Math Problem Statement

solution

Solve the trigonometric equations and simplify the expressions provided in the image.

Solution

The image contains trigonometric equations and expressions to be solved or simplified. I'll analyze the mathematical content and solve step by step:


1. Equation 1:

sinαcosα=28\sin \alpha \cos \alpha = \frac{\sqrt{2}}{8}

Solution:

Using the identity sinαcosα=12sin(2α)\sin \alpha \cos \alpha = \frac{1}{2} \sin(2\alpha): 12sin(2α)=28\frac{1}{2} \sin(2\alpha) = \frac{\sqrt{2}}{8} sin(2α)=24\sin(2\alpha) = \frac{\sqrt{2}}{4} Finding the general solution for sin(2α)=24\sin(2\alpha) = \frac{\sqrt{2}}{4}: 2α=arcsin(24)+2nπor2α=πarcsin(24)+2nπ2\alpha = \arcsin\left(\frac{\sqrt{2}}{4}\right) + 2n\pi \quad \text{or} \quad 2\alpha = \pi - \arcsin\left(\frac{\sqrt{2}}{4}\right) + 2n\pi α=12arcsin(24)+nπorα=π212arcsin(24)+nπ\alpha = \frac{1}{2} \arcsin\left(\frac{\sqrt{2}}{4}\right) + n\pi \quad \text{or} \quad \alpha = \frac{\pi}{2} - \frac{1}{2} \arcsin\left(\frac{\sqrt{2}}{4}\right) + n\pi


2. Equation 2:

2sin2(3θ)+sin2(6θ)=22 \sin^2(3\theta) + \sin^2(6\theta) = 2

Solution:

The maximum value of sin2(x)\sin^2(x) is 1. To satisfy the equation, both sin2(3θ)\sin^2(3\theta) and sin2(6θ)\sin^2(6\theta) must be equal to 1. Thus: sin2(3θ)=1andsin2(6θ)=1\sin^2(3\theta) = 1 \quad \text{and} \quad \sin^2(6\theta) = 1 This implies: 3θ=π2+nπand6θ=π2+mπ3\theta = \frac{\pi}{2} + n\pi \quad \text{and} \quad 6\theta = \frac{\pi}{2} + m\pi For sin2(6θ)\sin^2(6\theta), solving gives: θ=π6+nπ3for integers n.\theta = \frac{\pi}{6} + \frac{n\pi}{3} \quad \text{for integers } n.


3. Equation 3:

tan(θ)+tan(2θ)+tan(3θ)=0\tan(\theta) + \tan(2\theta) + \tan(3\theta) = 0

Solution:

Using the tangent sum formula and properties: Let x=tan(θ),y=tan(2θ),z=tan(3θ)x = \tan(\theta), y = \tan(2\theta), z = \tan(3\theta). Then: x+y+z=0x + y + z = 0 This is a transcendental equation, requiring numeric or analytic techniques for specific solutions.


4. Expression to Simplify:

P=cos2(A)+cos2(B)+cos2(C)2cos(A)cos(B)cos(C)1P = \cos^2(A) + \cos^2(B) + \cos^2(C) - 2 \cos(A) \cos(B) \cos(C) - 1

Solution:

Using the Pythagorean identity cos2(x)=1sin2(x)\cos^2(x) = 1 - \sin^2(x), rewrite PP to simplify. Substituting further will depend on the angle sum constraints of A,B,A, B, and CC.

Would you like step-by-step details or numerical solutions for a specific part? Let me know!


Further Questions:

  1. How can trigonometric identities simplify complex expressions like PP?
  2. What are the properties of solutions to sin(x)=k\sin(x) = k?
  3. How do periodic solutions appear in tangent equations like Equation 3?
  4. How does arcsin(x)\arcsin(x) behave for values beyond the principal range?
  5. Can PP simplify further under additional constraints on A,B,CA, B, C?

Tip:

Always visualize periodic functions like sin(x),cos(x),tan(x)\sin(x), \cos(x), \tan(x) on their respective intervals to better understand solutions!

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Math Problem Analysis

Mathematical Concepts

Trigonometric identities
Sine and cosine functions
Tangent properties
Simplification of trigonometric expressions

Formulas

sin(2α) = 2sin(α)cos(α)
Pythagorean identities: sin²(x) + cos²(x) = 1
Tangent addition formula: tan(A + B) = (tan(A) + tan(B))/(1 - tan(A)tan(B))

Theorems

Solutions of trigonometric equations using periodicity
Behavior of arcsin and arccos functions
Maximum value properties of trigonometric functions

Suitable Grade Level

Grades 10-12