Trigonometry - Solve a Trigonometric Function by Squaring Both Sides

Solve the following equation by squaring both sides: $1 + cot(\theta) = csc(\theta)$

$\theta = 0, \pi$

$\theta = 0$

$\theta = \frac{\pi}{2}$

$\theta = \frac{\pi}{2}, \frac{3\pi}{2}$

Explanation

We begin with our original equation:

$1 + cot(\theta) = csc(\theta)$

$(1+cot(\theta))^2 = (csc(\theta))^2$

$1 + 2cot(\theta) + cot^2(\theta) = csc^2(\theta)$

$1 + 2cot(\theta) + cot^2(\theta) = 1 + cot^2(\theta)$ (Pythagorean Identity)

$2cot(\theta) = 0$

$cot(\theta) = 0$

Looking at the unit circle we see that $cot(\theta) = 0$ at $\frac {\pi}{2}$ and $\frac{3\pi}{2}$ . We must plug these back into our original equation to validate them.

Checking $\theta = \frac{\pi}{2}$

$1 + cot(\frac{\pi}{2}) = csc(\frac{\pi}{2})$

Checking $\theta = \frac{3\pi}{2}$

$1 + cot(\frac{3\pi}{2}) = \frac{3\pi}{2}$

And so our only solution is $\theta = \frac{\pi}{2}$

Solve the following equation by squaring both sides: $cos(\theta) + sin(\theta) = 1$

$\theta = 3\pi$

$\theta = 0$

$\theta = 0, \pi$

$\theta = \pi$

Explanation

We begin with our original equation.

$cos(\theta) + sin(\theta) = 1$

$(cos(\theta) + sin(\theta))^2 = 1^2$

$cos^2(\theta) + 2cos(\theta)sin(theta) + sin^2(\theta) = 1$

$cos^2(\theta) + sin^2(\theta) + 2cos(\theta)sin(\theta) = 1$

$1 + 2cos(\theta)sin(\theta) = 1$ (Pythagorean Identity)

$2cos(\theta)sin(\theta) = 0$

$sin(2\theta) = 0$ (Double-Angle Formula)

We know that $sin(2\theta)$ will be equal to for when $\theta$ is any multiple of and when $\theta = 0$ . We need to check both solutions (we will simply check $\pi$ for simplicity) to make sure they are valid solutions.

Checking $\theta = 0$ :

Checking $\theta = \pi$

$cos(\pi) + sin(\pi) = 1$

By checking our solutions, we see the only solution to this equation is $\theta = 0$

Solve the following equation by squaring both sides: $\sqrt{5}sin(\theta) +cos(\theta) = 1$ .

$\theta = \pi$

$\theta = 0$

$\theta = 2.3$

$\theta = 2.3, 0$

Explanation

This one is not as straight-forward. We must manipulate the original equation before squaring both sides.

$\sqrt{5}sin(\theta) + cos(\theta) = 1$

$\sqrt{5}sin(\theta) = 1 - cos(\theta)$

$(\sqrt{5}sin(\theta))^2 = (1 - cos(\theta))^2$

$5sin^2(\theta) = 1 - 2cos(\theta) + cos^2(\theta)$

$5sin^2(\theta) = 1 + cos^2(\theta) - 2cos(\theta)$

$5(1 - cos^2(\theta) ) = 1 + cos^2(\theta) - 2cos(\theta)$ (Pythagorean Identity)

$5 - 5cos^2(\theta) = 1 +cos^2(\theta) - 2cos(\theta)$

$0 = 6cos^2(\theta) - 2cos(\theta) - 4$

$0 = 3cos^2(\theta) - cos(\theta) - 2$ (divide both sides by 2)

$0 = (3cos(\theta) + 2)(cos(\theta) -1)$

Solving for each:

$3cos(\theta) + 2 = 0$

$3cos(\theta) = -2$

$cos(\theta) = \frac{-2}{3}$

$\theta = cos^{-1}(\frac{-2}{3})$

$\theta = 2.3$ radians

$cos(\theta) - 1 = 0$

$cos(\theta) = 1$

From the unit circle we know that $cos(\theta) = 1$ when $\theta = 0, 2\pi$ .

So now we must go back and check all of our solutions.

Checking $\theta = 2.3$

$\sqrt{5}sin(2.3) + cos(2.3) = 1$

$\sqrt{5}(0.75) - 0.67 = 1$

Checking $\theta = 0$ (this is also equal to checking $\theta = 2\pi$ )

$\sqrt{5}sin(0) + cos(0) = 1$

Both of our solutions are correct.

