Trignometry - 5

Geometry Level 3

Given that, A n = sin ⁡ n θ + cos ⁡ n θ A_n = \sin^{n}\theta + \cos^{n}\theta where n is whole number and θ ϵ R \theta \epsilon \mathbb{R} .

If, A n − 2 − A n = ( sin ⁡ 2 θ cos ⁡ 2 θ ) A λ A_{n-2} - A_n =( \sin^2\theta \cos^2\theta ) A_{\lambda} . What is the value of λ \lambda ?

n - 2 n - 4 n - 3 n - 1

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Kunal Joshi by Kunal Joshi , 24, India

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1 solution

David Nasr
Oct 15, 2014

s i n n − 2 θ + cos ⁡ n − 2 θ − sin ⁡ n θ − cos ⁡ n θ = ( sin ⁡ 2 θ c o s 2 θ ) A λ sin^{n-2} \theta\ + \cos^{n-2} \theta\ - \sin ^{n} \theta\ - \cos^{n} \theta =(\sin^{2} \theta\\cos^{2} \theta)\ A_{\lambda}

A λ = sin ⁡ n − 4 θ cos ⁡ 2 θ + cos ⁡ n − 4 θ s i n 2 θ − sin ⁡ n − 2 θ cos ⁡ 2 θ − cos ⁡ n − 2 θ sin ⁡ 2 θ A_{\lambda}=\frac {\sin^{n-4}\theta}{\cos^2\theta}+\frac {\cos^{n-4}\theta}{sin^2\theta}-\frac {\sin^{n-2}\theta}{\cos^2\theta}-\frac {\cos^{n-2}\theta}{\sin^2\theta}

A λ = sin ⁡ n − 4 θ − sin ⁡ n − 2 θ c o s 2 θ + cos ⁡ n − 4 θ − cos ⁡ n − 2 θ s i n 2 θ A_{\lambda}=\frac {\sin^{n-4}\theta-\sin^{n-2}\theta}{cos^2\theta}+\frac {\cos^{n-4}\theta-\cos^{n-2}\theta}{sin^2\theta}

A λ = sin ⁡ n − 4 θ ( 1 − sin ⁡ 2 θ ) c o s 2 θ + cos ⁡ n − 4 θ ( 1 − cos ⁡ 2 θ ) s i n 2 θ A_{\lambda}=\frac {\sin^{n-4}\theta (1-\sin^2\theta)}{cos^2\theta}+\frac {\cos^{n-4}\theta (1-\cos^2\theta)}{sin^2\theta}

Knowing that 1 − sin ⁡ 2 θ = cos ⁡ 2 θ 1-\sin^2\theta=\cos^2\theta and 1 − cos ⁡ 2 θ = s i n 2 θ 1-\cos^2\theta=sin^2\theta ;

A λ = sin ⁡ n − 4 θ + cos ⁡ n − 4 θ A_{\lambda}=\sin^{n-4}\theta+\cos^{n-4}\theta

Then λ = n − 4 \lambda=\boxed{n-4}

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