If - 1x-x-2cos- then - x-n-1x-n- is equal to nbsp- 2- cos n - - 2- sin n - -cos n - -sin n - -

The correct option is A 2 cos n θ Given #160; x+ 1x =2cosθ #160; x^2 +1 =2xcosθ #160; x^2 2xcosθ +1=0 #160;By Quadratic form ...

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Applications of Dot Product

If 1x+x=...

Question

If $\frac{1}{x} + x = 2 c o s θ,$ then $x^{n} + \frac{1}{x^{n}}$ is equal to

2

cos n

θ

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2

sin n

θ

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cos n

θ

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sin n

θ

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Solution

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Similar questions

m s i n θ = n c o s θ

\frac{t a n θ + c o t θ}{t a n θ - c o t θ} = \frac{n s i n θ + m c o s θ}{n s i n θ - m c o s θ} = \frac{n^{2} + m^{2}}{n^{2} - m^{2}}

c o s θ c o s 2 θ c o s 2^{2} θ . . . c o s 2^{n} - 1 θ = \frac{sin 2^{n} θ}{2^{n} s i n θ},

n ϵ N .

cos θ - sin θ = m

sec θ - cos θ = n

θ

cos θ - sin θ = a^{3}

a^{2} b^{2} (a^{2} + b^{2}) = 1

Which of the following statements are correct?

1 . if sin θ = sin α ⇒ θ = nπ + $(- 1)^{n}$ α where α ∈ [ - $\frac{π}{2}$ , $\frac{π}{2}$ ] n ∈ I

2 . if cos θ = cos α ⇒ 2nπ±α where α ∈ [0,π] n ∈ I

3 . if tan θ = tan α ⇒ θ = nπ + α where α ∈ (- $\frac{π}{2}$ , $\frac{π}{2}$ ) n ∈ I

y = {cos}^{- 1} (\frac{x^{n} - x^{- n}}{x^{n} + x^{- n}})

y^{'} (x)

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