Which of the following is the main purpose of squaring both sides of a trigonometric equation?

To get rid of a radical

To produce a familiar identity/formula that we can use to solve the problem

To solve the problem, duh!

Working with square trigonometric functions is easier than those of the first power

Explanation

Our first line of defense when solving trigonometric functions is using a familiar identity/formula such as the Pythagorean Identities or the Double Angle Formulas. When we are unable to use an identity or formula we are able to square both sides of the equation and with further manipulation we are usually able to produce one of these identities thus simplifying our problem and making it easier to solve.

Solve the following equation by squaring both sides: $tan(\theta) + 1 = sec(\theta)$

$\theta = 0$

$\theta = \pi$

$\theta = 0, \pi$

$\theta = 1$

Explanation

We begin with our original equation

$tan(\theta) + 1 = sec(\theta)$

$(tan(\theta) + 1)^2 = (sec(\theta))^2$

$tan^2(\theta) + 2tan(\theta) + 1 = sec^2(\theta)$

$tan^2(\theta) + 2tan(\theta) + 1 = tan^2(\theta) + 1$ (Pythagorean Identity)

$2tan(\theta) = 0$

$tan(\theta) = 0$

$\frac{sin(\theta)}{cos(\theta)} = 0$ (substitution)

Using this form, we see we really only need to consider when $sin(\theta) = 0$ at and $\pi$ . Now we must plug these values into the original equation to check and see if they are both acceptable solutions to our problem.

Checking $\theta = 0$ :

Checking $\theta = \pi$

$tan(\pi) + 1 = sec(\pi)$

By checking our solutions we see the only solution to our equation is $\theta = 0$ .

Solve the following equation by squaring both sides: $-\sqrt{3} + tan(\theta) = sec(\theta)$

$\theta = \frac {\sqrt \pi} {6}$

$\theta = \frac{\sqrt7\pi} {6}$

$\theta = \frac {\sqrt\pi}{6}, \frac{7\pi}{6}$

$\theta = \frac{\sqrt11\pi}{6}$

Explanation

We begin with our original equation:

$-\sqrt{3} + tan(\theta) = sec(\theta)$

$(-\sqrt{3} + tan(\theta))^2$ $= (sec(\theta))^2 (sec(\theta))^2sec(\theta))^2c(\theta))^2(\theta))^2(\theta))^2(\tan))^2))^2^2$

$3 - 2\sqrt{3}tan(\theta) + tan^2(\theta) = sec^2(\theta)$

$3 - 2\sqrt{3}tan(\theta) + tan^2(\theta) = 1 + tan^2(\theta)$ (Pythagorean Identity)

$-2\sqrt{3}tan(\theta) = -2$

$tan(\theta) = \frac{1}{\sqrt{3}}$

From the unit circle, we see that $tan(\theta) = \frac{1}{\sqrt{3}}$$ when $$\theta = \frac{\pi}{6}, \frac{7\pi}{6}$ . We must check both of these solutions in the original equation.

Checking $\theta = \frac{\pi}{6}$

$-\sqrt{3} + tan(\frac{\pi}{6}) = sec(\frac{\pi}{6})$

$-\sqrt{3} + \frac{\sqrt{3}}{3} = \frac{2\sqrt{3}}{3}$

$\frac{\sqrt{3}}{3} eq \frac{5\sqrt{3}}{3}$

Checking $\theta = \frac{7\pi}{6}$

$-\sqrt{3} + tan(\frac{7\pi}{6}) = sec(\frac{7\pi}{6})$

$-\sqrt{3} + \frac{\sqrt{3}}{3} = -\frac{2\sqrt{3}}{3}$

$\frac{\sqrt{3}}{3} = \frac{\sqrt{3}}{3}$

So we see our only solution is $\theta = \frac{\sqrt{\pi}}{6}$

True or False: All solutions found from squaring both sides of a trigonometric function are valid should be given as a final answer.

True

False

Explanation

This is not true. This is because when squaring both sides and then plugging back into the original equations, some of our solutions may be extraneous solutions. Therefore, when solving a trigonometric equation by squaring both sides, all solutions found must be plugged back into the original equation and validated.

True or False: You should always solve for a trigonometric equation by squaring both sides. This will always be the most efficient method.

True

False

Explanation

We should only square by both sides when all other identities are not able to be used in an equation. Quite often, you will find that a trigonometric identity can be used to simplify an equation. Squaring both sides is ultimately trying to produce a trigonometric identity in order to solve for the equation.

Solve a Trigonometric Function by Squaring Both Sides

Help Questions

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation