MTH-112 Homework #4 Sections 7.1, 7.2, And 7.3

MTH-112 HOMEWORK #4 SECTIONS 7.1, 7.2, AND 7.3

This homework assignment covers sections 7.1, 7.2, and 7.3, focusing on verifying trigonometric identities and determining the validity of specific trigonometric expressions. The tasks involve simplifying expressions, verifying identities through algebraic manipulation, and providing counterexamples where appropriate. Students are encouraged to show full justifications and explanations using complete sentences.

Paper For Above instruction

Trigonometry is a fundamental branch of mathematics that explores the relationships between the angles and sides of triangles, and its identities are crucial for simplifying complex trigonometric expressions. This paper addresses the verification of several identities and the examination of specific expressions to establish whether they are identities or not, with detailed proofs and explanations.

Verification of Trigonometric Identities

In the first part, the goal is to verify given trigonometric identities by applying core identities such as the Pythagorean identities, reciprocal identities, quotient identities, and algebraic manipulations.

Part 1a: Verify that sec x csc x = tan x + cot x

Recall the definitions: sec x = 1/cos x, csc x = 1/sin x, tan x = sin x / cos x, cot x = cos x / sin x. The left-hand side (LHS) is

LHS = sec x · csc x = (1/cos x) · (1/sin x) = 1 / (sin x · cos x)

The right-hand side (RHS) is

RHS = tan x + cot x = (sin x / cos x) + (cos x / sin x) = (sin^2 x + cos^2 x) / (sin x · cos x) = 1 / (sin x · cos x)

Since LHS = RHS, the identity is verified.

Part 1b: Verify 1 + tan θ sin θ + cos θ = sec θ

Express tan θ as sin θ / cos θ and write sec θ as 1 / cos θ:

Left side = 1 + (sin θ / cos θ) · sin θ + cos θ = 1 + (sin^2 θ / cos θ) + cos θ

Rewrite as a common denominator:

= (cos θ / cos θ) + (sin^2 θ / cos θ) + (cos^2 θ / cos θ) = (cos θ + sin^2 θ + cos^2 θ) / cos θ

Using the Pythagorean identity sin^2 θ + cos^2 θ = 1:

= (cos θ + 1) / cos θ = 1 + 1 / cos θ = 1 + sec θ

This does not match the right side, which is explicitly sec θ. Since the expressions are not equal, the given is not an identity; thus, the original identity does not hold universally.

Part 1c: Verify (1 - sin θ)(tan θ + sec θ) = cos(−θ)

Recall that cos(−θ) = cos θ. Expand the left side:

(1 - sin θ)(tan θ + sec θ) = (1 - sin θ) [(sin θ / cos θ) + (1 / cos θ)]

Factor out 1 / cos θ:

= (1 - sin θ) [ (sin θ + 1) / cos θ ] = [ (1 - sin θ)(sin θ + 1) ] / cos θ

Note that (1 - sin θ)(sin θ + 1) = (1 - sin θ)^2 = 1 - 2 sin θ + sin^2 θ

Compare this to cos θ. Since the numerator simplifies to (1 - sin θ)^2, and it does not generally equal cos θ, the two sides are generally not equal. Therefore, this is not an identity, or it might be valid only under specific conditions, but not universally.

Part 1d: Verify (csc x - cot x)^2 = (1 - cos x) / (1 + cos x)

Recall that csc x = 1 / sin x, cot x = cos x / sin x. So,

(csc x - cot x)^2 = (1 / sin x - cos x / sin x)^2 = ( (1 - cos x) / sin x )^2 = (1 - cos x)^2 / sin^2 x

Using the Pythagorean identity sin^2 x + cos^2 x = 1, we have sin^2 x = 1 - cos^2 x. Rewrite the RHS:

RHS = (1 - cos x) / (1 + cos x)

Note that (1 - cos x) / (1 + cos x) can be rationalized by multiplying numerator and denominator by (1 - cos x):

= [(1 - cos x)(1 - cos x)] / [(1 + cos x)(1 - cos x)] = (1 - 2 cos x + cos^2 x) / (1 - cos^2 x)

But 1 - cos^2 x = sin^2 x, so RHS = (1 - 2 cos x + cos^2 x) / sin^2 x

Compare this to the numerator of the LHS, which is (1 - cos x)^2 = 1 - 2 cos x + cos^2 x. Therefore, both sides are equal, confirming the identity.

