Given the unit circle, if the point $(a, b)$ lies on the circle, find the value of $sin(2 heta)$, $cos(2 heta)$, and $ an(2 heta)$ where $ heta$ is the angle that corresponds to the point $(a, b)$. Verify that these values satisfy the double angle

Given the point (a, b) on the unit circle, we know:

$a = \cos \theta $

$b = \sin \theta $

We need to find $\sin(2\theta)$, $\cos(2\theta)$, and $\tan(2\theta)$. Using the double angle identities:

$ \sin(2\theta) = 2 \sin \theta \cos \theta $

$ \cos(2\theta) = \cos^2 \theta – \sin^2 \theta $

$ \tan(2\theta) = \frac{2\tan \theta}{1 – \tan^2 \theta} $

By substituting a = cos θ and b = sin θ:

$ \sin(2\theta) = 2ab $

$ \cos(2\theta) = a^2 – b^2 $

$ \tan(2\theta) = \frac{2b/a}{1 – b^2/a^2} = \frac{2b/a}{(a^2 – b^2)/a^2} = \frac{2ab}{a^2 – b^2} $

Verification of double-angle identities:

$ (2ab)^2 + (a^2 – b^2)^2 = 4a^2b^2 + a^4 – 2a^2b^2 + b^4 = a^4 + 2a^2b^2 + b^4 = (a^2 + b^2)^2 = 1 $

Given the point (a, b) on the unit circle, where:

$a = cos heta $

$b = sin heta $

Double angle identities are:

$ sin(2 heta) = 2 sin heta cos heta $

$ cos(2 heta) = cos^2 heta – sin^2 heta $

$ an(2 heta) = frac{2 an heta}{1 – an^2 heta} $

Substituting:

$ sin(2 heta) = 2ab $

$ cos(2 heta) = a^2 – b^2 $

$ an(2 heta) = frac{2b/a}{1 – b^2/a^2} = frac{2ab}{a^2 – b^2} $

Verification of identities:

$2ab
eq a^2 – b^2 $

Given (a, b) on the unit circle,

$sin(2 heta) = 2ab, cos(2 heta) = a^2 – b^2, an(2 heta) = frac{2ab}{a^2 – b^2}$

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