Trigonometric Equations

Trigonometric equations are algebraic equations that involve trigonometric functions of one or more unknown angles. The goal is to find all possible values of the angles that satisfy the equation. These equations can be simple, like sin(x) = 1/2, or complex, involving multiple trigonometric functions and arguments.

General Solution of Trigonometric Equations

When solving trigonometric equations, we are often interested in the general solution, which represents all possible values of the variable that satisfy the equation. The periodicity of trigonometric functions is key to finding the general solution. For any trigonometric function f(x), if f(x) = f(x + nT), where T is the period and n is an integer, then x + nT is also a solution if x is a solution.

1. Solutions of sin(x) = sin(α)

If sin(x) = sin(α), then the general solution is given by: x = nπ + (-1)ⁿα, where n ∈ ℤ (the set of integers).

Let's understand why. The sine function is positive in the first and second quadrants. If α is in the first quadrant, then π - α is in the second quadrant, and sin(π - α) = sin(α). So, if x = α is a solution, then x = π - α is also a solution. Due to the periodicity of sine (period 2π), solutions repeat every 2π. Considering n=0, x=α. Considering n=1, x=π-α. Considering n=2, x=2π+α. Considering n=3, x=3π-α, which is equivalent to π-α + 2π.

Example: Find the general solution of sin(x) = 1/2.

We know that sin(π/6) = 1/2. So, α = π/6.

Using the formula x = nπ + (-1)ⁿα:

x = nπ + (-1)ⁿ(π/6), where n ∈ ℤ.

2. Solutions of cos(x) = cos(α)

If cos(x) = cos(α), then the general solution is given by: x = 2nπ ± α, where n ∈ ℤ.

The cosine function is positive in the first and fourth quadrants. If α is in the first quadrant, then -α (or 2π - α) is in the fourth quadrant, and cos(-α) = cos(α). Due to the periodicity of cosine (period 2π), solutions repeat every 2π. So, if x = α is a solution, then x = -α is also a solution. The general form covers all these possibilities.

Example: Find the general solution of cos(x) = 1/2.

We know that cos(π/3) = 1/2. So, α = π/3.

Using the formula x = 2nπ ± α:

x = 2nπ ± π/3, where n ∈ ℤ.

3. Solutions of tan(x) = tan(α)

If tan(x) = tan(α), then the general solution is given by: x = nπ + α, where n ∈ ℤ.

The tangent function has a period of π. This means tan(x) = tan(x + nπ) for any integer n. If tan(x) = tan(α), then x = α is one solution. Due to the periodicity, x = α + π, x = α + 2π, etc., are also solutions. The formula x = nπ + α encapsulates all these solutions.

Example: Find the general solution of tan(x) = 1.

We know that tan(π/4) = 1. So, α = π/4.

Using the formula x = nπ + α:

x = nπ + π/4, where n ∈ ℤ.

4. Solutions of cot(x) = cot(α)

If cot(x) = cot(α), then the general solution is given by: x = nπ + α, where n ∈ ℤ.

Similar to the tangent function, the cotangent function also has a period of π. Thus, the general solution follows the same pattern.

Example: Find the general solution of cot(x) = √3.

We know that cot(π/6) = √3. So, α = π/6.

Using the formula x = nπ + α:

x = nπ + π/6, where n ∈ ℤ.

5. Solutions of sec(x) = sec(α)

If sec(x) = sec(α), then the general solution is given by: x = 2nπ ± α, where n ∈ ℤ.

Since sec(x) = 1/cos(x), this equation is equivalent to cos(x) = cos(α), and hence the solution is the same.

6. Solutions of csc(x) = csc(α)

If csc(x) = csc(α), then the general solution is given by: x = nπ + (-1)ⁿα, where n ∈ ℤ.

Since csc(x) = 1/sin(x), this equation is equivalent to sin(x) = sin(α), and hence the solution is the same.

