Sin n- - r-0 n b r -sin r - for every value of - then

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

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Standard XII

Mathematics

Functions

sin nθ =∑ r=0...

Question

$sin n θ = n \sum r = 0 b_{r} . {sin}^{r} θ$ 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 $b_{0} = 0, b_{1} = 1$
Given, $sin (n θ) = n \sum r = 1 b_{r} {sin}^{r} θ$
For $n = 0$
$0 = b_{0} {sin}^{0} θ \Rightarrow b_{0} = 0$
For $n = 1$
$sin θ = b_{0} {sin}^{0} θ + b_{1} {sin}^{1} θ \Rightarrow sin θ = 0 + b_{1} sin θ \Rightarrow b_{1} = 1$

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Similar questions

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Q. Let

f (x) = \frac{a_{2 k} x^{2 k} + a_{2 k - 1} x^{2 k - 1} + . . . . + a_{1} x + a_{0}}{b_{2 k} x^{2 k} + b_{2 k - 1} x^{2 k - 1} + . . . . + b_{1} x + b_{0}}

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sinnθ=n∑r=0br.sinrθ for every value of θ, then

The correct option is C b0=0,b1=1Given, sin(nθ)=n∑r=1brsinrθFor n=00=b0sin0θ⇒b0=0For n=1sinθ=b0sin0θ+b1sin1θ⇒sinθ=0+b1sinθ⇒b1=1

$sin n θ = n \sum r = 0 b_{r} . {sin}^{r} θ$ for every value of $θ$ , then

The correct option is C $b_{0} = 0, b_{1} = 1$
Given, $sin (n θ) = n \sum r = 1 b_{r} {sin}^{r} θ$
For $n = 0$
$0 = b_{0} {sin}^{0} θ \Rightarrow b_{0} = 0$
For $n = 1$
$sin θ = b_{0} {sin}^{0} θ + b_{1} {sin}^{1} θ \Rightarrow sin θ = 0 + b_{1} sin θ \Rightarrow b_{1} = 1$