:
Starting with the original half-angle identity for tangent:
\(\tan\left(\frac{x}{2}\right) = \sqrt{\frac{1 – \cos x}{1 + \cos x}} \)
Multiply the original root form by (\frac{\sqrt{1 – \cos x}}{\sqrt{1 – \cos x}}):
\(\tan\left(\frac{x}{2}\right) = \frac{1 – \cos x}{\sqrt{(1 + \cos x)(1 – \cos x)}} \)
Notice that:
\(1 + \cos x)(1 – \cos x) = 1 – \cos^2 x = \sin^2 x\)
So we are going to get:
\(\tan\left(\frac{x}{2}\right) = \frac{1 – \cos x}{\sin x} \)
Now multiply numerator and denominator by \(1 + \cos x\):
\(\tan\left(\frac{x}{2}\right) = \frac{1 – \cos x}{\sin x} \cdot \frac{1 + \cos x}{1 + \cos x} = \frac{(1 – \cos x)(1 + \cos x)}{\sin x(1 + \cos x)} \)
Again, use the identity:
\(1 – \cos x)(1 + \cos x) = \sin^2 x \)
So:
\(\tan\left(\frac{x}{2}\right) = \frac{\sin^2 x}{\sin x(1 + \cos x)} = \frac{\sin x}{1 + \cos x} \)
\(\tan\left(\frac{x}{2}\right) = \frac{\sin x}{1 + \cos x} \)
Sum and difference identity
\(\sin(a + b) = \sin(a)\cos(b) + \cos(a)\sin(b)\)
\(\sin(a – b) = \sin(a)\cos(b) – \cos(a)\sin(b)\)
\(\cos(a + b) = \cos(a)\cos(b) – \sin(a)\sin(b)\)
\(\cos(a – b) = \cos(a)\cos(b) + \sin(a)\sin(b)\)
\(\tan(a + b) = \frac{\tan(a) + \tan(b)}{1 – \tan(a)\tan(b)}\)
\(\tan(a – b) = \frac{\tan(a) – \tan(b)}{1 + \tan(a)\tan(b)}\)
Can you work out what \(\tan(75^\circ – 45^\circ)\) without using the calculator?
Solution:
Step 1: Use the tangent difference identity
\(\tan(a – b) = \frac{\tan a – \tan b}{1 + \tan a \cdot \tan b} \)
Let \(a = 75^\circ\), \(b = 45^\circ\):
\(\tan(75^\circ – 45^\circ) = \frac{\tan 75^\circ – \tan 45^\circ}{1 + \tan 75^\circ \cdot \tan 45^\circ} \)
Step 2: Substitute known values
\(\tan(45^\circ) = 1, \quad \tan(75^\circ) = 2 + \sqrt{3} \)
\(\tan(30^\circ) = \frac{(2 + \sqrt{3}) – 1}{1 + (2 + \sqrt{3}) \cdot 1} = \frac{1 + \sqrt{3}}{3 + \sqrt{3}} \)
Step 3: Rationalize the denominator
Multiply numerator and denominator by the conjugate of the denominator:
\(\frac{1 + \sqrt{3}}{3 + \sqrt{3}} \cdot \frac{3 – \sqrt{3}}{3 – \sqrt{3}}\)
\(= \frac{(1 + \sqrt{3})(3 – \sqrt{3})}{(3 + \sqrt{3})(3 – \sqrt{3})}\)
Numerator:
\(\ (1 + \sqrt{3})(3 – \sqrt{3}) = 3 – \sqrt{3} + 3\sqrt{3} – 3 = 2\sqrt{3} \)
Denominator:
\(3 + \sqrt{3})(3 – \sqrt{3}) = 9 – 3 = 6 \)
\(\rightarrow \tan(30^\circ) = \frac{2\sqrt{3}}{6} = \frac{\sqrt{3}}{3} \)
\(\tan(75^\circ – 45^\circ) = \frac{\sqrt{3}}{3}\)
Example two
\(\frac{1 + \sin\theta}{\cos\theta} + \frac{\cos\theta}{1 + \sin\theta} = 2\sec\theta \)
As you see from the question above, it’s insusiating that the expression on the left and the expression on the right are both the same, just different forms. But how? That’s what you need to verify through simplification and by making use of the trig identities you have been shown.
