Question

15a=π prove that cosacos2acos3acos4acos5acos6acos7a =1/2^7​

Answer 1

Answer:

Step-by-step explanation:

Given, 15A = π

CosACos2ACos3ACos4ACos5ACos6ACos7A

//Multiply and divide by 2SinA

=> 1/2SinA[(2SinACosA)Cos2ACos3ACos4ACos5ACos6ACos7A]

=> 1/2SinA[Sin2ACos2ACos3ACos4ACos5ACos6ACos7A]

//multiply and divide by 2

=> 1/2²SinA [ (2Sin2ACos2A)Cos3ACos4ACos5ACos6ACos7A]

=> 1/2²SinA [Sin4ACos3ACos4ACos5ACos6ACos7A]

//multiply and divide by 2

=> 1/2³SinA[(2Sin4ACos4A)Cos3ACos5ACos6ACos7A]

=>1/2³SinA[Sin8ACos3ACos5ACos6ACos7A]

We know that 15A = π; 8A = π - 7A

=> Sin8A = Sin(π - 7A) = Sin7A; (∵ SIn(180 - θ) = Sinθ)

Substituting above value in place of Sin8A  

=>1/2³SinA[Sin7ACos3ACos5ACos6ACos7A]

//multiply and divide by 2.

=> 1/2⁴SinA[(2Sin7ACos7A)Cos3ACos5ACos6A]

=> 1/2⁴SinA[Sin14ACos3ACos5ACos6A]

We know that 15A = π; 14A = π - A

=> Sin14A = Sin(π - A) = SinA; (∵ SIn(180 - θ) = Sinθ)

Substituting above value in place of Sin14A

=> 1/2⁴SinA[SinACos3ACos5ACos6A]

=>1/2⁴(Cos3ACos5ACos6A)

//Multiply and divide by 2Sin3A

=> 1/2⁵Sin3A[ (2Sin3ACos3A)Cos5ACos6A]

=> 1/2⁵Sin3A [Sin6ACos5ACos6A]

//multiply and divide by 2.

=> 1/2⁶Sin3A[(2Sin6ACos6A)Cos5A]

=>1/2⁶Sin3A (Sin12ACos5A)

We know that 15A = π; 12A = π - 3A

=> Sin12A = Sin(π - 3A) = Sin3A; (∵ SIn(180 - θ) = Sinθ)

Substituting above value in place of Sin12A

=> 1/2⁶Sin3A(Sin3ACos5A)

=> 1/2⁶ Cos5A

=> 1/2⁶ * Cos (5*π/15)  (∵ 15A = π => A = π/15)

=> 1/2⁶ * Cosπ/3

=> 1/2⁶ * 1/2

=> 1/2⁷

= R.H.S

Hence proved.

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