Verification of Non-identity Expressions and Counterexamples

In the second part, the focus shifts to verifying whether given expressions are identities or not. When an expression does not hold for all values, a counterexample demonstrates its invalidity.

Part 2a: (sin A − cos A)^2 = 1 − 2 sin^2 A cot A

Expand the left side:

(sin A − cos A)^2 = sin^2 A - 2 sin A cos A + cos^2 A = (sin^2 A + cos^2 A) - 2 sin A cos A = 1 - 2 sin A cos A

Recall the double-angle identity: sin 2A = 2 sin A cos A. So, the left side simplifies to 1 - sin 2A.

Now, examine the right side: 1 − 2 sin^2 A cot A. Recall that cot A = cos A / sin A:

RHS = 1 - 2 sin^2 A · (cos A / sin A) = 1 - 2 sin A · cos A = 1 - sin 2A

Since RHS = LHS, the expression is verified to be an identity.

Part 2b: sec^2 x − 1 = cos x csc x

Using the fundamental identity sec^2 x = 1 + tan^2 x, so sec^2 x - 1 = tan^2 x. The RHS is cos x · csc x = cos x / sin x. Since tan^2 x = sin^2 x / cos^2 x, and this generally does not equal cos x / sin x, the expressions are not equivalent for all x. For x = π/4, for example, sec^2(π/4) -1 = 2 - 1 = 1, while cos(π/4) csc(π/4) = (√2/2) · (√2/2) = 1/2. Since 1 ≠ 1/2, the given is not an identity.

Part 2c: sin^2 θ + cos^2 θ + tan^2 θ + sec^2 θ + cot^2 θ + csc^2 θ = 3

Substitute known identities:

sin^2 θ + cos^2 θ = 1, tan^2 θ + 1 = sec^2 θ, cot^2 θ + 1 = csc^2 θ. So, the sum becomes:

1 + tan^2 θ + sec^2 θ + cot^2 θ + csc^2 θ

But since sec^2 θ = tan^2 θ + 1 and csc^2 θ = cot^2 θ + 1, the sum simplifies to:

1 + tan^2 θ + (tan^2 θ + 1) + cot^2 θ + (cot^2 θ + 1) = 1 + tan^2 θ + tan^2 θ + 1 + cot^2 θ + cot^2 θ + 1

= (1 + 1 + 1) + (tan^2 θ + tan^2 θ) + (cot^2 θ + cot^2 θ) = 3 + 2 tan^2 θ + 2 cot^2 θ

Thus, the sum generally is greater than 3 unless tan θ and cot θ are zero, which is impossible. Therefore, the sum does not always equal 3, meaning it is not an identity.

Part 2d: sin x + cos x + tan x · sin x = sec x + cos x · tan x

Express tan x as sin x / cos x and sec x as 1 / cos x:

Left side = sin x + cos x + (sin x / cos x) · sin x = sin x + cos x + (sin^2 x / cos x)

Right side = (1 / cos x) + cos x · (sin x / cos x) = 1 / cos x + sin x

Compare both sides: left side has sin x + cos x + sin^2 x / cos x, whereas right side is 1 / cos x + sin x. Since sin^2 x / cos x + 1 / cos x = (sin^2 x + 1) / cos x, which is generally not equal to sin x + cos x, these expressions are not equal for all x. Therefore, this is not an identity.

Conclusion

Through algebraic verification and counterexamples, the analysis confirms which of the given trigonometric statements are identities. Equations like 1a and 1d are verified, while others such as 1b, 1c, 2b, 2c, and 2d do not hold universally. It is crucial in trigonometry to test expressions in multiple contexts and to understand fundamental identities to effectively manipulate and verify complex expressions.

References

  • Anton, H., Bivens, I., & Davis, S. (2013). Algebra and Trigonometry. Wiley.
  • Larson, R., & Hostetler, R. (2013). Precalculus with Limits. Cengage Learning.
  • Stewart, J. (2015). Calculus: Early Transcendentals. Cengage Learning.
  • Trigonometry. (n.d.). In Khan Academy. Retrieved from https://www.khanacademy.org/math/trigonometry
  • Math World. Trigonometric identities. Retrieved from https://mathworld.wolfram.com/TrigonometricIdentities.html

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