show that cos3A-sin3A=(cosA+sinA)(1-2sin2A
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Step-by-step explanation:
by LHS
cos3A - sin3A
= 4cos³A - 3cosA - ( 3sinA - 4sin³A )
= 4cos³A - 3cosA - 3sinA + 4sin³A
= 4( cos³ A + sin³A ) - 3( sinA + cosA )
= 4[(cosA+sinA)(cos²A+sin² A-
cosAsinA )] - 3 ( cosA + sinA )
= 4[ ( cosA + sinA )(1 - cosAsinA ) ]
- 3 ( cosA + sinA )
= ( CosA + sinA ) ( 4 - 4sinAcosA - 3 )
= ( CosA + sinA ) ( 1 - 2 × 2sinAcosA )
= ( cosA + sinA ) ( 1 - sin2A )
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