Solve for θ cos(theta)^2-sin(theta)^2=1

$\cos^{2} (θ) - \sin^{2} (θ) = 1$

Subtract $1$ from both sides of the equation.

$\cos^{2} (θ) - \sin^{2} (θ) - 1 = 0$

Simplify $\cos^{2} (θ) - \sin^{2} (θ) - 1$ .

Move $- 1$ .

$\cos^{2} (θ) - 1 - \sin^{2} (θ) = 0$

Reorder $\cos^{2} (θ)$ and $- 1$ .

$- 1 + \cos^{2} (θ) - \sin^{2} (θ) = 0$

Rewrite $- 1$ as $- 1 (1)$ .

$- 1 (1) + \cos^{2} (θ) - \sin^{2} (θ) = 0$

Factor $- 1$ out of $\cos^{2} (θ)$ .

$- 1 (1) - 1 (- \cos^{2} (θ)) - \sin^{2} (θ) = 0$

Factor $- 1$ out of $- 1 (1) - 1 (- \cos^{2} (θ))$ .

$- 1 (1 - \cos^{2} (θ)) - \sin^{2} (θ) = 0$

Rewrite $- 1 (1 - \cos^{2} (θ))$ as $- (1 - \cos^{2} (θ))$ .

$- (1 - \cos^{2} (θ)) - \sin^{2} (θ) = 0$

Apply pythagorean identity.

$- \sin^{2} (θ) - \sin^{2} (θ) = 0$

Subtract $\sin^{2} (θ)$ from $- \sin^{2} (θ)$ .

$- 2 \sin^{2} (θ) = 0$

Solve for $θ$ .

Divide each term in $- 2 \sin^{2} (θ) = 0$ by $- 2$ and simplify.

Divide each term in $- 2 \sin^{2} (θ) = 0$ by $- 2$ .

$\frac{- 2 \sin^{2} (θ)}{- 2} = \frac{0}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 \sin^{2} (θ)}{- 2} = \frac{0}{- 2}$

Divide $\sin^{2} (θ)$ by $1$ .

$\sin^{2} (θ) = \frac{0}{- 2}$

Simplify the right side.

Divide $0$ by $- 2$ .

$\sin^{2} (θ) = 0$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$\sin (θ) = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

Rewrite $0$ as $0^{2}$ .

$\sin (θ) = \pm \sqrt{0^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$\sin (θ) = \pm 0$

Plus or minus $0$ is $0$ .

$\sin (θ) = 0$

Take the inverse sine of both sides of the equation to extract $θ$ from inside the sine.

$θ = \arcsin (0)$

Simplify the right side.

The exact value of $\arcsin (0)$ is $0$ .

$θ = 0$

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from $π$ to find the solution in the second quadrant.

$θ = π - 0$

Subtract $0$ from $π$ .

$θ = π$

Find the period of $\sin (θ)$ .

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Divide $2 π$ by $1$ .

$2 π$

The period of the $\sin (θ)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$θ = 2 π n, π + 2 π n$ , for any integer $n$

Consolidate the answers.

$θ = π n$ , for any integer $n$

Solve for θ cos(theta)^2-sin(theta)^2=1

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