Question

express cos 5A in terms of Cos A

Answer 1

Answer:

From D Moivre's Theorem,

cos(5a) + i*sin(5a) = [cos(a) + i*sin(a)]^5

cos(5a) + i*sin(5a) = cos^5(a) + 5*cos^4(a)*i*sin(a) + 10*cos^3(a)*i^2*sin^2(a) + 10*cos^2(a)*i^3*sin^3(a) + 5*cos(a)*i^4*sin^4(a) + i^5*sin^5(a)

cos(5a) + i*sin(5a) = [cos^5(a) - 10*cos^3(a)*sin^2(a) + 5*cos(a)*sin^4(a)] + i[5*cos^4(a)*sin(a) - 10*cos^2(a)*sin^3(a) + sin^5(a)]

Answer 2

Cos5A = 4cos³A( 1 - 2sin²A ) - cosA( 3 -  8sin⁴A)
GIVEN: cos 5A
TO FIND: Representing cos5A in terms of cosA
SOLUTION:

As we are given in the question,

cos 5A

The following identities shall be used to expand the given expressions:

cos(A+B) = cosAcosB -sinAsinB

sin2A = 2sinAcosA

cos2A = 1-2sin²A

sin3A = 3sinA -4sin³A

cos3A = 4cos³A -3cosA

As we are given with,
cos 5A

= cos(3A+2A)

= cos3Acos2A -sin3Asin2A

= (4cos³A -3cosA)(1-2sin²A) - (3sinA -4sin³A)(2sinAcosA)

=[4cos³A - 8cos³Asin²A - 3cosA + 6cosAsin²A]-[6sin²AcosA - 8sin⁴AcosA]

= 4cos³A - 8cos³Asin²A - 3cosA + 6cosAsin²A - 6sin²AcosA + 8sin⁴AcosA

= 4cos³A - 8cos³Asin²A - 3cosA + 8sin⁴AcosA

= 4cos³A( 1 - 2sin²A ) - cosA( 3 -  8sin⁴A)

Therefore,

Cos5A = 4cos³A( 1 - 2sin²A ) - cosA( 3 -  8sin⁴A)

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