Begin by examining which of the sides is more “complicated”. Just by a look think about the possible trig identities you are familiar with that are applicable. If there’s nothing you can go ahead and see what operation you can work around to simplify or expand.
Let’s choose the LHS since it has two trig expressions. To add the two terms together we of course need them to have a common denominator; Multiply both the top and bottom (On both sides) by (1+\sin\theta)
\( \frac{(1 + \sin\theta)^2}{(1 + \sin\theta)\cos\theta} + \frac{\cos^2\theta}{(1 + \sin\theta)\cos\theta} = 2\sec\theta \)
As you can see, in the numerator you end up with a perfect square; Same binomial multiplied by itself.
\(\frac{1 + 2\sin\theta + \sin^2\theta + \cos^2\theta}{(1 + \sin\theta)\cos\theta} = 2\sec\theta \)
Now we have some trig identities that we can apply – you always want to make use of the pythagorean identities when possible
(\sin^2\theta + \cos^2\theta = 1)
\(\frac{1 + 2\sin\theta + 1}{(1 + \sin\theta)\cos\theta} = \frac{2 + 2\sin\theta}{(1 + \sin\theta)\cos\theta} \)
So now that we know that we need to keep \(\cos\theta\) in the denominator because secant is the reciprocal of cosine
To get rid of the \(1+\sin\theta\) we can factor out a 2:
\(\frac{2(1 + \sin\theta)}{(1 + \sin\theta)\cos\theta} = \frac{2}{\cos\theta} = 2\sec\theta \)
And just like that the LHS and RHS match confirming they are equal!
Example three
\(\tan\theta + \cot\theta = \sec\theta \csc\theta \)
Again, it’s either the RHS looks like the LHS or the LHS looks like the RHS.
So we start with taking what’s more complicated; Ask yourself if it’s easier to go from addition to multiplication or the opposite. When you think about it, we will be able to combine the terms in the LHS and write them as a product. Whereas, it’s hard to imagine going from multiplication to addition therefore we choose the LHS to work on
Step 2: At first glance you would think that we can’t add those two terms because it’s like adding x + y. What you need to do is write those trig functions in terms of sine and cosine (It’s a good tip to remember whenever you are stuck – express your trig functions in terms of sine and cosine when possible).
\( \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} = \sec\theta \csc\theta \)
Now that this is written as a rational expression we can create a common denominator by using scalars.
Identify the LCM (Lowest common multiple) – which is \(\cos\theta \times \sin\theta\)
(We simply took the product of our denominator):
\( \frac{\sin^2\theta + \cos^2\theta}{\sin\theta \cos\theta} = \sec\theta \csc\theta \)
Now that the denominators are exactly the same we can rewrite it under one denominator:
\(\frac{1}{\sin\theta \cos\theta} = \frac{1 \cdot 1}{\sin\theta \cos\theta} = \frac{1}{\sin\theta} \cdot \frac{1}{\cos\theta} \)
Next step is to recognize that the RHS in our expression is without a denominator, while the LHS has a numerator and denominator. That begs the question “how can we rewrite it as a single function?”
You should be able to recognize that we can use the reciprocal identity here:
\(= \csc\theta \sec\theta = \sec\theta \csc\theta \)
We’ve gotten our LHS to look the same as the RHS – except switched. Based on the commutative property of multiplication, we know that the order doesn’t matter so we can simply switch it.
