Fundamental Identities and Verifying Trigonometric Identities

Fundamental Identities

Learning Requirement: Students will be able to utilize fundamental, reciprocal, tangent and cotangent, and Pythagorean identities and verify trigonometric identities.
Even and Odd Functions:
- A function is even if f(x) = f(-x) for all x in the domain of f.
- A function is odd if f(-x) = -f(x) for all x in the domain of f.
Angle and its Negative:
- An angle \theta having the point (x, y) on its terminal side has a corresponding angle -\theta with the point (x, -y) on its terminal side.
Sine of Negative Angle:
- \sin(-\theta) = -\frac{y}{r} = -\sin(\theta)
Cosine of Negative Angle:
- \cos(-\theta) = \frac{x}{r} = \cos(\theta)
Tangent of Negative Angle:
- \tan(-\theta) = \frac{\sin(-\theta)}{\cos(-\theta)} = \frac{-\sin(\theta)}{\cos(\theta)} = -\tan(\theta)
Note: In trigonometric identities, \theta can represent an angle in degrees or radians, or a real number.

Fundamental Identities

Reciprocal Identities

\csc \theta = \frac{1}{\sin \theta}
\sec \theta = \frac{1}{\cos \theta}
\cot \theta = \frac{1}{\tan \theta}

Quotient Identities

\tan \theta = \frac{\sin \theta}{\cos \theta}
\cot \theta = \frac{\cos \theta}{\sin \theta}

Pythagorean Identities

\sin^2 \theta + \cos^2 \theta = 1
\tan^2 \theta + 1 = \sec^2 \theta
1 + \cot^2 \theta = \csc^2 \theta

Even-Odd Identities

\sin(-\theta) = -\sin \theta
\csc(-\theta) = -\csc \theta
\cos(-\theta) = \cos \theta
\sec(-\theta) = \sec \theta
\tan(-\theta) = -\tan \theta
\cot(-\theta) = -\cot \theta

Uses of the Fundamental Identities

Alternative forms of the fundamental identities can be used. For example:
- \sin^2 \theta = 1 - \cos^2 \theta
- \cos^2 \theta = 1 - \sin^2 \theta
These identities can be used to find the values of other trigonometric functions from the value of a given trigonometric function.
Example: If \tan \theta = -\frac{5}{3} and \theta is in quadrant II, find each function value.
- a) sec θ
  - \tan^2 \theta + 1 = \sec^2 \theta
  - (-\frac{5}{3})^2 + 1 = \sec^2 \theta
  - \frac{25}{9} + 1 = \sec^2 \theta
  - \sqrt{\frac{34}{9}} = \sqrt{\sec^2 \theta}
  - \sec \theta = \pm \frac{\sqrt{34}}{3}. Since \theta is in QII, sec is negative.
  - \sec \theta = -\frac{\sqrt{34}}{3}
- b) sin θ
  - \tan \theta = \frac{\sin \theta}{\cos \theta}
  - Since \sec \theta = -\frac{\sqrt{34}}{3}, then \cos \theta = -\frac{3}{\sqrt{34}}.
  - -\frac{5}{3} = \frac{\sin \theta}{-\frac{3}{\sqrt{34}}}
  - \sin \theta = -\frac{5}{3} \cdot -\frac{3}{\sqrt{34}} = \frac{5}{\sqrt{34}} = \frac{5\sqrt{34}}{34}
- c) cot(-θ)
  - \cot(-\theta) = -\cot(\theta)
  - \cot(-\theta) = -\frac{1}{\tan(\theta)}
  - \cot(-\theta) = -\frac{1}{-\frac{5}{3}} = \frac{3}{5}
Caution: When taking the square root, be sure to choose the sign based on the quadrant of \theta and the function being evaluated.

