Jump to a key chapter

In fact, a vast number of identities can be formed from fundamental identities. To verify such identities, we use the same type of procedure that we use to solve different equations i.e. we solve for the common identity.

**Trigonometric Identities Verification Rules**

If there is an identity consisting of different Trigonometric functions such as an identity consisting of tangent and sine, try to separate them.

Try to get tangent on one side and sine on the other hand of the equation, this makes it easier to apply the basic identities.

Now the most important thing is to know which fundamental identity one has to apply.

Convert the equation into the same function i.e. for instance, either convert all the tangents to sine or all the sines to tangents.

After doing all the above steps, if the right-hand side (RHS) of the equation is equal to the left-hand side (LHS) then the identity is true: $LHS=RHS$

## How to prove a Trigonometric Identity Algebraically?

We can prove the trigonometric identities algebraically by solving LHS and RHS separately. Here first we solve the complex side of the equation and check if both the LHS and RHS are equal or not. Let's look at a few examples.

Verify that${\mathrm{cot}}^{2}\mathrm{\theta}-{\mathrm{cos}}^{2}\mathrm{\theta}={\mathrm{cot}}^{2}\theta {\mathrm{cos}}^{2}\mathrm{\theta}$ is a consistent identity.

**Solution: **

**Step 1:** Adding ${\mathrm{cos}}^{2}\mathrm{\theta}$ on both sides of the equation, we get,

${\mathrm{cot}}^{2}\mathrm{\theta}={\mathrm{cos}}^{2}\mathrm{\theta}+{\mathrm{cos}}^{2}\mathrm{\theta}{\mathrm{cot}}^{2}\mathrm{\theta}\phantom{\rule{0ex}{0ex}}={\mathrm{cos}}^{2}\mathrm{\theta}(1+{\mathrm{cot}}^{2}\mathrm{\theta})\phantom{\rule{0ex}{0ex}}$

**Step 2:** It can be seen that $1+{\mathrm{cot}}^{2}\mathrm{\theta}$ is a fundamental identity, which is

**Step 3: **Substituting for this identity, we get

${\mathrm{cot}}^{2}\mathrm{\theta}={\mathrm{cos}}^{2}\mathrm{\theta}{\mathrm{cosec}}^{2}\mathrm{\theta}$

**Step 4: **Now, we will use a reciprocal identity for cosecant,$\mathrm{cosec\theta}=\frac{1}{\mathrm{sin\theta}}$ and squaring on both sides, we get

**Step 5: **Substituting the above identity, we get

${\mathrm{cot}}^{2}\mathrm{\theta}=\frac{{\mathrm{cos}}^{2}\mathrm{\theta}}{{\mathrm{sin}}^{2}\mathrm{\theta}}$

**Step 6:** We already know that $\mathrm{cot\theta}=\frac{\mathrm{cos\theta}}{\mathrm{sin\theta}}$, and after squaring it, the above step is obtained.

Hence, $LHS=RHS$. Thus the identity is verified.

Verify the identity $\frac{1+\mathrm{cot\varphi}}{\mathrm{cosec\varphi}}=\mathrm{sin\varphi}+\mathrm{cos\varphi}$ where $\mathrm{\varphi}$ is defined over the appropriate domain.

**Solution: **

The LHS of the equation seems a bit complicated compared to the RHS of the equation. It is always easier to simplify the more complicated side of the equation. So let us simplify the LHS to get the RHS.

**Step 1:** Writing the denominator separately, we get

LHS =$\frac{1}{\mathrm{cosec\varphi}}+\frac{\mathrm{cot\varphi}}{\mathrm{cosec\varphi}}$

**Step 2:** Using the reciprocal identity,

$\mathrm{sin\varphi}=\frac{1}{\mathrm{cosec\varphi}}$

we get

LHS =$\mathrm{sin\varphi}+\frac{\mathrm{cot\varphi}}{\mathrm{cosec\varphi}}$

**Step 3: **Using the same identity again, and $\mathrm{cot}\varphi =\frac{\mathrm{cos\varphi}}{\mathrm{sin\varphi}}$, we get

LHS =$\mathrm{sin\varphi}+\mathrm{cos\varphi}$

Hence we can see that $LHS=RHS$.

