Let I n -2 cos nx - n - N- then I 1 - I n -1- I n is equal toA- I n -3B- I n-2C- I n -1 - I 2D- I n -1 - I 2

The correct option is B I(n+2) I(n) = 2cosnx ⇒ I(1).I(n+1) I(n) = 4cosx cos(n+1)x 2cosnx = 2[2cos(n+1)x cosx cosnx] = 2[cos(n+2)x + cosnx cosnx] = 2 co ...

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Byju's Answer

Standard XII

Mathematics

Principal Solution of Trigonometric Equation

Let I n =2 co...

Question

Let I(n) = 2cos nx $\forall n \in N$ , then I (1).I(n+1) - I(n) is equal to

I(n+3)

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I(n+2)

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I(n+1).I(2)

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I(n+1).I(2)

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Solution

The correct option is B
I(n+2)

I(n) = 2cosnx $\Rightarrow$ I(1).I(n+1) - I(n)

= 4cosx cos(n+1)x - 2cosnx

= 2[2cos(n+1)x cosx - cosnx]

= 2[cos(n+2)x + cosnx - cosnx]

= 2 cos(n+2)x = I(n+2)

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Let I(n) = 2cos nx ∀n∈N, then I (1).I(n+1) - I(n) is equal to

The correct option is B I(n+2) I(n) = 2cosnx ⇒ I(1).I(n+1) - I(n) = 4cosx cos(n+1)x - 2cosnx = 2[2cos(n+1)x cosx - cosnx] = 2[cos(n+2)x + cosnx - cosnx] = 2 cos(n+2)x = I(n+2)

Let I(n) = 2cos nx $\forall n \in N$ , then I (1).I(n+1) - I(n) is equal to

The correct option is B
I(n+2)

I(n) = 2cosnx $\Rightarrow$ I(1).I(n+1) - I(n)

= 4cosx cos(n+1)x - 2cosnx

= 2[2cos(n+1)x cosx - cosnx]

= 2[cos(n+2)x + cosnx - cosnx]

= 2 cos(n+2)x = I(n+2)