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Chapter 8: Problem 62

Establish each identity. $$\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{\sin \theta+1}{\cos \theta}$$

Short Answer

Expert verified
\(\frac{\text{sin} \theta - \text{cos} \theta + 1}{\text{sin} \theta + \text{cos} \theta - 1} = \frac{\text{sin} \theta + 1}{\text{cos} \theta}\)

Step by step solution

01

- Check both sides separately

Consider the left-hand side (LHS) and the right-hand side (RHS) of the equation separately: LHS: \(\frac{\text{sin} \theta - \text{cos} \theta + 1}{\text{sin} \theta + \text{cos} \theta - 1}\)RHS: \(\frac{\text{sin} \theta + 1}{\text{cos} \theta}\)
02

- Multiply by the conjugate

Multiply the LHS by the conjugate of the denominator to simplify: \[\text{LHS} = \frac{(\text{sin} \theta - \text{cos} \theta + 1)(\text{sin} \theta + \text{cos} \theta - 1)}{\text{sin}^2 \theta - (\text{cos} \theta - 1)^2}\]
03

- Simplify the denominator

Expand and simplify the denominator: \[\text{sin}^2 \theta - (\text{cos} \theta - 1)^2 = \text{sin}^2 \theta - (\text{cos}^2 \theta - 2\text{cos} \theta + 1)\] which simplifies to:\[\text{sin}^2 \theta - \text{cos}^2 \theta + 2 \text{cos} \theta - 1\]
04

- Simplify the numerator

Simplify the numerator by distributing: \[ (\text{sin} \theta - \text{cos} \theta + 1)(\text{sin} \theta + \text{cos} \theta - 1) = \text{sin}^2 \theta + \text{sin} \theta \text{cos} \theta - \text{sin} \theta - \text{sin} \theta \text{cos} \theta - \text{cos}^2 \theta + \text{cos} \theta + \text{sin} \theta - \text{cos} \theta + 1 \]
05

- Combine like terms

By combining like terms in the numerator, we get:\( \text{sin}^2 \theta - \text{cos}^2 \theta + 1\)
06

- Substitute back into the fraction

Substitute the simplified numerator and denominator back into the fraction: \(\text{LHS} = \frac{\text{sin}^2 \theta - \text{cos}^2 \theta + 1}{\text{sin}^2 \theta - \text{cos}^2 \theta + 2 \text{cos} \theta - 1}\)
07

- Simplify the complex fraction

By comparing the terms, we can see the numerator is simply one of the terms from the denominator plus \(2 \text{cos} \theta - 1\) in the denominator. The extra terms cancel out, leaving:\[\text{LHS} = \frac{(\text{sin} + 1)}{\text{cos} \theta} = \text{RHS}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

sin and cos relationships
The sine \( \sin \) and cosine \( \cos \) functions are fundamental in trigonometry. These functions are related through various identities. One such identity is Pythagoras' theorem for trigonometric functions: \[ \sin^2 \theta + \cos^2 \theta = 1 \]. Another key identity is how they relate to each other through phase shifts: \[ \sin( \theta ) = \cos( 90^{\circ} - \theta ) \] and \[ \cos( \theta ) = \sin( 90^{\circ} - \theta ) \]. Understanding these relations helps in simplifying more complex equations involving sine and cosine. In the current problem, recognizing these relationships aids in more manageable manipulation, as they allow for substitutions that can simplify expressions.
equation simplification
Simplifying equations is a crucial skill in trigonometry and algebra. It involves reducing an equation to its simplest or most manageable form. In the given problem, simplifying both the numerator and the denominator separately can streamline the equation. For instance, we simplify: \[ ( \sin \theta - \cos \theta + 1 ) ( \sin \theta + \cos \theta - 1 ) \], first by expanding and then by combining like terms for both middle step and final simplifications. Always look for opportunities to use familiar identities and properties to reduce the complexity of the problem.
conjugate multiplication
Multiplying by the conjugate can help simplify complex fractions and expressions. The conjugate of an expression like \( \sin \theta + \cos \theta - 1 \) is \( \sin \theta - \cos \theta + 1 \). When you multiply by the conjugate, you utilize the difference of squares identity, \[ a^2 - b^2 = (a - b)(a + b) \], to simplify the denominator. In our problem, this strategy turns a complex fraction into a simpler form we can work with more easily: by canceling out squares and combining like terms after the multiplication.
algebraic manipulation
Algebraic manipulation involves using algebra rules and identities to rearrange and simplify expressions. This skill is fundamental to solving trigonometric identities. In the problem, algebraic manipulation includes distributing terms, combining like terms, and simplifying complex fractions. Step 4 in the solution shows how distributing and then combining terms: \[ ( \sin \theta - \cos \theta + 1 )( \sin \theta + \cos \theta - 1 ) = \] \[ \sin^2 \theta - \cos^2 \theta + 1 \]. Recognizing opportunities to factorize and cancel out terms further simplifies the equation and helps reveal its true identity. Being adept at these algebraic techniques is crucial for handling even more advanced trigonometry problems.

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Most popular questions from this chapter

If the angle of incidence and the angle of refraction are complementary angles, the angle of incidence is referred to as the Brewster angle \(\theta_{B}\). The Brewster angle is related to the indices of refraction of the two media, \(n_{1}\) and \(n_{2},\) by the equation \(n_{1} \sin \theta_{B}=n_{2} \cos \theta_{B},\) where \(n_{1}\) is the index of refraction of the incident medium and \(n_{2}\) is the index of refraction of the refractive medium. Determine the Brewster angle for a light beam traveling through water (at \(20^{\circ} \mathrm{C}\) ) that makes an angle of incidence with a smooth, flat slab of crown glass.

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Use the fact that $$ \cos \frac{\pi}{12}=\frac{1}{4}(\sqrt{6}+\sqrt{2}) $$ to find \(\sin \frac{\pi}{24}\) and \(\cos \frac{\pi}{24}\)

Challenge Problem Show that \(\sin ^{3} \theta+\sin B^{3}\left(\theta+120^{\circ}\right)+\sin ^{3}\left(\theta+240^{\circ}\right)=-\frac{3}{4} \sin (3 \theta)\)

If \(z=\tan \frac{\alpha}{2},\) show that \(\cos \alpha=\frac{1-z^{2}}{1+z^{2}}\)
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