Solving More Complex Trigonometric Equations

Many trigonometric equations are not in the standard forms above. To solve them, we often need to use trigonometric identities and algebraic manipulations. Here are some common strategies:

a) Equations Reducible to Quadratic Form

Some equations can be transformed into quadratic equations in terms of a single trigonometric function (e.g., sin(x), cos(x), tan(x)).

Example: Solve 2sin²(x) + 3sin(x) - 2 = 0.

Let y = sin(x). The equation becomes 2y² + 3y - 2 = 0.

Factor the quadratic equation: (2y - 1)(y + 2) = 0.

This gives two possible values for y: y = 1/2 or y = -2.

Substitute back y = sin(x):

Case 1: sin(x) = 1/2. The general solution is x = nπ + (-1)ⁿ(π/6), n ∈ ℤ.

Case 2: sin(x) = -2. This equation has no real solutions because the range of sin(x) is [-1, 1].

Therefore, the general solution is x = nπ + (-1)ⁿ(π/6), n ∈ ℤ.

b) Equations Involving Different Trigonometric Functions

Use trigonometric identities (like sin²(x) + cos²(x) = 1, tan(x) = sin(x)/cos(x), etc.) to express the equation in terms of a single trigonometric function.

Example: Solve sin(x) + cos(x) = 1.

We can solve this by squaring both sides, but we must check for extraneous solutions. Squaring gives (sin(x) + cos(x))² = 1².

sin²(x) + cos²(x) + 2sin(x)cos(x) = 1

1 + sin(2x) = 1

sin(2x) = 0

The general solution for sin(θ) = 0 is θ = kπ, where k ∈ ℤ.

So, 2x = kπ, which means x = kπ/2, k ∈ ℤ.

Now, we must check these solutions in the original equation sin(x) + cos(x) = 1.

If k=0, x=0: sin(0) + cos(0) = 0 + 1 = 1. (Valid)

If k=1, x=π/2: sin(π/2) + cos(π/2) = 1 + 0 = 1. (Valid)

If k=2, x=π: sin(π) + cos(π) = 0 + (-1) = -1. (Invalid)

If k=3, x=3π/2: sin(3π/2) + cos(3π/2) = -1 + 0 = -1. (Invalid)

If k=4, x=2π: sin(2π) + cos(2π) = 0 + 1 = 1. (Valid, same as k=0)

The valid solutions repeat every 2π. The solutions are of the form x = 2mπ and x = π/2 + 2mπ, where m ∈ ℤ.

Alternatively, we can use the R-cos(x-α) or R-sin(x+α) form.

Consider a sin(x) + b cos(x) = c. Here, a=1, b=1, c=1.

We can write a sin(x) + b cos(x) = √(a² + b²) [ (a/√(a²+b²)) sin(x) + (b/√(a²+b²)) cos(x) ].

Let cos(α) = a/√(a²+b²) and sin(α) = b/√(a²+b²). Then the expression becomes:

√(a² + b²) [ cos(α) sin(x) + sin(α) cos(x) ] = √(a² + b²) sin(x + α).

In our example, a=1, b=1, so √(a² + b²) = √2.

cos(α) = 1/√2, sin(α) = 1/√2. This implies α = π/4.

So, sin(x) + cos(x) = √2 sin(x + π/4).

The equation becomes √2 sin(x + π/4) = 1.

sin(x + π/4) = 1/√2.

Let θ = x + π/4. Then sin(θ) = 1/√2.

The general solution for sin(θ) = sin(π/4) is θ = nπ + (-1)ⁿ(π/4), n ∈ ℤ.

Substitute back θ = x + π/4:

x + π/4 = nπ + (-1)ⁿ(π/4).

x = nπ + (-1)ⁿ(π/4) - π/4.

If n is even, let n = 2m. Then x = 2mπ + π/4 - π/4 = 2mπ.

If n is odd, let n = 2m + 1. Then x = (2m+1)π - π/4 - π/4 = (2m+1)π - π/2 = 2mπ + π - π/2 = 2mπ + π/2.

