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14 and 15 no pls Multiple angle. .........................................................

14 and 15 no pls
Multiple angle.                                          .........................................................

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Grade:11

1 Answers

Aditya Gupta
2081 Points
5 years ago
 
32 sin^6x = 4.(2sin^2 x)^3 = 4(1- cos2x)^3 = 4(1-cos^3 2x -3cos 2x +3cos^2 2x) = 4- 12cos 2x + 6(1+cos 4x) – (cos 6x+ 3cos2x) = 10 – 15cos 2x +6cos 4x -cos6x
then,
10- 15cos 2x +6cos 4x – cos 6x =10 – 15cos2x + b cos4x+a cos6x
or (6-b)cos4x = (1+a)cos6x
cos4x/cos6x = (1+a)/(6-b)
componendo ad dividendo
or (cos4x – cos 6x)/(cos 4x + cos6x) = (1+a-6+b)/(1+a+6-b)
or tan5x tanx =(a+b-5)/(7+a-b)

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