Sines And Cosines Are Better

Geometry Level 2

tan ⁡ 2 θ − 1 tan ⁡ 2 θ + 1 = ? \large \dfrac { \tan^2 \theta - 1 }{ \tan^2 \theta + 1 } = \, ?

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cos ⁡ 2 θ \cos ^{ 2 }{ \theta } 1 − 2 cos ⁡ 2 θ 1 - 2\cos ^{ 2 }{ \theta } − 1 -1 sin ⁡ 2 θ \sin ^{ 2 }{ \theta }

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Angela Fajardo by Angela Fajardo , 19, Philippines

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2 solutions

Michael Fuller
Mar 13, 2016

tan ⁡ 2 θ − 1 tan ⁡ 2 θ + 1 \frac { \tan^2 \theta - 1 }{ \tan^2 \theta + 1 }

= tan ⁡ 2 θ + 1 − 2 tan ⁡ 2 θ + 1 = \frac { \tan^2 \theta + 1 - 2 }{ \tan^2 \theta + 1 }

= 1 − 2 tan ⁡ 2 θ + 1 = 1 - \frac { 2 }{ \tan^2 \theta + 1 }

= 1 − 2 sec ⁡ 2 θ = 1 - \frac { 2 }{ \sec^2 \theta }

= 1 − 2 cos ⁡ 2 θ = \large \color{#20A900}{\boxed{1 - 2 \cos^2 \theta }}

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tan ⁡ 2 θ − 1 tan ⁡ 2 θ + 1 = s i n 2 θ − c o s 2 θ cos ⁡ 2 θ sin ⁡ 2 θ + cos ⁡ 2 θ cos ⁡ 2 = sin ⁡ 2 θ − cos ⁡ 2 θ = \frac{\tan^2 \theta - 1}{\tan^2 \theta + 1} = \frac{\frac{sin^2 \theta - cos^2 \theta}{\cos^2 \theta}}{\frac{\sin^2 \theta + \cos^2 \theta}{\cos^2}} = \sin^2 \theta - \cos^2 \theta = = sin ⁡ 2 θ + cos ⁡ 2 θ − cos ⁡ 2 θ − cos ⁡ 2 θ = 1 − 2 cos ⁡ 2 θ = \sin^2 \theta + \cos^2 \theta - \cos^2 \theta - \cos^2 \theta = 1 - 2\cos^2 \theta

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In your last step, we could have instead written it as sin ⁡ 2 θ − cos ⁡ 2 θ = − cos ⁡ ( 2 θ ) , \sin^2\theta - \cos^2 \theta = -\cos(2\theta),

which could be more useful, depending on the context where this expression comes up.

Eli Ross Staff - 5 years, 2 months ago

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