Find all solutions to the equation.(cos⁡θ+sin⁡θ)(cos⁡θ−sin⁡θ)=−12\left(\cos\theta+\sin\theta\right)\left(\cos\theta-\sin\theta\right)=-\frac12(cosθ+sinθ)(cosθ−sinθ)=−21

θ = π 3 + π n , 2 π 3 + π n \theta=\frac{\pi}{3}+\pi n,\frac{2\pi}{3}+\pi n θ = 3 π + πn , 3 2 π + πn

Find all solutions to the equation. ( cos ⁡ θ + sin ⁡ θ ) ( cos ⁡ θ − sin ⁡ θ ) = − 1 2 \left(\cos\theta+\sin\theta\right)\left(\cos\theta-\sin\theta\right)=-\frac12 ( cos θ + sin θ ) ( cos θ − sin θ ) = − 2 1

Hey, everyone. In this problem, we're asked to find all solutions to the equation. Cosine of theta plus sine of theta times cosine of theta minus sine of theta is equal to negative one half. So let's go ahead and get started. Now looking at that left side of my equation here I have the cosine plus the sine times the cosine and minus the sine. So that tells me that this is a difference of squares and if I multiply this out this will give me the cosine squared of theta minus the sine squared of theta And this is equal to that right side negative one half. Now, remember, we want to be constantly scanning for identities here. And here, this cosine squared of theta minus the sine squared of theta is really just my double angle identity. So my cosine double angle identity tells me that the cosine of 2 theta is equal to exactly that. So now I just have the cosine of 2 theta is equal to negative one half. Now this is just a linear trig equation so we can go ahead and solve this because we only have one trig function here. Now I want to find my values for which the cosine is equal to negative one half, setting my argument equal to that. Now here, remember that your argument is 2 theta. It's not just theta. So here our answers are that 2 theta is equal to the value for which the cosine is equal to negative one half, which is 2 pi over 3 and also 4pi over 3. So these are my 2 solutions here, but we're not done yet because remember that we were asked to find all solutions. So I need to go ahead and add 2pi in to each of these. Now we still aren't done because our arguments are 2 theta, not just theta. So we have one final step here, and that's to isolate our argument as theta. So we want to go ahead and divide both sides by 2 for both of these solutions here. Now those twos will cancel and we will have isolated theta. Now we'll also be able to cancel a bunch of twos over here and that first answer is that theta is equal to pi over 3 plus pi n having canceled and divided everything by 2. And then our final answer is that theta is equal to 2 pi over 3 plus pi n having canceled that 2 and also divided that 4 by 2. So these are final answers here with theta fully isolated having represented all of our solutions with that extra factor of pi n. Thanks for watching, and let me know if you have questions.

Find all solutions to the equation. ( cos ⁡ θ + sin ⁡ θ ) ( cos ⁡ θ − sin ⁡ θ ) = − 1 2 \left(\cos\theta+\sin\theta\right)\left(\cos\theta-\sin\theta\right)=-\frac12 ( cos θ + sin θ ) ( cos θ − sin θ ) = − 2 1

θ = π 3 + π n , 2 π 3 + π n \theta=\frac{\pi}{3}+\pi n,\frac{2\pi}{3}+\pi n θ = 3 π + πn , 3 2 π + πn

θ = 5 π 12 + 2 π n , 7 π 12 + 2 π n \theta=\frac{5\pi}{12}+2\pi n,\frac{7\pi}{12}+2\pi n θ = 12 5 π + 2 πn , 12 7 π + 2 πn

θ = 2 π 3 + 2 π n , 4 π 3 + 2 π n \theta=\frac{2\pi}{3}+2\pi n,\frac{4\pi}{3}+2\pi n θ = 3 2 π + 2 πn , 3 4 π + 2 πn

θ = π 3 + 2 π n , 2 π 3 + 2 π n \theta=\frac{\pi}{3}+2\pi n,\frac{2\pi}{3}+2\pi n θ = 3 π + 2 πn , 3 2 π + 2 πn

12. Trigonometric Identities

Solving Trigonometric Equations Using Identities

Precalculus

12. Trigonometric Identities

Solving Trigonometric Equations Using Identities

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Multiple Choice

Find all solutions to the equation.
$\left(\cos\theta+\sin\theta\right)\left(\cos\theta-\sin\theta\right)=-\frac12$

$\theta=\frac{5\pi}{12}+2\pi n,\frac{7\pi}{12}+2\pi n$

$\theta=\frac{2\pi}{3}+2\pi n,\frac{4\pi}{3}+2\pi n$

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

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

Verified step by step guidance

Start by expanding the expression $(\cos\theta + \sin\theta)(\cos\theta - \sin\theta)$. This is a difference of squares, which can be simplified using the identity $a^2 - b^2 = (a+b)(a-b)$.

Apply the identity: $(\cos\theta + \sin\theta)(\cos\theta - \sin\theta) = \cos^2\theta - \sin^2\theta$.

Recognize that $\cos^2\theta - \sin^2\theta$ can be rewritten using the identity $\cos^2\theta - \sin^2\theta = \cos(2\theta)$.

Set the equation $\cos(2\theta) = -\frac{1}{2}$ and solve for $2\theta$. The solutions for $\cos(\alpha) = -\frac{1}{2}$ are $\alpha = \frac{2\pi}{3} + 2\pi n$ and $\alpha = \frac{4\pi}{3} + 2\pi n$.

Divide the solutions for $2\theta$ by 2 to find $\theta$. This gives $\theta = \frac{\pi}{3} + \pi n$ and $\theta = \frac{2\pi}{3} + \pi n$. These are the solutions to the original equation.

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Struggling with Precalculus?

Find all solutions to the equation.(cos⁡θ+sin⁡θ)(cos⁡θ−sin⁡θ)=−12\left(\cos\theta+\sin\theta\right)\left(\cos\theta-\sin\theta\right)=-\frac12(cosθ+sinθ)(cosθ−sinθ)=−21​

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Find all solutions to the equation.
$\left(\cos\theta+\sin\theta\right)\left(\cos\theta-\sin\theta\right)=-\frac12$