Question

༒ Bʀᴀɪɴʟʏ Sᴛᴀʀs

༒Mᴏᴅᴇʀᴀᴛᴇʀs
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Proof that
$\cos^{2} \: 2x - \cos {}^{2} \: 6x = \sin \: 4x \: sin \: 8x$
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Answer 1

Step-by-step explanation:

LHS = cos²(2x) - cos²(6x)

= (cos2x - cos6x)(cos2x + cos6x)

[as we know from algebraic identity a² - b² = (a - b)(a + b)]

now, use formula, cosC - cosD = 2sin(C + D)/2. sin(D-C)/2 cosC + cosD = 2cos(C + D)/2.cos(C-D)/2.

cos2x - cos6x = 2sin(2x + 6x)/2.sin(6x - 2x)/2 = 2sin4x.sin2x

cos2x+cos6x = 2cos(2x + 6x)/2.cos(2x - 6x)/2.

= 2cos4x.cos(-2x)

now,

(cos2x - cos6x)(cos2x + cos6x) = (2sin4x.sin2x)(2cos4x.cos2x)

= (2sin4x.cos4x)(2sin2x.cos2x)

use formula, 2sinA.cosA = sin2A.

= 2cos4x.cos2x

sin8x.sin4x = RHS

hence proved.

Answer 2

Step-by-step explanation:

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