This gives the solutions x = 2mπ and x = π/2 + 2mπ, m ∈ ℤ, which matches the previous method.

c) Equations Involving Identities with Multiple Angles

Use identities like sin(2x), cos(2x), tan(2x), sin(3x), cos(3x) etc.

Example: Solve cos(2x) = sin(x).

We know cos(2x) = 1 - 2sin²(x).

So, 1 - 2sin²(x) = sin(x).

Rearrange into a quadratic equation in terms of sin(x):

2sin²(x) + sin(x) - 1 = 0.

Let y = sin(x). Then 2y² + y - 1 = 0.

Factor the quadratic: (2y - 1)(y + 1) = 0.

This gives y = 1/2 or y = -1.

Case 1: sin(x) = 1/2. General solution: x = nπ + (-1)ⁿ(π/6), n ∈ ℤ.

Case 2: sin(x) = -1. We know sin(3π/2) = -1. So, α = 3π/2.

General solution: x = kπ + (-1)^k(3π/2), k ∈ ℤ.

Alternatively, for sin(x) = -1, the principal value is x = -π/2. The general solution is x = 2mπ - π/2, m ∈ ℤ.

Combining both cases, the solutions are x = nπ + (-1)ⁿ(π/6) and x = 2mπ - π/2, where n, m ∈ ℤ.

Standard Transformations in Trigonometry

Trigonometric transformations involve rewriting trigonometric expressions into different, often simpler or more useful, forms. These are essential for solving equations, simplifying complex expressions, and in calculus.

1. Sum-to-Product Formulas

These formulas convert sums or differences of sine and cosine into products. They are derived from the sum and difference identities for sine and cosine.

Derivation Sketch:

We know:

sin(A + B) = sin A cos B + cos A sin B (1)

sin(A - B) = sin A cos B - cos A sin B (2)

Adding (1) and (2): sin(A + B) + sin(A - B) = 2 sin A cos B.

Let A + B = X and A - B = Y. Then A = (X+Y)/2 and B = (X-Y)/2.

Substituting these into the equation gives the first sum-to-product formula.

Formulas:

i) sin C + sin D = 2 sin((C+D)/2) cos((C-D)/2)

ii) sin C - sin D = 2 cos((C+D)/2) sin((C-D)/2)

iii) cos C + cos D = 2 cos((C+D)/2) cos((C-D)/2)

iv) cos C - cos D = -2 sin((C+D)/2) sin((C-D)/2)

Example: Simplify sin(75°) + sin(15°).

Using the formula sin C + sin D = 2 sin((C+D)/2) cos((C-D)/2):

C = 75°, D = 15°.

(C+D)/2 = (75° + 15°)/2 = 90°/2 = 45°.

(C-D)/2 = (75° - 15°)/2 = 60°/2 = 30°.

So, sin(75°) + sin(15°) = 2 sin(45°) cos(30°).

We know sin(45°) = 1/√2 and cos(30°) = √3/2.

Result = 2 * (1/√2) * (√3/2) = √3/√2 = √(3/2).

2. Product-to-Sum Formulas

These are essentially the reverse of the sum-to-product formulas. They are derived directly from the angle addition and subtraction formulas.

Formulas:

i) 2 sin A cos B = sin(A + B) + sin(A - B)

ii) 2 cos A sin B = sin(A + B) - sin(A - B)

iii) 2 cos A cos B = cos(A + B) + cos(A - B)

iv) 2 sin A sin B = cos(A - B) - cos(A + B)

Example: Express sin(5x) cos(3x) as a sum or difference.

We use the formula 2 sin A cos B = sin(A + B) + sin(A - B). Here, A = 5x, B = 3x.

First, multiply and divide by 2: sin(5x) cos(3x) = (1/2) * [2 sin(5x) cos(3x)].

Applying the formula:

sin(5x) cos(3x) = (1/2) * [sin(5x + 3x) + sin(5x - 3x)].

sin(5x) cos(3x) = (1/2) * [sin(8x) + sin(2x)].