Thus the identity is proved.

### When is a Trigonometric identity invalid?

A trigonometric identity is true if and only if it satisfies all the values for which the function is defined. To demonstrate this, consider the identity,

${\mathrm{sin}}^{2}\mathrm{x}=\mathrm{cosx}$

This identity does not hold for all values of x in the domain of sine and cosine. It is only true for certain values of x. To prove this identity is false, one can algebraically show that it is invalid or give a counterexample. Here one counterexample is

$LHS={\mathrm{sin}}^{2}45\xb0=\frac{1}{2}$

$RHS=\mathrm{cos}45\xb0=\frac{1}{\sqrt{2}}$

Where it can be seen that $LHS\ne RHS$.

So far we have seen how a Trigonometric identity can be proved algebraically using the fundamental identities. It is important to note that certain identities may seem true but actually be false. This has to do with one function being undefined for a certain value in the domain while the other function is defined. In that case, when we divide by something, the denominator may tend to 0 and that identity would be false. This is a rare scenario but one should be aware about it.

Solving an identity algebraically is not the only way to prove it, an alternative way is to prove it is graphically, lets take a look at it.

## How to prove a Trigonometric Identity Graphically?

For proving an identity graphically, we verify LHS and RHS separately. Firstly, the graph of LHS is plotted, then the graph of RHS is plotted. After examining the graphs, if the graphs are identical i.e. exactly the same for every value in its domain, then we say that $LHS=RHS$. The main drawback of using this method is that only simple identities can be proved. Identities with multiple and higher powers can be very difficult to graph. This is the reason why the algebraic method is preferred over this one. Let's prove an identity to get a grasp of this method.

Prove that $\mathrm{sinx}=\frac{1}{\mathrm{cosecx}}$.

**Solution: **

**Step 1: **Plot the graph of LHS, here it is sinx, and its graph looks like:

**Step 2:** Plot the graph of$y=\frac{1}{\mathrm{cos}ecx}$versus x, which looks like:

It can be seen that the above graphs are exactly the same and hence we deduce that $\mathrm{sin}x=\frac{1}{\mathrm{cos}ecx}$.

## Examples on Verifying Trigonometric Identities

Prove that ${\mathrm{cos}}^{4}\mathrm{\theta}-{\mathrm{sin}}^{4}\mathrm{\theta}=\mathrm{cos}2\mathrm{\theta}$ is a valid trigonometric identity.

**Solution: **

**Step 1:** In this particular example, the LHS seems more complicated. Hence we will try to simplify LHS in order to arrive at RHS.

**Step 2: **The term${\mathrm{cos}}^{4}\mathrm{\theta}-{\mathrm{sin}}^{4}\mathrm{\theta}$ can also be written as ${\left({\mathrm{cos}}^{2}\mathrm{\theta}\right)}^{2}-{\left({\mathrm{sin}}^{2}\mathrm{\theta}\right)}^{2}$ and now it can be written as a product of two terms using the algebraic identity ${\mathrm{a}}^{2}-{\mathrm{b}}^{2}=\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right)$. Hence, we have:

${\left({\mathrm{cos}}^{2}\mathrm{\theta}\right)}^{2}-{\left({\mathrm{sin}}^{2}\mathrm{\theta}\right)}^{2}=\left({\mathrm{cos}}^{2}\mathrm{\theta}+{\mathrm{sin}}^{2}\mathrm{\theta}\right)\left({\mathrm{cos}}^{2}\mathrm{\theta}-{\mathrm{sin}}^{2}\mathrm{\theta}\right)$

**Step 3: **Now, we can apply the Pythagorean identity, ${\mathrm{cos}}^{2}\mathrm{\theta}+{\mathrm{sin}}^{2}\mathrm{\theta}=1$,

