# Proving Trigonometric Identities

Definition: An identity is an equation or statement that is always true. To prove trigonometric identities, it is necessary manipulate on one side of a given equation to demonstrate that it is equivalent to the other side.

There are various techniques we can use to achieve this, such as factoring, rewriting trigonometric functions in terms of sine or cosine, finding common denominators, multiplying by a conjugate or by a fraction equal to 1, or using fundamental trigonometric identities.

Example 1:$\dpi{120} \bg_white \frac{sin^2\,x+4 sin\,x+3}{cos^2\, x}=\frac{3+sin\,x}{1-sin\,x}$

Solution:

Step 1: We always work on the more complicated side to get it to look like the other. Thus, we will work on the left side.

As the left side is already expressed in terms of sine and cosine, we can start by simplifying. The numerator looks especially complicated, but factoring allows us to rewrite it as:

$\dpi{120} \bg_white \\ {\,\;\;\;\;\;} \frac{sin^2\,x+4 sin\,x+3}{cos^2\, x}\\\\\\ = \;\frac{(sin\,x+3)(sin\,x + 1)}{cos^2\, x}$

Step 2: Comparing the sides, we can see that the right side contains sine, while both sine and cosine are used on the left side. Using the Pythagorean identity, we can rewrite the cosine in the denominator in terms of sine:$\dpi{120} \bg_white \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\textup{Pythagorean identity:\;\;\;\;} sin^2x + cos^2x = 1 \\\\\\ {\,\;\;\;\;}\frac{(sin\,x+3)(sin\,x + 1)}{cos^2\, x}\\\\\\ = \; \frac{(sin\,x+3)(sin\,x + 1)} {1 - sin^2\, x}$

Step 3: Simplify the left side further. The numerator is already factored, so let’s factor the denominator to match:

$\dpi{120} \bg_white \\= \; \frac{(sin\,x+3)(sin\,x + 1)} {1 - sin^2\, x}\\\\\\ = \; \frac{(sin\,x+3)(sin\,x + 1)} {(1 - sin\, x)(1 + sin\,x)}$

Step 4: Reducing the left hand side gives us an expression equal to that on the right side, which successfully proves the given identity to be true:

$\dpi{120} \bg_white \\\frac{(sin\,x+3)(sin\,x + 1)} {(1 - sin\, x)(1 + sin\,x)} = \;\frac{sin\,x+3} {1 - sin\, x}\\\\\\\ {\color{DarkGreen}\frac{sin\,x+3} {1 - sin\, x} = \frac{3+sin\,x} {1 - sin\, x}}$

Example 2:  $\dpi{120} \bg_white (sin\,x+cos\,x)(tan\,x+cot\,x)=sec\,x+csc\,x$

Solution:

Step 1: Keeping in mind that we always start with the more complicated side of the equation, for this example we will start on the left side.

Rewrite the left side to express everything is in terms of sine and cosine:$\dpi{120} \bg_white (sin\,x+cos\,x)(tan\,x+cot\,x) = (sin\,x + cos\,x)\left (\frac{sin\,x}{cos\,x}+\frac{cos\,x}{sin\,x} \right )$

Step 2: Now we do what the left side is telling us to do: add and multiply.

We’ll add the fractions first:

$\dpi{120} \bg_white \\ {\;\;\;\;}(sin\,x + cos\,x)\left (\frac{sin\,x}{cos\,x}+\frac{cos\,x}{sin\,x} \right ) \\\\= (sin\,x + cos\,x)\left (\frac{sin^2\,x}{cos\,x\,sin\,x}+\frac{cos^2\,x}{sin\,x\,cos\,x} \right ) \\\\= (sin\,x + cos\,x)\left (\frac{sin^2\,x + cos^2\,x}{sin\,x\,cos\,x} \right )$

Applying the Pythagorean identity to the numerator:

$\dpi{120} \bg_white = (sin\,x + cos\,x)\left (\frac{1}{sin\,x\,cos\,x} \right )$

Step 3: Multiply by distributing:$\dpi{120} \bg_white \\\\(sin\,x + cos\,x)\!\!\left(\frac{1}{sin\,x\,cos\,x} \right ) = \left (\frac{sin\,x}{sin\,x\,cos\,x}\right )\!+\!\left (\frac{cos\,x}{sin\,x\,cos\,x}\right ) = \frac{1}{cos\,x}\;+\;\frac{1}{sin\,x}$

Which is now equivalent to the right side:

$\dpi{120} \bg_white \\\frac{1}{cos\,x}\;+\;\frac{1}{sin\,x}\;=\;sec\,x + csc\,x \\\\\\ {\color{DarkGreen} sec\,x + csc\,x = \;sec\,x + csc\,x}$

Example 3: $\dpi{120} \bg_white \frac{cos\,x+1}{sin^3\,x}\;=\;\frac{csc\,x}{1-cos\,x}$

Solution:

Step 1: Starting with the more complicated right side, express everything in terms of sine and cosine:

$\dpi{120} \bg_white \frac{csc\,x}{1-cos\,x} = \frac{\frac{1}{sin\,x}}{1 - cos\,x}$

Simplify by doing what the right side is telling us to do, divide:$\dpi{120} \bg_white \frac{\frac{1}{sin\,x}}{1 - cos\,x} = \frac{1}{sin\,x\,(1 - cos\,x)}$
Step 2
: Comparing both sides, we notice that the left side has cosine in the numerator and sine in the denominator. We can put cosine in the numerator of the right side by multiplying by 1 + cos(x) on top and bottom:

$\dpi{120} \bg_white \frac{1}{sin\,x\,(1 - cos\,x)}\;\cdot\;\;\frac{1+cos\,x}{1+cos\,x}\\\\\\\;\; =\frac{1 + cos\,x}{sin\,x\,(1 - cos\,^2\,x)}$

Remember, there was only sine in the denominator of the left side, so we can rewrite (1 – cos² x) on the right side using the Pythagorean identity:$\dpi{120} \bg_white \frac{1 + cos\,x}{sin\,x\,(1 - cos\,^2\,x)} = \frac{1 + cos\,x}{sin\,x\,(sin\,^2\,x)}$

Which is now equivalent to the left side:

$\dpi{120} \bg_white \\\frac{1 + cos\,x}{sin\,x\,(sin\,^2\,x)} = \frac{1 + cos\,x}{sin\,^3\,x}\\\\\\ {\color{DarkGreen} {\;\;}\frac{1 + cos\,x}{sin\,^3\,x} = \frac{1 + cos\,x}{sin\,^3\,x} }$

Trigonometric identities can be tricky! Watch and see how a Yup tutor helps a student like you verify the trig identity of csc(x) – sin(x) = cos(x)cot(x).

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