This transformation is useful when integrating products of trigonometric functions.

3. Multiple Angle Formulas (Double, Triple Angles)

These formulas express trigonometric functions of multiple angles (like 2θ, 3θ) in terms of trigonometric functions of the single angle (θ).

Double Angle Formulas:

i) sin(2θ) = 2 sin θ cos θ

ii) cos(2θ) = cos² θ - sin² θ = 2 cos² θ - 1 = 1 - 2 sin² θ

iii) tan(2θ) = 2 tan θ / (1 - tan² θ)

Example: If sin θ = 3/5 and θ is in the first quadrant, find sin(2θ) and cos(2θ).

First, find cos θ. Since sin² θ + cos² θ = 1, (3/5)² + cos² θ = 1.

9/25 + cos² θ = 1 => cos² θ = 1 - 9/25 = 16/25.

Since θ is in the first quadrant, cos θ = √(16/25) = 4/5.

Now, calculate:

sin(2θ) = 2 sin θ cos θ = 2 * (3/5) * (4/5) = 24/25.

cos(2θ) = cos² θ - sin² θ = (4/5)² - (3/5)² = 16/25 - 9/25 = 7/25.

Alternatively, using cos(2θ) = 2 cos² θ - 1 = 2 * (16/25) - 1 = 32/25 - 1 = 7/25.

Or, using cos(2θ) = 1 - 2 sin² θ = 1 - 2 * (9/25) = 1 - 18/25 = 7/25.

Triple Angle Formulas:

i) sin(3θ) = 3 sin θ - 4 sin³ θ

ii) cos(3θ) = 4 cos³ θ - 3 cos θ

iii) tan(3θ) = (3 tan θ - tan³ θ) / (1 - 3 tan² θ)

Derivation Sketch for sin(3θ):

sin(3θ) = sin(2θ + θ)

= sin(2θ) cos θ + cos(2θ) sin θ

= (2 sin θ cos θ) cos θ + (1 - 2 sin² θ) sin θ

= 2 sin θ cos² θ + sin θ - 2 sin³ θ

= 2 sin θ (1 - sin² θ) + sin θ - 2 sin³ θ

= 2 sin θ - 2 sin³ θ + sin θ - 2 sin³ θ

= 3 sin θ - 4 sin³ θ.

Example: Find the value of sin(30°) using the triple angle formula.

We know sin(3θ) = 3 sin θ - 4 sin³ θ. Let θ = 10°.

Then sin(30°) = 3 sin(10°) - 4 sin³(10°).

This is more useful for expressing sin(3θ) in terms of sin(θ), not for finding a specific value like sin(30°) unless sin(10°) is known.

A more practical application: If sin(10°) is known, we can find sin(30°). Or, if sin(30°) is known, we can sometimes solve for sin(10°).

4. Half-Angle Formulas

These formulas express trigonometric functions of an angle θ/2 in terms of trigonometric functions of the angle θ. They are derived from the double angle formulas for cosine.

Formulas:

From cos(2A) = 2 cos² A - 1, let 2A = θ, so A = θ/2.

cos(θ) = 2 cos²(θ/2) - 1 => cos²(θ/2) = (1 + cos θ) / 2.

So, cos(θ/2) = ±√[(1 + cos θ) / 2].

From cos(2A) = 1 - 2 sin² A, let 2A = θ, so A = θ/2.

cos(θ) = 1 - 2 sin²(θ/2) => sin²(θ/2) = (1 - cos θ) / 2.

So, sin(θ/2) = ±√[(1 - cos θ) / 2].

From tan(2A) = 2 tan A / (1 - tan² A), let 2A = θ, so A = θ/2.

tan(θ/2) = sin(θ/2) / cos(θ/2)

= [±√[(1 - cos θ) / 2]] / [±√[(1 + cos θ) / 2]]

= ±√[(1 - cos θ) / (1 + cos θ)]

Multiplying numerator and denominator inside the square root by (1 - cos θ):

tan(θ/2) = ±√[(1 - cos θ)² / (1 - cos² θ)] = ±√[(1 - cos θ)² / sin² θ]

= ±(1 - cos θ) / sin θ.

