Trigonometry #86

Geometry Level 1

sin ⁡ θ − 2 sin ⁡ 3 θ 2 cos ⁡ 3 θ − cos ⁡ θ \frac{\sin \theta - 2\sin^{3} \theta}{2\cos^{3} \theta - \cos \theta} equals

This problem is part of the set Trigonometry .

sin ⁡ 2 θ cos ⁡ θ \frac{\sin^{2}\theta}{\cos\theta} cos ⁡ 2 θ sin ⁡ θ \frac{\cos^{2}\theta}{\sin\theta} cot ⁡ θ \cot\theta tan ⁡ θ \tan \theta

Omkar Kulkarni by Omkar Kulkarni , 22, India

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3 solutions

Caleb Townsend
Feb 4, 2015

First, factor out tan ⁡ θ \tan \theta to get tan ⁡ θ 1 − 2 sin ⁡ 2 θ 2 cos ⁡ 2 θ − 1 \tan \theta \frac{1 - 2\ \sin ^2 \theta}{2\ \cos ^2 \theta - 1} Next, use the identity sin ⁡ 2 θ + cos ⁡ 2 θ = 1 \sin ^2 \theta + \cos ^2 \theta = 1 to get tan ⁡ θ 1 − 2 sin ⁡ 2 θ 1 − 2 sin ⁡ 2 θ = tan ⁡ θ \tan \theta \frac{1 - 2\ \sin ^2 \theta}{1 - 2\ \sin ^2 \theta} = \boxed{\tan \theta}

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Lu Chee Ket
Feb 5, 2015

Sin q (1 - 2 Sin^2 q)/ [Cos q (2 Cos^2 q - 1)] = Sin q/ Cos q = Tan q

Either knowing Cos 2 q or

1 - 2 Sin^2 q = 1 - 2 (1 - Cos^2 q) = 2 Cos^2 q - 1

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Parag Zode
Feb 4, 2015

We common the factors s i n θ sin\theta and c o s θ cos\theta in Numerator and denominator respectively ,i.e.

s i n θ ( 1 − s i n 2 θ ) c o s θ ( 2 cos ⁡ 2 θ − 1 ) \dfrac{sin \theta(1-sin^2 {\theta})}{cos \theta(2\ \cos^2{\theta}-1)} = s i n θ ( 2 cos ⁡ 2 θ − 1 ) c o s θ ( 2 cos ⁡ 2 θ − 1 ) =\dfrac{sin \theta(2\ \cos^{2} \theta-1)}{cos \theta(2\ \cos^{2}\theta-1)} = s i n θ c o s θ =\dfrac{sin \theta}{cos \theta} = t a n θ =\boxed{tan \theta}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

can you further explain why 1 - sin^2 theta = 2 cos^2 theta - 1

atharv toraskar - 5 years, 8 months ago

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