Simplification using Trigonometric Substitutions

Example: Write each expression in terms of \sin \theta and \cos \theta, and then simplify the expression so that no quotients appear and all functions are of \theta only.
- a) \frac{1 + \cot^2 \theta}{1 - \csc^2 \theta}
  - \frac{1 + \frac{\cos^2 \theta}{\sin^2 \theta}}{1 - \frac{1}{\sin^2 \theta}} = \frac{\frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta}}{\frac{\sin^2 \theta - 1}{\sin^2 \theta}} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta - 1} = \frac{1}{-\cos^2 \theta} = -\sec^2 \theta
  - Shorter method:
    - \frac{1 + \cot^2 \theta}{1 - \csc^2 \theta} = \frac{\csc^2 \theta}{-\cot^2 \theta} = \frac{\frac{1}{\sin^2 \theta}}{-\frac{\cos^2 \theta}{\sin^2 \theta}} = -\frac{1}{\cos^2 \theta} = -\sec^2 \theta
- b) \tan \theta \cos \theta
  - \frac{\sin \theta}{\cos \theta} \cdot \cos \theta = \sin \theta
- c) (\sec \theta - 1)(\sec \theta + 1)
  - (\sec \theta - 1)(\sec \theta + 1) = \sec^2 \theta - 1 = \tan^2 \theta
  - Shorter Method:
    - (\sec \theta - 1)(\sec \theta + 1) = \sec^2 \theta - 1 = \tan^2 \theta
- d) \csc \theta \sin \theta
  - \frac{1}{\sin \theta} \cdot \sin \theta = 1
- e) \frac{1 - \sin^2 \theta}{\sin \theta} \cdot \frac{1 + \cot^2 \theta}{\cot \theta}
  - \frac{\cos^2 \theta}{\sin \theta} \cdot \frac{\csc^2 \theta}{\cot \theta} = \frac{\cos^2 \theta}{\sin \theta} \cdot \frac{\frac{1}{\sin^2 \theta}}{\frac{\cos \theta}{\sin \theta}} = \frac{\cos^2 \theta}{\sin \theta} \cdot \frac{1}{\sin^2 \theta} \cdot \frac{\sin \theta}{\cos \theta} = \frac{\cos \theta}{\sin^2 \theta} = \cos \theta \cdot \cot \theta
  - Alternative simplification:
    - \frac{\cos^2 \theta}{\sin \theta} \cdot \frac{1 + \frac{\cos^2 \theta}{\sin^2 \theta}}{\frac{\cos \theta}{\sin \theta}} = \frac{\cos^2 \theta}{\sin \theta} \cdot \frac{\frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta}}{\frac{\cos \theta}{\sin \theta}} = \frac{\cos^2 \theta}{\sin \theta} \cdot \frac{1}{\sin^2 \theta} \cdot \frac{\sin \theta}{\cos \theta} = \frac{\cos \theta}{\sin \theta} = \cot \theta

Strategies for Verifying Identities

One of the skills required for more advanced work in mathematics, especially in calculus, is the ability to use identities to write expressions in alternative forms.
We develop this skill by using the fundamental identities to verify that a trigonometric equation is an identity (for those values of the variable for which it is defined).

Hints for Verifying Identities

Learn the fundamental identities. Whenever you see either side of a fundamental identity, the other side should come to mind. Also, be aware of equivalent forms of the fundamental identities. For example, \sin^2 \theta = 1 - \cos^2 \theta is an alternative form of \sin^2 \theta + \cos^2 \theta = 1.
Try to rewrite the more complicated side of the equation so that it is identical to the simpler side.
It is sometimes helpful to express all trigonometric functions in the equation in terms of sine and cosine and then simplify the result.
Usually, any factoring or indicated algebraic operations should be performed. These algebraic identities are often used in verifying trigonometric identities.
- x^2 + 2xy + y^2 = (x + y)^2
- x^2 - 2xy + y^2 = (x - y)^2
- x^3 - y^3 = (x - y)(x^2 + xy + y^2)
- x^3 + y^3 = (x + y)(x^2 - xy + y^2)
- x^2 - y^2 = (x + y)(x - y)
- For example, the expression \sin^2 x - 2 \sin x + 1 can be factored as (\sin x - 1)^2.
- The sum or difference of two trigonometric expressions can be found in the same way as any other rational expression. For example,
  - \frac{1}{\sin \theta} + \frac{1}{\cos \theta} = \frac{1 \cdot \cos \theta}{\sin \theta \cos \theta} + \frac{1 \cdot \sin \theta}{\cos \theta \sin \theta} = \frac{\cos \theta + \sin \theta}{\sin \theta \cos \theta}
When selecting substitutions, keep in mind the side that is not changing, because it represents the goal. For example, to verify that the equation \tan^2 x + 1 = \frac{1}{\cos^2 x} is an identity, think of an identity that relates \tan x to \cos x. In this case, because \sec x = \frac{1}{\cos x} and \sec^2 x = \tan^2 x + 1, the secant function is the best link between the two sides.
If an expression contains 1 + \sin x, multiplying both numerator and denominator by 1 - \sin x would give 1 - \sin^2 x, which could be replaced with \cos^2 x. Similar procedures apply for 1 - \sin x, 1 + \cos x, and 1 - \cos x.

