Math Problem Statement

Simplify each trigonometric expression shown below using a single trig function with no fractions, using a POSITIVE angle in the argument. If an exponent is necessary, write it in the form of trig²(α). 1. \(\frac{\cos(-\alpha) - \sin(-\alpha)}{\tan(\alpha)}\) 2. \(\tan(-\alpha)\sin(-\alpha) + \cos(-\alpha)\)

Solution

Let's solve these expressions step-by-step.

Problem 1:

cos(α)sin(α)tan(α)\frac{\cos(-\alpha) - \sin(-\alpha)}{\tan(\alpha)}

Solution:

  1. Use trigonometric identities:

    • cos(α)=cos(α)\cos(-\alpha) = \cos(\alpha) (cosine is an even function).
    • sin(α)=sin(α)\sin(-\alpha) = -\sin(\alpha) (sine is an odd function).
    • tan(α)=sin(α)cos(α)\tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)}.
  2. Substitute these identities:

    = \frac{\cos(\alpha) + \sin(\alpha)}{\frac{\sin(\alpha)}{\cos(\alpha)}}$$
  3. Simplify by multiplying both numerator and denominator by cos(α)\cos(\alpha):

    = \cos^2(\alpha) + \cos(\alpha)\sin(\alpha)$$

However, the problem asks to simplify using a single trig function. Further simplification may involve recognizing that this expression can be difficult to reduce to a single function.

Let's verify the specific context or ask for clarification if there are specific guidelines on the simplification level.

Problem 2:

tan(α)sin(α)+cos(α)\tan(-\alpha)\sin(-\alpha) + \cos(-\alpha)

Solution:

  1. Use trigonometric identities:

    • tan(α)=tan(α)\tan(-\alpha) = -\tan(\alpha).
    • sin(α)=sin(α)\sin(-\alpha) = -\sin(\alpha).
    • cos(α)=cos(α)\cos(-\alpha) = \cos(\alpha).
  2. Substitute these identities: (tan(α))(sin(α))+cos(α)(-\tan(\alpha)) \cdot (-\sin(\alpha)) + \cos(\alpha) Simplify by multiplying: =tan(α)sin(α)+cos(α)= \tan(\alpha) \sin(\alpha) + \cos(\alpha)

  3. Substitute tan(α)=sin(α)cos(α)\tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)}:

    = \frac{\sin^2(\alpha)}{\cos(\alpha)} + \cos(\alpha)$$
  4. Combine terms using a common denominator: =sin2(α)+cos2(α)cos(α)= \frac{\sin^2(\alpha) + \cos^2(\alpha)}{\cos(\alpha)} Using the Pythagorean identity sin2(α)+cos2(α)=1\sin^2(\alpha) + \cos^2(\alpha) = 1: =1cos(α)=sec(α)= \frac{1}{\cos(\alpha)} = \sec(\alpha)

So, the answer for Problem 2 is: sec(α)\sec(\alpha)

Would you like further details on any of these steps or explanations?

Here are some related questions:

  1. What are even and odd trigonometric functions, and why do they matter?
  2. How do you apply Pythagorean identities to simplify trigonometric expressions?
  3. What are the properties of tangent and secant functions?
  4. Why is tan(α)=sin(α)cos(α)\tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)} essential in simplifying expressions?
  5. How can complex trigonometric expressions sometimes be simplified to a single trigonometric function?

Tip: Knowing the basic identities like Pythagorean, even-odd, and reciprocal identities can significantly simplify many trigonometric expressions.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Trigonometric Identities

Formulas

\(\cos(-\alpha) = \cos(\alpha)\)
\(\sin(-\alpha) = -\sin(\alpha)\)
\(\tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)}\)
Pythagorean identity: \(\sin^2(\alpha) + \cos^2(\alpha) = 1\)

Theorems

Even-Odd Identities
Pythagorean Identity

Suitable Grade Level

Grades 10-12