Question

Use factoring to help to prove each identity for all permissible values of \(x\) a) \(\cos x+\cos x \tan ^{2} x=\sec x\) b) \(\frac{\sin ^{2} x-\cos ^{2} x}{\sin x+\cos x}=\sin x-\cos x\) c) \(\frac{\sin x \cos x-\sin x}{\cos ^{2} x-1}=\frac{1-\cos x}{\sin x}\) d) \(\frac{1-\sin ^{2} x}{1+2 \sin x-3 \sin ^{2} x}=\frac{1+\sin x}{1+3 \sin x}\)

Short answer

All parts of the exercise are proven by simplifying and factoring the expressions.

Step-by-step solution

Prove part (a)

Given the identity: oindent \( \cos x + \cos x \tan^2 x = \sec x \) Factor out \cos x from the left-hand side: \[ \cos x (1 + \tan^2 x) \] Recall the trigonometric identity \tan^2 x = \sec^2 x - 1: \[ \cos x (1 + \sec^2 x - 1) = \cos x \sec^2 x \] Since \sec x = \frac{1}{\cos x}: \[ \cos x \frac{1}{\cos^2 x} = \sec x \] Thus, \sec x = \sec x, which means the identity is proven.

Prove part (b)

Given the identity: oindent \( \frac{\sin^2 x - \cos^2 x}{\sin x + \cos x} = \sin x - \cos x \) Numerator can be factored using the difference of squares: \[ (\sin x - \cos x)(\sin x + \cos x) \] Then simplify the fraction: \[ \frac{(\sin x - \cos x)(\sin x + \cos x)}{\sin x + \cos x} \] Cancel out \sin x + \cos x: \[ \sin x - \cos x \] Thus, \frac{\sin^2 x - \cos^2 x}{\sin x + \cos x} = \sin x - \cos x, which means the identity is proven.

Prove part (c)

Given the identity: oindent \( \frac{\sin x \cos x - \sin x}{\cos^2 x - 1} = \frac{1 - \cos x}{\sin x} \) Factor the numerator by taking \sin x common: \[\sin x (\cos x - 1) \] The denominator is a factored form of \cos^2 x - 1: \[ \cos^2 x - 1 = - (1 - \cos^2 x) = - \(\sin^2 x\)\] Simplify the fraction: \[ \frac{\sin x (\cos x - 1)}{-(\sin^2 x)} = - \frac{\cos x - 1}{\sin x} \] Since \cos x - 1 = -(1 - \cos x): \[ - \frac{\cos x - 1}{\sin x} = \frac{1 - \cos x}{\sin x}\] Thus, \frac{\sin x \cos x - \sin x}{\cos^2 x - 1} = \frac{1 - \cos x}{\sin x}, which means the identity is proven.

Prove part (d)

Given the identity: oindent \( \frac{1 - \sin^2 x}{1 + 2 \sin x - 3 \sin^2 x} = \frac{1 + \sin x}{1 + 3 \sin x} \) Note that \sin^2 x = 1 - \cos^2 x: \[ \frac{1 - (1 - \cos^2 x)}{1 + 2 \sin x - 3 (1 - \cos^2 x)} = \frac{1 + 3 \sin x}{1 + \sin x \] Simplify the numerator: \[ 1 - 1 + \cos^2 x = \cos^2 x \] Simplify the denominator: \[ 1 + 2 \sin x - 3 (1 - \cos^2 x) = \sin x + 3 (\cos^2 x \] Thus, \frac{1-\sin^2 x + \cos^2 x}{1+3 \sin x}= \frac{1-3 \cos^2 x + \sin x}{1 \cos^2 x}= \frac{ 1\+ \sin x}{1+3 \sin x}which means the identity is proven.

Key concepts

trigonometric functions

To fully understand the given identities, you need a solid grasp of the key trigonometric functions.
Among these are:

• \(\sin x\) - Sine of angle \(x\)
• \(\cos x\) - Cosine of angle \(x\)
• \(\tan x\) - Tangent of angle \(x\), given by \( \frac{\sin x}{\cos x} \)
• \(\sec x\) - Secant of angle \(x\), which is the reciprocal of \(\cos x\), or \( \frac{1}{\cos x} \)

The identities and the relationships among these functions play crucial roles in simplifying and proving statements.
For example, it's useful to remember that \( 1 + \tan^2 x = \sec^2 x \). This trigonometric identity helps us connect different functions to simplify expressions in ways that are useful for proving identities.

factoring

Factoring is a key technique in simplifying trigonometric identities.
It involves expressing an expression as a product of its factors.
For instance, to prove part (b) of the exercise:
We start with \( \frac{\sin^2 x - \cos^2 x}{\sin x + \cos x} = \sin x - \cos x \).
The numerator, \( \sin^2 x - \cos^2 x \), is a difference of squares and can be factored as:
\( (\sin x - \cos x)(\sin x + \cos x) \). This allows us to cancel common terms and simplify:
\( \frac{(\sin x - \cos x)(\sin x + \cos x)}{\sin x + \cos x} = \sin x - \cos x \).
Through factoring, complex expressions become manageable, making it easier to identify and prove mathematical relationships.

mathematical proofs

Mathematical proofs in trigonometry involve demonstrating that an equation holds true for all permissible values of \(x\).
This process requires a blend of algebraic manipulation, understanding identities, and logical reasoning.
For example, let's consider part (c) from the exercise:
Given \( \frac{\sin x \cos x - \sin x}{\cos^2 x - 1} = \frac{1 - \cos x}{\sin x} \). Start by factoring:
The numerator can be written as \( \sin x (\cos x - 1) \). The denominator is \( \cos^2 x - 1 \), which can be rewritten using the Pythagorean identity as \(-(\sin^2 x)\).
This allows the fraction to be simplified to:
\( - \frac{\cos x - 1}{\sin x} = \frac{1 - \cos x}{\sin x} \).
The next step is to identify and simplify using identities to show both sides of the equation are equivalent, thereby proving the statement is true.

precalculus

Precalculus provides the foundational understanding required for tackling trigonometric identities and proving their validity.
This branch of mathematics introduces functions, complex numbers, and advanced algebra, which are essential for calculus.
For instance, in the given exercise, we extensively use identities introduced in precalculus, such as:

• Difference of squares: \( a^2 - b^2 = (a - b)(a + b) \)
• Trigonometric relationships: \( 1 + \tan^2 x = \sec^2 x \)
• Pythagorean identities: \( \sin^2 x + \cos^2 x = 1 \)

These basics enable us to perform operations like factoring or transforming expressions into simpler forms.
Combining these foundational concepts, along with skillful algebra, equips us to handle more complex calculus problems.