Verifying Identities by Working with One Side

Avoid the temptation to use algebraic properties of equations to verify identities. One strategy is to work with one side and rewrite it to match the other side.
Example: Verify that the following equations are identities
- a) \cot \theta + 1 = \csc \theta (\cos \theta + \sin \theta)
  - \csc \theta (\cos \theta + \sin \theta) = \csc \theta \cdot \cos \theta + \csc \theta \cdot \sin \theta = \frac{1}{\sin \theta} \cdot \cos \theta + \frac{1}{\sin \theta} \cdot \sin \theta = \frac{\cos \theta}{\sin \theta} + 1 = \cot \theta + 1
- b) \tan^2 x (1 + \cot^2 x) = \frac{1}{\cos^2 x}
  - \tan^2 x (\csc^2 x) = \frac{\sin^2 x}{\cos^2 x} \cdot \frac{1}{\sin^2 x} = \frac{1}{\cos^2 x}
- c) \frac{\tan x}{\tan x - \cot x} = \frac{\sin x \cos x}{\sin^2 x - \cos^2 x}
  - \frac{\tan x}{\tan x - \cot x} = \frac{\frac{\sin x}{\cos x}}{\frac{\sin x}{\cos x} - \frac{\cos x}{\sin x}} = \frac{\frac{\sin x}{\cos x}}{\frac{\sin^2 x - \cos^2 x}{\sin x \cos x}} = \frac{\sin x}{\cos x} \cdot \frac{\sin x \cos x}{\sin^2 x - \cos^2 x} = \frac{\sin x \cos x}{\sin^2 x - \cos^2 x}
- d) \frac{\cos x}{1 + \sin x} = \frac{1 - \sin x}{\cos x}
  - \frac{\cos x}{1 + \sin x} = \frac{\cos x}{1 + \sin x} \cdot \frac{1 - \sin x}{1 - \sin x} = \frac{\cos x (1 - \sin x)}{1 - \sin^2 x} = \frac{\cos x (1 - \sin x)}{\cos^2 x} = \frac{1 - \sin x}{\cos x}

Verifying Identities by Working with Both Sides

If both sides of an identity appear to be equally complex, the identity can be verified by working independently on the left side and on the right side, until each side is changed into some common third result. Each step, on each side, must be reversible.
Example: Verify that the following equations are identities.
- a) \frac{\sec x + \tan x}{\sec x - \tan x} = \frac{(1 + \sin x)^2}{\cos^2 x}
  - Left Side:
    - \frac{\sec x + \tan x}{\sec x - \tan x} = \frac{\frac{1}{\cos x} + \frac{\sin x}{\cos x}}{\frac{1}{\cos x} - \frac{\sin x}{\cos x}} = \frac{\frac{1 + \sin x}{\cos x}}{\frac{1 - \sin x}{\cos x}} = \frac{1 + \sin x}{\cos x} \cdot \frac{\cos x}{1 - \sin x} = \frac{1 + \sin x}{1 - \sin x}
  - Right Side:
    - \frac{(1 + \sin x)^2}{\cos^2 x} = \frac{(1 + \sin x)(1 + \sin x)}{1 - \sin^2 x} = \frac{(1 + \sin x)(1 + \sin x)}{(1 + \sin x)(1 - \sin x)} = \frac{1 + \sin x}{1 - \sin x}
- b) (1 - \cos^2 x)(1 + \cos^2 x) = 2 \sin^2 x - \sin^4 x
  - Left Side:
    - (\sin^2 x)(1 + \cos^2 x) = \sin^2 x (1 + 1 - \sin^2 x) = \sin^2 x (2 - \sin^2 x) = 2 \sin^2 x - \sin^4 x
  - Right Side:
    - 2 \sin^2 x - \sin^4 x