Question

cos7x+cos5x+cos3x+cosx=4cosx.cos2xcos 4x


✏✏✏Ans these que plz



❎❎No spam❎❎

Answer 1

Rewriting the Left hand side as:

LHS: cos7x + cosx + cos5x + cos3x

Apply the formula : cos a + cos b = 2{cos(a+b)/2}{cos(a-b)/2} on first two terms and other two terms.

LHS: 2{cos(8x)/2}{cos(6x)/2} + 2{cos(8x)/2}{cos(2x)/2}

LHS: 2cos 4x cos 3x + 2 cos 4x cos x

Taking out 2 cos4x common

LHS: 2cos 4x (cos 3x + cos x)

Now again apply the same formula

LHS: 2cos 4x ( 2{cos (4x)/2}{cos(2x)/2})

LHS: 2cos 4x (2cos 2x cos x)

LHS: 4cos 4x cos 2x cos x = RHS

HENCE PROVED

Answer 2

Hello!!

Rewriting the Left hand side as:

LHS: cos7x + cosx + cos5x + cos3x

Apply the formula : cos a + cos b = 2{cos(a+b)/2}{cos(a-b)/2}

On first two terms and other two terms.

LHS: 2{cos(8x)/2}{cos(6x)/2} + 2{cos(8x)/2}{cos(2x)/2}

LHS: 2cos 4x cos 3x + 2 cos 4x cos x

Taking out 2 cos4x common

LHS: 2cos 4x (cos 3x + cos x)

Now again apply the same formula

LHS: 2cos 4x ( 2{cos (4x)/2}{cos(2x)/2})

LHS: 2cos 4x (2cos 2x cos x)

LHS: 4cos 4x cos 2x cos x = RHS

HENCE PROVED