Question

p=cosπ/20× cos 3π/20× cos7π/20× cos9π/20
q=cosπ/11× cos2π/11× cos4π/11× cos8π/11× cos16π/11
Then Find p/q.

Answer 1

||✪✪ QUESTION ✪✪||

p=cosπ/20× cos 3π/20× cos7π/20× cos9π/20

q=cosπ/11× cos2π/11× cos4π/11× cos8π/11× cos16π/11

Then Find p/q.

|| ✰✰ ANSWER ✰✰ ||

Solving p First :-

→ p = cosπ/20 × cos 3π/20 × cos7π/20 × cos9π/20

→ p = cosπ/20 × cos 3π/20 × cos(π/2 - 3π/20) × cos(π/2 - π/20)

Using cos(π/2 - A) = SinA now,

→ p = cosπ/20 × cos 3π/20 × sin3π/20 × sinπ/20

→ p = cosπ/20 × sinπ/20 × cos 3π/20 × sin3π/20

Multiply and Divide by 4 now,

→ p = 1/4[4×cosπ/20 × sinπ/20 × cos 3π/20 × sin3π/20]

→ p = 1/4[(2×cosπ/20 × sinπ/20) × (2×cos 3π/20 × sin3π/20)]

Using 2sinA×cosA = sin2A now,

→ p = 1/4[ (sin2π/20) × (sin6π/20) ]

→ p = 1/4 × sinπ/10 × sin3π/10

→ p = 1/4 × sin(180°/10°) × sin(3×180°/10°)

→ p = 1/4 × sin18° × sin54°

Now, putting sin18° = (√5-1)/4 and sin54° = (√5+1)/4

→ p = 1/4 × (√5 - 1)/4 × (√5 + 1)/4

→ p = (5-1)/(4×4×4)

→ p = 1/16 ------------------------ Equation (1)

____________________________

Now, Solving Value of q :-

→ q = cosπ/11× cos2π/11× cos4π/11× cos8π/11× cos16π/11

using cosA × cos2A × cos4A × ______ cosnA = (1/2^n)[sin2^nA/sinA] we get,

Here n = 5 terms.

So,

→ q = (1/2^5)[sin2^5(π/11)/sin(π/11)]

→ q = (1/32) [ sin(32π/11) /sin(π/11) ]

→ q = (1/32) [ sin(3π - π/11) / sin(π/11) ]

∵ sin(π - A) = sinA

→ q = (1/32) [ sin(π/11) / sin(π/11) ]

→ q = 1/32 -------------------------- Equation (2)

___________________________

Putting both Value from both Equations we get,

→ p/q = (1/16) / (1/32)

→ p/q = (1/16) × (32/1)

→ p/q = 2 (Ans).

Hence, value of p/q will be 2.

Answer 2

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