How do you simplify $\sqrt{(1 - cos θ) (1 + cos θ)}$ $sqrt((1-costheta)(1+costheta))$ ?

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How do you simplify $\sqrt{(1 - cos θ) (1 + cos θ)}$ $sqrt((1-costheta)(1+costheta))$ ?

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Answer 1 · 2016-07-23T20:51:22

It is

$\sqrt{(1 - cos θ) (1 + cos θ)} = \sqrt{1 - {cos}^{2} θ} = \sqrt{{sin}^{2} θ} = | sin θ |$ $sqrt((1-costheta)(1+costheta))=sqrt[1-cos^2 theta]= sqrt[sin^2 theta]=abs(sin theta)$

where $| |$ $abs$ is the absolute value.

We used the fact that ${sin}^{2} θ + {cos}^{2} θ = 1$ $sin^2 theta+cos^2 theta=1$

Answer 2 · 2016-07-23T20:58:01

$\sqrt{(1 - cos θ) (1 + cos θ)} = | sin θ |$ $sqrt((1-costheta)(1+costheta)) = |sintheta|$

Explanation:

For this problem, we can use the difference of squares methods, which tells us that

$(a - b) (a + b) = a^{2} - b^{2}$ $(a-b)(a+b) = a^2-b^2$

Applying this method, we then get

$\sqrt{(1 - cos θ) (1 + cos θ)} = \sqrt{1 - {cos}^{2} θ}$ $sqrt((1-costheta)(1+costheta)) = sqrt(1-cos^2theta)$

Also note that ${sin}^{2} θ + {cos}^{2} θ = 1 \to {sin}^{2} θ = 1 - {cos}^{2} θ$ $sin^2theta + cos^2theta = 1 -> sin^2theta = 1-cos^2theta$

Thus we can simplify this even further, giving us

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

We have to be careful here, because the root implies that we should have to answers. In fact, if we would have made the mistake of saying that $\sqrt{{sin}^{2} θ} = sin x$ $sqrt(sin^2theta) = sin x$ , we would have the following graphs:

Graph of $\sqrt{{sin}^{2} θ}$ $sqrt(sin^2theta)$ :
graph{sqrt(sin x * sinx) [-5.504, 5.596, -2.153, 3.396]}
Graph of $sin θ$ $sin theta$ :
graph{sin x [-5.504, 5.596, -2.153, 3.396]}

These functions are noticeably different, but they're only off by a little bit. In this problem, we could say that

$\sqrt{(1 - cos θ) (1 + cos θ)}$ $sqrt((1-costheta)(1+costheta))$

$= \sqrt{{sin}^{2} θ} = - sin θ$ $= sqrt(sin^2theta) = -sintheta$ and $sin θ \to | sin θ |$ $sin theta -> |sintheta|$

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How do you simplify $\sqrt{(1 - cos θ) (1 + cos θ)}$ $sqrt((1-costheta)(1+costheta))$ ?

2 Answers

Explanation:

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