How do you find the exact value of $cos 2 θ + {sin}^{2} θ = 1$ $cos2theta+sin^2theta=1$ in the interval $0 \leq θ < 360$ $0<=theta<360$ ?

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Answer 1 · 2016-12-01T16:10:15

In the interval $[0, 2 π)$ $[0,2pi)$ the solutions are $θ = 0$ $theta =0$ and $θ = π$ $theta = pi$ .

Use the formula for double angle:

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

so that:

$cos (2 θ) + {sin}^{2} θ = 1$ $cos(2theta) + sin^2theta =1$

becomes:

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

${cos}^{2} θ = 1$ $cos^2 theta = 1$

$cos θ = \pm 1$ $cos theta =+-1$

In the interval $[0, 2 π)$ $[0,2pi)$ the solutions are $θ = 0$ $theta =0$ and $θ = π$ $theta = pi$ .

How do you find the exact value of cos2theta+sin^2theta=1cos2θ+sin2θ=1cos2theta+sin^2theta=1 in the interval 0<=theta<3600≤θ<3600<=theta<360?