Verifying trigonometric identities is demonstrating that one side of an equation is equal to the other side of the equation. There are seven steps to this process. These are not necessarily black and white rules. These are the steps that were taught to me, and they are as follows.
- Know the Identitities
- Start with the more complicated side
- Express the trig functions in terms of sine and cosine
- Complete any factoring
- Write the sum of two quotients as one quotient
- Keep in mid the other side (goal) as you’re working with the other side
- Multiply by the conjugate of the denominator (below is an example of what a conjugate is)
1 + \text{cos}x \rightarrow1-\text{cos}xExamples
1.
a)
x^2 - y^2 = (x+y)(x-y)
b)
x^3-y^3 = (x-y)(x^2+xy+y^2)
c)
x^3+y^3 = (x+y)(x^2-xy+y^2)
d)
x^2+2xy+y^2 = (x+y)(x+y) \ \ \ \ \text{OR} \ \ \ \ (x+y)^22.
\text{cot}x(\text{tan}x-\text{sec}x) \\ \text{cot}x\text{tan}x-\text{cot}x\text{sec}x \\ = \frac{1}{\text{tan}x} \cdot \frac{\text{tan}x}{1} - \frac{\text{cos}x}{\text{sin}x} \cdot \frac{1}{\text{cos}x} \\ = 1 - \frac{1}{\text{sin}x} \\ \text {OR} \\ =1-\text{csc}x3.
Factor:
\text{sin}^2\alpha-\text{cos}^2\alpha \\ = (\text{sin}\alpha+\text{cos}\alpha)(\text{sin}\alpha-\text{cos}\alpha)4.
Factor:
(\text{cot}x+\text{tan}x)^2 - (\text{cot}x-\text{tan}x)^2 \\ (\text{cot}x+\text{tan}x+\text{cot}x-\text{tan}x)(\text{cot}x+\text{tan}x-(\text{cot}x-\text{tan}x)) \\ (2\text{cot}x)(\text{cot}x+\text{tan}x-\text{cot}x+{tan}x) \\ (2\text{cot}x)(2\text{tan}x) \\ =\frac{2}{1} \cdot\frac{1}{\text{tan}x}\cdot\frac{2}{1}\cdot\frac{\text{tan}x}{1} \\ =45.
Factor:
25^4z - 60\text{tan}^2z +36 \\ =(5\text{tan}^2z-6)(5\text{tan}^2-6) \\=(5\text{tan}^2z-6)^26.
Simplify:
\frac{\text{cos}^2x}{\text{sin}^2x} + \text{csc}x\text{sin}x=\text{cot}^2x+\frac{1}{\text{sin}x}\cdot\frac{\text{sin}x}{1} \\ =\text{cot}^2x+1 \\ =\text{csc}^2xWe know that cosine over sine is cotangent, then we are going to express the rest of the equation in terms of sine so that sinx can cancel out, and then we are left with a Pythagorean identity.
7.
Simplify:
\text{tan}^2\theta-\frac{1}{\text{cos}^2\theta}************************************REMEMBER***********************************
\text{tan}^2\theta+1=\text{sec}^2\theta***********************************************************************************
=\text{tan}^2\theta-\text{sec}^2\theta \\ =-1Here we can see that we are going to use one of our Pythagorean identities to simplify this problem. 1 over cosine squared theta is simply secant squared theta, and then when we rearrange the problem, we simply get -1.
8.
Verify Identity:
\text{sin}x-\text{csc}x=-\text{cos}x\text{cot}xWe are going to work with the left side of the identity in this instance.
\text{sin}x-\text{csc}x Write in terms of sine.
=\frac{\text{sin}x}{1}-\frac{1}{\text{sin}x}Multiply by the conjugate.
=\frac{\text{sin}x}{\text{sin}x}\cdot\frac{\text{sin}x}{1}-\frac{1}{\text{sin}x}=\frac{\text{sin}^2x}{\text{sin}x}-\frac{1}{\text{sin}x}=\frac{\text{sin}^2x-1}{\text{sin}x} = \frac{-\text{cos}^2x}{\text{sin}x}=\frac{-\text{cos}^2\theta x}{1}\cdot\frac{\text{cos}x}{\text{sin}x}=-\text{cos}x\text{cot}x \checkmarkThis identity is valid and verified.
9.
Verify Identity:
\text{cot}^4x-\text{csc}^4x=-2\text{cot}^2x-1\text{cot}^4-\text{csc}^4x \\ =(\text{cot}^2x+\text{csc}^2x)(\text{cot}^2x-\text{csc}^2x) \\ =-1(\text{cot}^2x+\text{cot}^2x+1) \\ = -1 (2\text{cot}^2x+1) \\ =-2\text{cot}^2x-1 \checkmarkThis identity checks out.
10.
Verify Identity:
\frac{1}{1-\text{cos}x}+\frac{1}{1+\text{cos}x}=\frac{2}{\text{sin}^2x}\frac{1}{(1-\text{cos}x)}+\frac{1}{(1+\text{cos}x)} \\ \frac{(1+\text{cos}x)}{(1+\text{cos}x)}\cdot\frac{1}{(1-\text{cos}x)}+\frac{1}{(1+\text{cos}x)}\cdot\frac{(1-\text{cos}x)}{(1-\text{cos}x)} \\ \frac{1+\text{cos}x}{1-\text{cos}^2x}+\frac{1-\text{cos}x}{1-\text{cos}^2x}
\\ = \frac{1+\text{cos}x+1-\text{cos}x}{1-\text{cos}^2x} = \frac{2}{1-\text{cos}^2x} = \frac{2}{\text{sin}^2x} \checkmarkThis identity is verified.