Let n be a natural number- If sin n-r-0n br sinr- for every value of - then

The correct option is C b_0=0, b_1=1 sin( nθ #160;) =∑ _ r=1 ^ n b _ r sin ^ r θ #160; For n=0 0= b _ 0 sin^ 0 θ⇒ b _ 0 =0 For n ...

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Standard XII

Physics

Vector Component

Let n be a ...

Question

Let $n$ be a natural number. If $sin n θ = n \sum r = 0 b_{r} {sin}^{r} θ$ , for every value of $θ$ , then

b_{0} = 1, b_{1} = 3

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b_{0} = 0, b_{1} = 1

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b_{0} = - 1, b_{1} = n

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b_{0} = 0, b_{1} = n^{2} - 3 n + 3

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Solution

The correct option is C $b_{0} = 0, b_{1} = 1$
$s i n (n θ) = \sum_{r = 1}^{n} b_{r} {s i n}^{r} θ$

For $n = 0$
$0 = b_{0} {sin}^{0} θ \Rightarrow b_{0} = 0$
For $n = 1$
$sin θ = b_{0} {s i n}^{0} θ + b_{1} {sin}^{1} θ$

$\Rightarrow sin θ = 0 + b_{1} sin θ$

$\Rightarrow b_{1} = 1$

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Write a general polynomial $q (z)$ of degree n with coefficients that are
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Q. If

(1 + 2 x + 3 x^{2})^{10} = b_{0} + b_{1} x + b_{2} x^{2} . . . . + b_{20} x^{20}

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Q. Assertion :

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and

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Let n be a natural number. If sinnθ=n∑r=0brsinrθ, for every value of θ, then

The correct option is C b0=0,b1=1sin(nθ)=∑nr=1br sinrθFor n=00=b0sin0θ⇒b0=0For n=1sinθ=b0 sin0θ+b1sin1θ⇒sinθ=0+b1sinθ⇒b1=1

Let $n$ be a natural number. If $sin n θ = n \sum r = 0 b_{r} {sin}^{r} θ$ , for every value of $θ$ , then

The correct option is C $b_{0} = 0, b_{1} = 1$
$s i n (n θ) = \sum_{r = 1}^{n} b_{r} {s i n}^{r} θ$

For $n = 0$
$0 = b_{0} {sin}^{0} θ \Rightarrow b_{0} = 0$
For $n = 1$
$sin θ = b_{0} {s i n}^{0} θ + b_{1} {sin}^{1} θ$

$\Rightarrow sin θ = 0 + b_{1} sin θ$

$\Rightarrow b_{1} = 1$