cos3theta - sin 3theta=cos theta
+sintheta*1-2sin^2THETA
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prove :----
cos3A - sin3A = (cosA+sinA)(1-2sin2A)
Formula used :------
cos3A = 4cos³A - 3cosA
sin3A = 3sinA - 4sin³A
sin²A + cos²A = 1
2sinAcosA = sin2A
a³+b³ = (a+b)(a²+b²-ab)
Solution :-------
Putting all values in LHS and solving further we get,
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 ) =
RHS
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