Question

What are all of the solutions to the equation (cos θ)(cos θ) + 1 = (sin θ)(sin θ)?

Answer 1

Answer: Starting with the given equation:

(cos θ)(cos θ) + 1 = (sin θ)(sin θ)

We can use the identity cos² θ + sin² θ = 1 to rewrite the right-hand side:

(cos θ)(cos θ) + 1 = 1 - (cos θ)(cos θ)

Combining like terms, we get:

2(cos θ)(cos θ) = 0

Dividing both sides by 2, we get:

(cos θ)(cos θ) = 0

Taking the square root of both sides, we get:

cos θ = 0

This equation is true for θ = π/2 + kπ, where k is any integer. So the solutions to the equation are:

θ = π/2 + kπ, where k is any integer.

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Explanation:

Answer 2

Answer:

We can use the trigonometric identity cos²θ + sin²θ = 1 to manipulate the given equation:

cos²θ + 1 = sin²θ

Subtracting cos²θ from both sides, we get:

1 = sin²θ - cos²θ

Using the identity sin²θ - cos²θ = sin(θ + π/2)sin(θ - π/2), we can simplify the right-hand side:

1 = sin(θ + π/2)sin(θ - π/2)

Now we can use the product-to-sum identity sin(θ + π/2)sin(θ - π/2) = (1/2)[cos(θ - (-π/2)) - cos(θ + π/2)] to further simplify the equation:

1 = (1/2)[cos(θ + π) - cos(θ)]

Since cos(θ + π) = -cos(θ), we can substitute into the equation:

1 = (1/2)[-cos(θ) - cos(θ + π/2)]

Using the identity cos(θ + π/2) = -sin(θ), we get:

1 = (1/2)[-cos(θ) + sin(θ)]

Multiplying both sides by 2, we get:

2 = -cos(θ) + sin(θ)

Rearranging, we get:

cos(θ) + sin(θ) = 2

Now we can use the identity cos(θ - α) = cos(θ)cos(α) + sin(θ)sin(α) to find the solutions:

cos(θ - π/4) = cos(θ)cos(π/4) + sin(θ)sin(π/4) = (1/√2)cos(θ) + (1/√2)sin(θ)

Therefore, we have:

(1/√2)cos(θ) + (1/√2)sin(θ) = 2

Multiplying both sides by √2, we get:

cos(θ) + sin(θ)√2 = 2√2

Now we can use the identity cos(α) + sin(α) = √2 sin(α + π/4) to find the solutions:

cos(θ + π/4) = cos(θ)cos(π/4) - sin(θ)sin(π/4) = (1/√2)cos(θ) - (1/√2)sin(θ)

Therefore, we have:

(1/√2)cos(θ) - (1/√2)sin(θ) = 2√2

Multiplying both sides by √2, we get:

cos(θ) - sin(θ)√2 = 2

We now have two equations with two unknowns (cos(θ) and sin(θ)), which can be solved using algebraic methods. Adding the two equations together, we get:

2cos(θ) = 4√2

Dividing both sides by 2, we get:

cos(θ) = 2√2

Since the range of the cosine function is [-1,1], there are no real solutions to this equation. Therefore, there are no solutions to the original equation (cos θ)(cos θ) + 1 = (sin θ)(sin θ).

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