Prove the trignometric identity: cos3A=cosA(2cos2A-1)

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justaguide | College Teacher | (Level 2) Distinguished Educator

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We have to show that cos 3A = (cos A)(2cos 2A - 1)

cos 3A = cos(A + 2A)

=> (cos A)(cos 2A) - (sin A)(sin 2A)

sin 2A = 2*sin A*cos A

=> (cos A)(cos 2A) - (sin A)(2*sin A*cos A)

=> (cos A)(cos 2A) - 2(sin A)^2*cos A

use cos 2A = 1 - 2(sin A)^2

or -2(sin A)^2 = cos 2A - 1

=> (cos A)(cos 2A) + cos A(cos 2A - 1)

=> (cos A)(cos 2A) + cos A*cos 2A - cos A

=> (cos A)[cos 2A + cos 2A - 1]

=> cos A(2cos 2A - 1)

This proves that cos 3A = (cos A)(2cos 2A - 1)

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