The sign depends on the quadrant of θ/2. However, we can also write:

tan(θ/2) = (1 - cos θ) / sin θ

tan(θ/2) = sin θ / (1 + cos θ)

These two forms are always correct without the ± sign because the sign of tan(θ/2) is implicitly handled by the signs of sin θ and (1 - cos θ) or (1 + cos θ).

Example: Find the value of sin(22.5°).

We want to find sin(θ/2) where θ/2 = 22.5°, so θ = 45°.

cos(45°) = 1/√2.

sin(22.5°) = ±√[(1 - cos 45°) / 2] = ±√[(1 - 1/√2) / 2].

Since 22.5° is in the first quadrant, sin(22.5°) is positive.

sin(22.5°) = √[( (√2 - 1)/√2 ) / 2] = √[(√2 - 1) / (2√2)].

To simplify, multiply numerator and denominator inside the root by √2:

sin(22.5°) = √[( (√2 - 1)√2 ) / (2√2 * √2)] = √[(2 - √2) / 4] = (1/2)√(2 - √2).

5. Identities involving tan(θ/2) (Weierstrass Substitution)

This substitution is very powerful for solving trigonometric equations and integration. It expresses sin(θ), cos(θ), and tan(θ) in terms of t = tan(θ/2).

Let t = tan(θ/2).

sin(θ) = 2t / (1 + t²)

cos(θ) = (1 - t²) / (1 + t²)

tan(θ) = 2t / (1 - t²)

Derivation Sketch:

sin(θ) = sin(2 * θ/2) = 2 sin(θ/2) cos(θ/2).

Divide numerator and denominator by cos²(θ/2):

sin(θ) = [2 sin(θ/2) cos(θ/2)] / [cos²(θ/2) + sin²(θ/2)]

= [2 (sin(θ/2)/cos(θ/2))] / [1 + (sin²(θ/2)/cos²(θ/2))]

= 2 tan(θ/2) / (1 + tan²(θ/2)) = 2t / (1 + t²).

cos(θ) = cos(2 * θ/2) = cos²(θ/2) - sin²(θ/2).

Divide numerator and denominator by cos²(θ/2):

cos(θ) = [cos²(θ/2) - sin²(θ/2)] / [cos²(θ/2) + sin²(θ/2)]

= [1 - (sin²(θ/2)/cos²(θ/2))] / [1 + (sin²(θ/2)/cos²(θ/2))]

= (1 - tan²(θ/2)) / (1 + tan²(θ/2)) = (1 - t²) / (1 + t²).

Example: Solve the trigonometric equation sin(x) + cos(x) = 1 using the t-substitution.

Let t = tan(x/2).

Substitute the t-formulas into the equation:

[2t / (1 + t²)] + [(1 - t²) / (1 + t²)] = 1.

Multiply both sides by (1 + t²):

2t + 1 - t² = 1 + t².

Rearrange the terms:

2t² - 2t = 0.

Factor out 2t:

2t(t - 1) = 0.

This gives two possibilities: t = 0 or t = 1.

Case 1: t = 0.

tan(x/2) = 0.

x/2 = nπ, where n ∈ ℤ.

x = 2nπ.

Case 2: t = 1.

tan(x/2) = 1.

x/2 = mπ + π/4, where m ∈ ℤ.

x = 2mπ + π/2.

The solutions are x = 2nπ and x = π/2 + 2mπ, which matches the result obtained earlier.

Important Note: The t-substitution method might miss solutions where tan(x/2) is undefined. This occurs when x/2 = π/2 + kπ, i.e., x = π + 2kπ. We must check if these values satisfy the original equation.

In sin(x) + cos(x) = 1, if x = π, sin(π) + cos(π) = 0 + (-1) = -1 ≠ 1. So, no solutions were missed.