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# Factoring Trigonometric Expressions – Explanation and Examples

*Factoring trigonometric expressions involves dividing terms in a trigonometric expression by a common term. This is often done with the goal of making a trigonometric identity more explicit.*

This factoring process works in much the same way as factoring polynomials.

When working with trigonometric identities, factoring is important. It can help simplify complex expressions. This, in turn, makes them easier to differentiate and integrate.

Since this article focuses on factoring, review by reading about factoring polynomials.

## How To Factor Trigonometric Expressions

To factor trigonometric expressions, recall how to factor polynomials. This requires first finding the greatest common factor of all the terms in an expression. Then factor out the GCF by dividing each term by the GCF. Finally, put the GCF outside of parentheses.

For a polynomial $4x^3-6x^2+2x$, the GCF is $2x$. First, divide each term in this expression by $2x$ to get $2x^2-3x+1$. Therefore, the factored expression is $2x(2x^2-3x+1)$.

Now, factoring trigonometric expressions works in much the same way. In this case, however, trig functions can also be factored out.

Sometimes, when all of the terms in an expression are trig functions, one trig function can be divided by another to make a different function. For example, $\frac{sinx}{cosx} = tanx$ and $\frac{sinx}{tanx} = cosx$.

Factoring trig expressions can make them easier to solve, differentiate, or integrate. This is also true when factoring polynomials. Another upside, however, of factoring trig functions is that it can reveal trig identities. This then makes the functions simpler.

### Factoring Trigonometric Expressions and Trigonometric Identities

As noted above, factoring trig expressions can reveal trigonometric identities. This is because the part that is factored out and/or the part leftover may be part of a trig identity. These are often a double angle or Pythagorean identity.

For example, consider $\frac{sin^4x}{cos^2x}+sin^2x$. Now, the GCF of this expression is $sin^2x$. Factoring this out yields $sin^2x(tan^2x+1)$. By the Pythagorean identity, $tan^2x+1 = sec^2x$. Therefore, this is $sin^2x(sec^2x) = \frac{sin^2x}{cos^2x} = tan^2x$.

### Factoring by Grouping

In some cases, several of the terms in an expression may contain a common factor, but other expressions may not. These leftover terms have a different greatest common factor.

Sometimes factoring these groups separately works to simplify the problem. This is especially true when the remaining terms (the parts in parentheses) for both groups are the same.

For example, consider the expression $2sinxcos^3x+3cos^2x-2sin^3xcosx-3sin^2x$. The first two terms have a GCF of $cos^2x$, and the second two have a GCF of $-sin^2x$. Then, factor this as $cos^2x(2sinxcosx+3)-sin^2x(2sinxcosx+3)$.

Now, there are two terms with a GCF of $2sinxcosx+3$. Factoring this out yields, $(2sinxcosx+3)(cos^2x-sin^2x)$. Both parentheses contain trig identities though, so this is $(sin(2x)+3)(cos(2x))$.

## Examples

This section goes over common examples of problems involving factoring trigonometric expressions and their step-by-step solutions.

### Example 1

Find the GCF of $tan^2xsinx+cos^2xsin^2x+cot^2xsin^3x$.

### Solution

The greatest common factor of this expression is actually $sinx$. This becomes clearer after employing the quotient identities.

$tan^2x=\frac{sin^2x}{cos^2x}$, and $cot^2x=\frac{cos^2x}{sin^2x}$.

Therefore, the original expression is equal to $\frac{sin^3x}{cos^2x}+cos^2xsin^2x+\frac{cos^2xsin^3x}{sin^2x}$. Simplifying this makes it $\frac{sin^3x}{cos^2x}+cos^2xsin^2x+cos^2xsinx$.

Therefore, the GCF is $sinx$, and factoring this out yields $sinx(tan^2x+cos^2xsinx+cos^2x).

### Example 2

Factor out the GCF of $cot^2xtanx+cotxsinx+cos^2x$.

### Solution

As with the first example, it helps to begin by simplifying. Use reciprocal and quotient identities to do this.

The expression then becomes $\frac{cos^2x}{sin^2x}\frac{sinx}{cosx}+\frac{cosx}{sinx}sinx+cos^2x$.

Simplifying, this is:

$\frac{cosx}{sinx}+cosx+cos^2x$.

Here, it is clearer that the GCF is $cosx$. Factoring this out yields:

$cosx(\frac{1}{sinx}+1+cosx) = cosx(cscx+1+cosx)$.

### Example 3

Use factoring and trigonometric identities to simplify $tan^2xcosx(1+cot^2x+2sinxcosx) = $tan^2xcosx+tan^2xcosxcot^2x+2sinxcos^2xtan^2x$.

### Solution

Even before simplifying, there are a few obvious common terms. It would help to factor these out. Specifically, $tan^2xcosx$ is common to all of the terms in the expression.

Factoring this out yields:

$tan^2xcosx(1+cot^2x+2sinxcosx)$.

Now, however, the inside of the parenthesis includes two identities. This expression then becomes:

$tan^2xcosx(sec^2x+sin(2x))$.

Note that the external term can also be simplified to $\frac{sin^2x}{cosx}$.

### Example 4

Factor the following expression to simplify.

$tan^4x+2tan^2x+1$

### Solution

This expression looks a lot like a quadratic, so it makes sense to factor it similarly.

This expression becomes two binomials: $(tan^2x+1)(tan^2x+1) = (tan^2x+1)$.

But, by the Pythagorean identity, $tan^2x+1= sec^2x$. Therefore, the entire expression is equal to $sec^4x$.

### Example 5

Simplify the expression $2(1-cos^2x)cot^2x-2sin^2xcot^2x$.

### Solution

The GCF of both terms is $2cot^2x$. Factoring this out yields:

$2cot^2x((1-cos^2x)-sin^2x)$.

But, by the Pythagorean identity, $1-cos^2x=sin^2x$. Therefore, the term in parenthesis is equal to $0$, and the whole expression is equal to zero.