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(d^(20)y)/(dx^(20))(2cosxcos3x)is equal ...

`(d^(20)y)/(dx^(20))(2cosxcos3x)is equal to ` `2^(20)(cos2x-2^(20)os3x)` `2^(20)(cos2x+2^(20)cos4x)` `2^(20)(sin2x+2^(20)sin4x)` `2^(20)(sin2x-2^(20)sin4x)`

A

`2^(20)( cos 2x -2^(20) cos 3x)`

B

`2^(20)( cos 2x +2^(20) cos 4x)`

C

`2^(20)( sin 2x +2^(20) sin 4x)`

D

`2^(20)( sin 2x - 2^(20) sin 4x)`

Text Solution

AI Generated Solution

To solve the problem \(\frac{d^{20}}{dx^{20}}(2\cos x \cos 3x)\), we will use the trigonometric identity for the product of cosines. ### Step-by-Step Solution: 1. **Apply the Trigonometric Identity**: We start with the identity: \[ 2 \cos a \cos b = \cos(a + b) + \cos(a - b) ...
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(d^(20)y)/(dx^(20))(2cosxcos3x)i se q u a l to a)2^(20)(cos2x-2^(20)os3x) b)2^(20)(cos2x+2^(20)cos4x) c)2^(20)(sin2x+2^(20)sin4x) d)2^(20)(sin2x-2^(20)sin4x)

(d^(20))/(dx^(20))(2cosxcos3x) is equal to (a) 2^(20)(cos2x-2^(20)cos3x) (b) 2^(20)(cos2x+2^(20)cos4x) (c) 2^(20)(sin2x+2^(20)sin4x) (d) 2^(20)(sin2x-2^(20)sin4x)

(d^(20))/(dx^(20))(2cosxcos3x) is equal to (a) 2^(20)(cos2x-2^(20)cos3x) (b) 2^(20)(cos2x+2^(20)cos4x) (c) 2^(20)(sin2x+2^(20)sin4x) (d) 2^(20)(sin2x-2^(20)sin4x)

(d^(20))/(dx^(20))(2cosxcos3x)= 2^(20)(cos2x-2^(20)cos4x) (b) 2^(20)(cos2x+2^(20)cos4x) (c) 2^(20)(sin2x+2^(20)sin4x) (d) 2^(20)(sin2x-2^(20)sin4x)

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