58th Problem 2016

Geometry Level 1

sin ⁡ θ − sin ⁡ θ cos ⁡ 2 θ = ? \large \sin{ \theta - \sin{ \theta \cos{ ^{ 2 }\theta } } } = \, ?

Check out the set: 2016 Problems

1 sin ⁡ 3 θ \sin ^{ 3 }{ \theta } 1 cos ⁡ 2 θ \frac { 1 }{ \cos{ ^{ 2 }\theta } } 1 sin ⁡ 2 θ \frac { 1 }{ \sin{ ^{ 2 }\theta } }

Angela Fajardo by Angela Fajardo , 19, Philippines

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1 solution

Ralph James
Mar 12, 2016

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

Extract the common factor, sin ⁡ θ \sin \theta .

sin ⁡ θ ⋅ ( 1 − cos ⁡ 2 θ ) \sin \theta \cdot ( 1 - \cos^{2} \theta) .

Use the identity: sin ⁡ 2 θ + cos ⁡ 2 θ = 1 \sin^{2} \theta + \cos^{2} \theta = 1 . sin ⁡ 2 θ + cos ⁡ 2 θ = 1 ⟹ 1 − cos ⁡ 2 θ = sin ⁡ 2 θ \sin^{2} \theta + \cos^{2} \theta = 1 \implies 1 - \cos^{2} \theta = \sin^{2} \theta .

Substituting: sin ⁡ θ ⋅ sin ⁡ 2 θ = sin ⁡ 3 θ \sin \theta \cdot \sin^{2} \theta = \boxed{\sin^{3} \theta} .

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