Question

32) Prove that cos2x-cos6x=sin4x sin8x
33) Express the complex number z= 13-i in polar form.​

Answer 1

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Step-by-step explanation:

32.

LHS =cos²(2x) - cos²(6x)

Use the formula,

a² - b² = (a - b)(a + b)

= (cos2x - cos6x)(cos2x + cos6x)

Use the formula

cosC + cosD = 2cos(C+D)/2.cos(C-D)/2

cosC - cosD = 2sin(C+D)/2.sin(D-C)/2

= {2sin(2x + 6x)/2.sin(6x-2x)/2}{2cos(2x+6x)/2.cos(6x -2x)/2}

={2sin2x.sin4x}{2cos4x.cos2x}

={2sin2x.cos2x}{2sin4x.cos4x}

Now,

Use the formula,

sin2A = 2sinA.cosA

= sin2(2x).sin2(4x)

= sin4x.sin8x = RHS