Question

cosx.cos(x/2)-cos3x.cos(9x/2)=sin7x.sin8x(prove)

Answer 1

HELLO DEAR,

we know:-

• cosa.cosb = 1/2[cos(a + b) + cos(a - b)]
• cos(-∅) = cos∅
• cosa - cosb = 2sin(a + b)/2 * sin(b - a)/2

now,

L.H.S:- cosx.cos(x/2) - cos3x.cos(9x/2)

≈» 1/2[{cos(3x/2) + cos(x/2)} - {cos(15x/2) + cos(3x/2)}]

≈» 1/2[cos(3x/2) + cos(x/2) - cos(15x/2) - cos(3x/2)]

≈» 1/2[cos(x/2) - cos(15x/2)]

≈> 1/2*2[sin{(x/2) + (15x/2)}/2 * sin{(15x/2) - (x/2)}/2]

≈» [sin(8x/2) * sin(7x/2)]

≈» sin4x * sin7x/2

I think some misprints in questions may be but according to solution we can say that sin4x.sin7x/2 is the correct answer

I HOPE IT'S HELP YOU DEAR,

THANKS

Answer 2

Step-by-step explanation:

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