Math, asked by madhavanrajagopalan0, 10 months ago

cos3theta - sin 3theta=cos theta
+sintheta*1-2sin^2THETA

Answers

Answered by SyedNomanShah
0

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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