What is sin^2(θ)-1+2cos(θ)sin(θ) ?

The solution to sin^2(θ)-1+2cos(θ)sin(θ) is sin(2θ)-cos^2(θ)

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simplify sin^2(θ)-1+2cos(θ)sin(θ)

simplify

sin^{2} (θ) - 1 + 2 cos (θ) sin (θ)

sin (2 θ) - cos^{2} (θ)

Solution steps

sin^{2} (θ) - 1 + 2 cos (θ) sin (θ)

Use the Double Angle identity:

2 sin (x) cos (x) = sin (2 x)

= sin^{2} (θ) - 1 + sin (2 θ)

Use the Pythagorean identity:

1 = cos^{2} (x) + sin^{2} (x)

1 - sin^{2} (x) = cos^{2} (x)

= sin (2 θ) - cos^{2} (θ)

1 - sin^{2} (- x)

sin (- y) - sin (2 π - y)

49 tan^{2} (x) + 49

- sin (27 0^{\circ} - x)

\frac{d v}{cot ( v )}

What is sin^2(θ)-1+2cos(θ)sin(θ) ?
The solution to sin^2(θ)-1+2cos(θ)sin(θ) is sin(2θ)-cos^2(θ)