For sin(α + β), we use the formula: sin(α + β) = sin α cos β + cos α sin β

For cos(α + β), we use the formula: cos(α + β) = cos α cos β - sin α sin β

We use the formula: sin(α - β) = sin α cos β - cos α sin β

Given: sin α = 3/5 and cos β = 5/13

First, we need to find cos α and sin β:

From sin α = 3/5, we can find cos α using the identity sin²α + cos²α = 1

cos²α = 1 - sin²α = 1 - (3/5)² = 1 - 9/25 = 16/25

cos α = 4/5 (positive since 0 < α < π/2)

From cos β = 5/13, we can find sin β:

sin²β = 1 - cos²β = 1 - (5/13)² = 1 - 25/169 = 144/169

sin β = 12/13 (positive since 0 < β < π/2)

Now we can calculate sin(α - β):

sin(α - β) = sin α cos β - cos α sin β

sin(α - β) = (3/5)(5/13) - (4/5)(12/13)

sin(α - β) = 15/65 - 48/65 = -33/65

Let's expand the given expression:

E = 3sin(x + π/6) - 2sin(x - π/3)

Using the formula sin(A + B) = sin A cos B + cos A sin B:

sin(x + π/6) = sin x cos(π/6) + cos x sin(π/6)

sin(x + π/6) = sin x · (√3/2) + cos x · (1/2)

sin(x + π/6) = (√3/2)sin x + (1/2)cos x

Similarly, sin(x - π/3) = sin x cos(π/3) - cos x sin(π/3)

sin(x - π/3) = sin x · (1/2) - cos x · (√3/2)

sin(x - π/3) = (1/2)sin x - (√3/2)cos x

Substituting these values:

E = 3[(√3/2)sin x + (1/2)cos x] - 2[(1/2)sin x - (√3/2)cos x]

E = (3√3/2)sin x + (3/2)cos x - sin x + √3cos x

E = (3√3/2 - 1)sin x + (3/2 + √3)cos x

Let a = 3√3/2 - 1 and b = 3/2 + √3

E = a·sin x + b·cos x

To express this in the form A·sin(x + C), we need to find A and C such that:

A·sin(x + C) = a·sin x + b·cos x

A·sin x·cos C + A·cos x·sin C = a·sin x + b·cos x

Comparing coefficients:

A·cos C = a

A·sin C = b

Squaring and adding:

A²cos²C + A²sin²C = a² + b²

A² = a² + b²

A = √(a² + b²) = √[(3√3/2 - 1)² + (3/2 + √3)²]

A = √[9·3/4 - 3√3 + 1 + 9/4 + 3√3 + 3]

A = √[9·3/4 + 9/4 + 4] = √[27/4 + 9/4 + 16/4] = √[52/4] = √13

Now we find C:

cos C = a/A = (3√3/2 - 1)/√13

sin C = b/A = (3/2 + √3)/√13

Since both sin C and cos C are positive, C is in the first quadrant.

C = arctan((3/2 + √3)/(3√3/2 - 1))

Let's express 3sin θ + 4cos θ in the form r·sin(θ + α).

Let a = 3 and b = 4

The expression can be rewritten as:

3sin θ + 4cos θ = r·sin(θ + α)

where r = √(a² + b²) and α = arctan(b/a)

r = √(3² + 4²) = √(9 + 16) = √25 = 5

α = arctan(4/3) ≈ 53.13°

So, 3sin θ + 4cos θ = 5·sin(θ + α)

The maximum value of sin(θ + α) is 1, which occurs when θ + α = π/2.

Therefore, the maximum value of 5·sin(θ + α) is 5·1 = 5.

We use the formula: tan(A + B) = (tan A + tan B)/(1 - tan A·tan B)

Substituting the given values:

tan(A + B) = ((1/4) + (2/3))/(1 - (1/4)·(2/3))

Finding a common denominator in the numerator:

tan(A + B) = ((3/12) + (8/12))/(1 - (2/12))

tan(A + B) = (11/12)/(1 - 1/6)

tan(A + B) = (11/12)·(6/5)

tan(A + B) = 11/10

We have the equation: sin(x + π/4) = cos(x - π/4)

Using the identity cos(α) = sin(α + π/2), we can rewrite:

cos(x - π/4) = sin(x - π/4 + π/2) = sin(x + π/4)

So, the equation becomes: sin(x + π/4) = sin(x + π/4)

This is an identity that is true for all values of x.