${\mathrm{cos}}^{4}\mathrm{\theta}-{\mathrm{sin}}^{4}\mathrm{\theta}={\mathrm{cos}}^{2}\mathrm{\theta}-{\mathrm{sin}}^{2}\mathrm{\theta}$

**Step 4: **Applying the Half angle formula $\mathrm{cos}2\mathrm{\theta}={\mathrm{cos}}^{2}\mathrm{\theta}-{\mathrm{sin}}^{2}\mathrm{\theta}$, we get

${\mathrm{cos}}^{4}\mathrm{\theta}-{\mathrm{sin}}^{4}\mathrm{\theta}=\mathrm{cos}2\mathrm{\theta}$

which is what we are asked to prove.

Thus, $LHS=RHS$.

Is $\left(1+\mathrm{sinx}\right)\left(1-\mathrm{sinx}\right)=\mathrm{cosx}$ a valid trigonometric identity?

**Solution: **

**Step 1:** The LHS seems trickier than the RHS so let’s simplify LHS and check whether we arrive at RHS or not.

**Step 2:** It can be seen that LHS can be simplified using the algebraic identity.

$\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right)={\mathrm{a}}^{2}-{\mathrm{b}}^{2}$

Applying this identity respectively, we get

$\left(1+\mathrm{sinx}\right)\left(1-\mathrm{sinx}\right)=1-{\mathrm{sin}}^{2}\mathrm{x}$

**Step 3:** Now applying the Pythagorean identity for sine and cosine, we get

$1-{\mathrm{sin}}^{2}\mathrm{x}={\mathrm{cos}}^{2}\mathrm{x}$

which cannot be simplified further.

It can be observed that $LHS\ne RHS$ for every value of x. But for an identity to be valid, it should satisfy each value for which the function is defined.

Thus, the given trigonometric identity is false.

## Verifying Trigonometric Identities - Key takeaways

- A Trigonometric identity is true if and only if it satisfies all the values for which the domain is defined.
- If an identity is only true for certain values then those values are called the solutions to that equation.
- There are two ways to prove a Trigonometric identity: Algebraically and Graphically.
- To prove an identity algebraically, simplify one of the sides of the identity by reducing it to simpler terms with the help of fundamental identities.
- For the graphical method, plot the graphs of both sides of the equation and if both of them are exactly the same, then the identity is true.
- It is always more convenient to prove an identity using the Algebraic method as it is more effective and easy.

###### Learn with 1 Verifying Trigonometric Identities flashcards in the free StudySmarter app

We have **14,000 flashcards** about Dynamic Landscapes.

Already have an account? Log in

##### Frequently Asked Questions about Verifying Trigonometric Identities

How many ways are there to verify trigonometric identities?

There are two ways to verify a trigonometric identity: Algebraically and Graphically.

How to graphically verify trigonometric identities?

Plot both the sides of the identity as different graphs. If both the graphs are exactly the same for all values in its domain then the identity is true.

What are the rules when verifying trigonometric identities?

Ensure that the identity is true for each and every value in its domain. Start with the side of the identity which is complicated and try to reduce it to simpler terms.

How can trigonometric identities be verified mathematically?

By using different fundamental identities and carefully simplifying the terms, an identity can be proved mathematically.

##### About StudySmarter

StudySmarter is a globally recognized educational technology company, offering a holistic learning platform designed for students of all ages and educational levels. Our platform provides learning support for a wide range of subjects, including STEM, Social Sciences, and Languages and also helps students to successfully master various tests and exams worldwide, such as GCSE, A Level, SAT, ACT, Abitur, and more. We offer an extensive library of learning materials, including interactive flashcards, comprehensive textbook solutions, and detailed explanations. The cutting-edge technology and tools we provide help students create their own learning materials. StudySmarter’s content is not only expert-verified but also regularly updated to ensure accuracy and relevance.

Learn more