If A - r -0 n n C r cos n - r x -sin rx - B -2 n sin nx and C -1- 0 - n -1-1- 3 -n-3-1- 5 - n -5-1-n-1 - 1- thenA- A - t -2B- A -C- A -33 In -2 a -1 BCD- A - BC

The correct options are A A=B/2 B A=! nCB/2^nC=1/! n (1/! 0 ! n 1+1/! 3 ! n 3+1/! 5 ! n 5+⋯ +1/! n 1 ! 1 ) =1/! n(^nC_1+^nC_3+⋯+^nC_n 1 ) C=2^n 2/n ! B=2^n s ...

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Byju's Answer

Standard XII

Mathematics

Basic Inverse Trigonometric Functions

If A =∑ r =0 ...

Question

If $A = \sum_{r = 0}^{n} (^{n} C_{r} c o s (n - r) x . s i n r x), B = 2^{n} s i n n x$ and $C = \frac{1}{! 0! n - 1} + \frac{1}{! 3! n - 3} + \frac{1}{! 5! n - 5} + \dots + \frac{1}{! n - 1! 1}$ , then

A = \frac{B}{2}

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A = \frac{! n C B}{2^{n}}

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A = \frac{3! n}{2^{n - 1}} B C

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A = B C

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Solution

The correct options are
A $A = \frac{B}{2}$
B $A = \frac{! n C B}{2^{n}}$
$C = \frac{1}{! n} (\frac{1}{! 0! n - 1} + \frac{1}{! 3! n - 3} + \frac{1}{! 5! n - 5} + \dots + \frac{1}{! n - 1! 1}) = \frac{1}{! n} (^{n} C_{1} +^{n} C_{3} + \dots +^{n} C_{n - 1}) C = \frac{2^{n} - 2}{n!} B = 2^{n} s i n n x A =^{n} C_{1} c o s (n - 1) x s i n x +^{n} C_{2} c o s (n - 2) x s i n (2 x) + \dots +^{n} C_{n - 1} c o s x s i n (n - 1) x +^{n} C_{n} s i n (n x) . A =^{n} C_{n} s i n (n x) +^{n} C_{n - 1} c o s x s i n (n - 1) x +^{n} C_{n - 2} c o s (n - 2) x s i n (2 x) + \dots +^{n} C_{1} c o s (n - 1) x s i n x 2 A = 2^{n} C_{n} s i n (n x) +^{n} C_{1} (c o s (n - 1) x s i n x + c o s x s i n (n - 1) x) +^{n} C_{2} (c o s (n - 2) x s i n (2 x) + c o s (2 x) s i n (n - 2) x) + \dots +^{n} C_{n - 1} (c o s x s i n (n - 1) x + c o s (n - 1) x s i n x) = 2 s i n (n x) + (s i n (n x)) (^{n} C_{1} +^{n} C_{1} + \dots +^{n} C_{n - 1}) 2 A = 2 s i n (n x) + (s i n (n x)) (2^{n} - 2) 2 A = 2^{n} s i n (n x) A = 2^{n - 1} s i n (n x) H e n c e A = \frac{B}{2} a n d A = \frac{! n}{2^{n}} 2^{n} s i n (n x) \frac{2^{n - 1}}{! n} \Rightarrow A = \frac{! n C B}{2^{n}}$

Suggest Corrections

If A=∑nr=0(nCrcos(n−r)x.sinrx),B=2n sin nx and C=1!0!n−1+1!3!n−3+1!5!n−5+⋯+1!n−1!1, then

If $A = \sum_{r = 0}^{n} (^{n} C_{r} c o s (n - r) x . s i n r x), B = 2^{n} s i n n x$ and $C = \frac{1}{! 0! n - 1} + \frac{1}{! 3! n - 3} + \frac{1}{! 5! n - 5} + \dots + \frac{1}{! n - 1! 1}$ , then