Therefore, the solution is: x ∈ R (all real numbers)

We use the formula: sin α - sin β = 2sin((α-β)/2)cos((α+β)/2)

Therefore, sin α - sin β = 2sin((α-β)/2)cos((α+β)/2)

We use the identity sin²x + cos²x = 1, so cos²x = 1 - sin²x

f(x) = 3sin²x + 4cos²x = 3sin²x + 4(1 - sin²x) = 3sin²x + 4 - 4sin²x

f(x) = 4 - sin²x

To find the minimum value, we need to maximize sin²x.

The maximum value of sin²x is 1, which occurs when sin x = ±1.

When sin²x reaches its maximum value of 1:

f(x) = 4 - 1 = 3

Given: α + β + γ = π

Let's rewrite each term using the sine of the sum formula.

From α + β + γ = π, we get γ = π - α - β

Therefore:

sin(α + β) = sin(π - γ) = sin π cos γ - cos π sin γ = 0·cos γ - (-1)·sin γ = sin γ

Similarly:

sin(β + γ) = sin(β + π - α - β) = sin(π - α) = sin π cos α - cos π sin α = sin α

sin(γ + α) = sin(π - β) = sin β

So, the expression becomes:

sin γ·sin γ + sin α·sin α + sin β·sin β = sin²γ + sin²α + sin²β

Now, from the identity sin²A + cos²A = 1, we have:

sin²α = 1 - cos²α

sin²β = 1 - cos²β

sin²γ = 1 - cos²γ

So, sin²α + sin²β + sin²γ = 3 - (cos²α + cos²β + cos²γ)

From α + β + γ = π, we can derive that cos²α + cos²β + cos²γ = 1 + 2cos α cos β cos γ

Therefore, sin²α + sin²β + sin²γ = 3 - (1 + 2cos α cos β cos γ) = 2 - 2cos α cos β cos γ

From the condition α + β + γ = π, we get cos α cos β cos γ = 1

Thus, sin²α + sin²β + sin²γ = 2 - 2 = 0

We'll use the following formulas:

sin α + sin β = 2sin((α + β)/2)cos((α - β)/2)

cos α + cos β = 2cos((α + β)/2)cos((α - β)/2)

Substituting these in the given expression:

(sin α + sin β)/(cos α + cos β) = 2sin((α + β)/2)cos((α - β)/2) / 2cos((α + β)/2)cos((α - β)/2)

Simplifying:

(sin α + sin β)/(cos α + cos β) = sin((α + β)/2) / cos((α + β)/2) = tan((α + β)/2)

The given expression can be rewritten as R·cos(θ - α) for some R and α.

We know that a·sin θ + b·cos θ = R·cos(θ - α) where R = √(a² + b²) and tan α = a/b

Here, a = 5 and b = 12

R = √(5² + 12²) = √(25 + 144) = √169 = 13

tan α = 5/12, so α = arctan(5/12) ≈ 22.62°

Therefore, 5sin θ + 12cos θ = 13·cos(θ - 22.62°)

We will express f(θ) = sin θ + cos θ in the form R·cos(θ - α)

Let's compare: sin θ + cos θ and R·cos(θ - α)

sin θ + cos θ = 1·sin θ + 1·cos θ

Using a·sin θ + b·cos θ = R·cos(θ - α) where R = √(a² + b²) and tan α = a/b:

a = 1, b = 1

R = √(1² + 1²) = √2

tan α = 1/1 = 1, so α = π/4

Therefore, sin θ + cos θ = √2·cos(θ - π/4)

Alternatively, we can express this as: sin θ + cos θ = √2·sin(θ + π/4)

The maximum value of cos(θ - π/4) is 1, which occurs when θ - π/4 = 0, or θ = π/4.

The minimum value of cos(θ - π/4) is -1, which occurs when θ - π/4 = π, or θ = 5π/4.

Thus, the maximum value of f(θ) = sin θ + cos θ is √2·1 = √2.

The minimum value of f(θ) = sin θ + cos θ is √2·(-1) = -√2.