Solving Trigonometric Equations Using Identities - Online Tutor, Practice Problems & Exam Prep

To solve complex linear trigonometric equations, utilize trigonometric identities to simplify expressions into a single function. For example, the equation sec2⁢θ-1 over tan(θ) equals 1 can be simplified using the Pythagorean identity to find tan(θ) equals 1. Similarly, for sin(2θ) over cos(-θ) equals 1, apply double angle and even-odd identities to isolate sin(θ) equals 1/2, yielding solutions θ=π6+2πn and θ=5π6+2πn.

Solve Trig Equations Using Identity Substitutions

Hey, everyone. When solving linear trig equations, we wanted to find an angle theta that made our equation true. And now that we're tasked with solving some more complicated trig equations, we still want to do the same thing. We want to find an angle theta that makes our equation true. But now that these equations have more than just one trig function, it's not quite so simple because how are we going to find an angle theta for which the secant squared of it minus 1 over the tangent of it is equal to 1? I don't know how to do that just using the unit circle. But now that we know a bunch of different trig identities, we can now use those identities in order to rewrite these equations in terms of just one trig function. Then we're just left with a linear trig equation that we already know how to solve. So here, I'm going to walk you through exactly how to do that. Let's go ahead and get started.

Now, here we have the equation the sec⁡θ2-1sec⁡θ²⁡2-1tan⁡θ=1. Seeing that I have multiple trig functions here tells me that I need to go ahead and use my trig identities in order to rewrite this in terms of just one trig function. Using the Pythagorean identity, we can rewrite this, giving that this is just tan⁡θ2/tan⁡θ=1. Now from here, I can cancel because one tangent in the numerator will cancel with the tangent in the denominator, leaving me with just the tangent of theta is equal to 1.

So, we started with multiple trig functions and then used identities to get down to just one trig function. From here, we can find our angles theta on the unit circle for which this is true and then add this factor of pi in in order to account for all possible solutions. Now not all of these equations are going to be quite so simple and we're not always just going to use our Pythagorean identities. So let's go ahead and walk through another example together.

Here, we're asked to find all solutions to the equation sin⁡(2θ)cos⁡(-θ)=1. Now, note that we have a sine over a cosine, but these arguments are not the same. In the numerator, I have 2θ, and in the denominator, I have -θ. We can't directly use the tangent identity here, so we need to think of another way to simplify this. Using the double angle identity, we can rewrite the top as 2·sin⁡θ·cos⁡θ over the cosine of negative theta. Since cosine is an even function, this simplifies further to 2 times the sine of theta equals 1. Now, I just need to solve this resulting linear trig equation.

By dividing both sides by 2, we get the sine of theta equals 1/2. The angles for which this is true are theta equals pi/6 and theta equals 5pi/6. But we need to find all solutions, so we'll add 2pi n to each of these. Thus, theta equals pi/6 + 2pi n and theta equals 5pi/6 + 2pi n. Now that we know how to use identities to solve some more complicated trig equations, let's continue practicing together. Thanks for watching, and I'll see you in the next one.

Find all solutions to the equation.

$\left(\cos\theta+\sin\theta\right)\left(\cos\theta-\sin\theta\right)=-\frac12$

• A. $\theta=\frac{5\pi}{12}+2\pi n,\frac{7\pi}{12}+2\pi n$
• B. $\theta=\frac{2\pi}{3}+2\pi n,\frac{4\pi}{3}+2\pi n$
• C. $\theta=\frac{\pi}{3}+2\pi n,\frac{2\pi}{3}+2\pi n$
• D. $\theta=\frac{\pi}{3}+\pi n,\frac{2\pi}{3}+\pi n$

Find all solutions to the equation where 0 ≤ $\theta$ ≤ $2\pi$.

$\sin\theta\cos\left(2\theta\right)-\sin\left(2\theta\right)\cos\theta=\frac{\sqrt2}{2}$

• A. $\theta=\frac{5\pi}{4},\frac{7\pi}{4}$
• B. $\theta=\frac{\pi}{4},\frac{3\pi}{4}$
• C. $\theta=\frac{5\pi}{4}+2\pi n,\frac{7\pi}{4}+2\pi n$
• D. $\theta=\frac{3\pi}{4},\frac{7\pi}{4}$

Here’s what students ask on this topic:

How do you solve trigonometric equations using Pythagorean identities?

To solve trigonometric equations using Pythagorean identities, first identify parts of the equation that match known identities. For example, if you have ${\sec }^2\theta -\; 1$, recognize that it can be rewritten using the identity ${\sec }^2\theta =\; 1\; +{\tan }^2\theta$. This simplifies to ${\tan }^2\theta$. Substitute this back into the equation to reduce it to a single trigonometric function, which can then be solved using standard methods.

What are double angle identities and how are they used in solving trigonometric equations?

Double angle identities express trigonometric functions of double angles (2θ) in terms of single angles (θ). For example, $\sin \left(2\theta \right)=\; 2\sin \theta \cos \theta$. To use them in solving equations, replace the double angle term with its identity. For instance, in the equation $\frac{\sin }{\left(}=\; 1$, use the double angle identity to rewrite the numerator, then simplify and solve for θ.

How do you use even-odd identities to solve trigonometric equations?

Even-odd identities help simplify trigonometric functions with negative arguments. For example, $\cos (-\theta )=\cos \left(\theta \right)$ and $\sin (-\theta )=\; -\sin \left(\theta \right)$. To use them, replace the negative argument with its equivalent positive form. For instance, in the equation $\frac{\sin }{\left(}=\; 1$, rewrite $\cos (-\theta )$ as $\cos \left(\theta \right)$ to simplify the equation.

What are the steps to solve the equation $\frac{\sin }{\left(}=\; 1$?

To solve $\frac{\sin }{\left(}=\; 1$, follow these steps:

1. Use the double angle identity: $\sin \left(2\theta \right)=\; 2\sin \theta \cos \theta$.
2. Rewrite the equation: $\frac{2}{\sin }=\; 1$.
3. Use the even-odd identity: $\cos (-\theta )=\cos \left(\theta \right)$.
4. Simplify: $\frac{2}{\sin }=\; 1$ becomes $2\sin \theta =\; 1$.
5. Solve for $\sin \theta$: $\sin \theta =\frac{1}{2}$.
6. Find all solutions: $\theta =\frac{\pi }{6}+\; 2\pi n$ and $\theta =\frac{5}{\pi }+\; 2\pi n$.

How do you find all solutions to a trigonometric equation?

To find all solutions to a trigonometric equation, first solve for the principal solutions within one period. For example, if $\sin \theta =\frac{1}{2}$, the principal solutions are $\theta =\frac{\pi }{6}$ and $\theta =\frac{5}{\pi }$. To find all solutions, add the period of the function multiplied by an integer $n$. For sine and cosine, the period is $2\pi$, so the general solutions are $\theta =\frac{\pi }{6}+\; 2\pi n$ and $\theta =\frac{5}{\pi }+\; 2\pi n$.