Example four
\(\frac{\cos\theta \cot\theta}{1 – \sin\theta} – 1 = \csc\theta \)
In this third example you can see that right away we’ll be working with the LHS because the RHS is already simplified
First thing to do is to apply the operation; Make sure it has the same denominator by multiplying 1 with (1-\sin\theta) (Both in the bottom and top):
\(\frac{\cos\theta \cot\theta}{1 – \sin\theta} – \frac{1}{1} = \csc\theta \)
\( \frac{\cos\theta \cot\theta – (1 – \sin\theta)}{1 – \sin\theta} = \csc\theta \)
Next, let’s simplify the numerator by distributing the negative and
rewriting \(\cot\) as \(\frac{\cos\theta}{\sin\theta}\)
\(\frac{\cos\theta \cdot \left( \frac{\cos\theta}{\sin\theta} \right) – 1 + \sin\theta}{1 – \sin\theta} = \csc\theta \)
Now the only way we can combine all of these is if the numerator all has
the same denominator; Multiply everything by \(\sin\theta\):
\( \frac{\frac{\cos^2\theta}{\sin\theta} – \frac{\sin\theta}{\sin\theta} + \frac{\sin^2\theta}{\sin\theta}}{1 – \sin\theta} = \frac{\frac{\cos^2\theta – \sin\theta + \sin^2\theta}{\sin\theta}}{1 – \sin\theta} \)
\(= \frac{\frac{1 – \sin\theta}{\sin\theta}}{1 – \sin\theta} \)
Remember the pythagorean identity where \(\cos^2\theta + \sin^2\theta = 1\)
To get rid of the \(1-\sin\theta\) off the denominator we can multiply by its reciprocal:
\(= \frac{1 – \sin\theta}{\sin\theta} \cdot \frac{1}{1 – \sin\theta} \)
For the numerator: You can cancel out the \(1-\sin\theta\) from the
denominator and numerator
\(= \frac{1}{\sin\theta} = \csc\theta \)
Hence proving that LHS and RHS are equal!
Example five
\(\cos x – \left( \frac{\cos x}{1 – \tan x} \right) = \frac{\sin x \cos x}{\sin x – \cos x} \)
Let’s take the LHS to simplify since it has two terms and that give us an opportunity to combine them.
Starting by getting them to have the same denominator; Multiply top and bottom with (1-\tan x)
\(\frac{\cos x}{1} – \frac{\cos x}{1 – \tan x} = \frac{(1 – \tan x)\cos x – \cos x}{1 – \tan x} \)
\(= \frac{-\tan x \cos x}{1 – \tan x} \)
We are almost there but this is where it gets a little confusing.
Let’s convert things to sine and cosine and see what happens since we don’t have any clear identity we can use
\(= \frac{ – \left( \frac{\sin x}{\cos x} \right) \cos x }{1 – \frac{\sin x}{\cos x}} = \frac{ -\sin x }{ \frac{\cos x – \sin x}{\cos x} } \)
\(= -\sin x \cdot \frac{\cos x}{\cos x – \sin x} = \frac{ -\sin x \cos x }{ \cos x – \sin x } \)
It’s closer now but the signs do not match; Factor out a negative in
the denominator:
\(= \frac{ -\sin x \cos x }{ -(\sin x – \cos x) } = \frac{\sin x \cos x}{\sin x – \cos x} \)
Practice questions
If you understood the solutions for the previous examples try solving the following three examples as practice:
Example Questions
\(\frac{1 – \sin A}{1 + \sin A} = (\sec A – \tan A)^2\)
\(\frac{\cos\theta}{1 + \sin\theta} = \frac{1 – \sin\theta}{\cos\theta}\)
\(\cos 3x = 4\cos^3 x – 3\cos x\)
They should be fairly easy.
Like any topic in mathematics, the key to getting better at proving these identities is practice. It is very important that you don’t neglect this part of trigonometry as you will see it come up a lot in small areas with different complex questions. For instance, when solving integration of non-trigonometric functions you’ll likely need to use the substitution rule with a trigonometric function technique, and then simplifying the resulting integral with a trigonometric identity. An additional benefit to learning trig identities is how it sharpens your problem solving skills making you prominent in other areas of calculation all across the science field so keep learning!
