Question

prove

Cos^6 A+ Sin^6 A=1-3Cos²A×Sin²A​

Answer 1

hope you understand

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Answer 2

Answer:

Step-by-step explanation:

LHS = cos^6 A + sin^6 A

= (cos²A)^3 + (sin²A)^3

Using identity (a + b)^3 = a^3 +b^3 + 3ab ( a+b)

=> a^3 +b^3 = ( a+ b)^3 -3ab ( a+ b)

So, (cos²A)^3 + ( sin² A)^3

=(cos²A+ sin²A)^3 - 3 cos²A sin²A(cos²A+sin² A)

= 1- 3cos² A sin² A ( since sin² A + cos² A = 1 ) ……………….. LHS

Now, RHS = { 1 + 3cos² ( 2A)} / 4

= { 1 + 3 ( cos 2A )² } /4

= {1+ 3 ( cos² A - sin² A )² } /4 { since cos 2A = cos² A - sin² A)

= { 1+ 3( cosA + sinA)² ( cosA - sinA)² }/4

= {1+ 3( 1 + 2sinA cosA) ( 1 - 2sinA cosA)} /4

= { 1 + 3 ( 1 - 4 sin² A cos² A) } / 4

= { 1 + 3 - 12 sin² A cos² A} /4

= (4 - 12 sin² A cos² A) /4

=> 1 - 3 sin²A cos ² A ………….. RHS

Hence proved that LHS = RHS

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186 Views · · Answer requested by

LHS=(cosA)^6+(sins)^6=(cos²A)³+(sin²A)³={(cos²A+sin²A)³-3(cos²A.sin²A)(cos²A+sin²A)

=1–3sin²A.cos²A=1–3(sinA.cosA)²

=1–3{(2sinA.cosA)/2}²=1–3{(sin2A/2)²}

=1–(3/4)sin²2A=1-(3/4){(1-cos²2A)}

=1-(3/4)+(3/4)cos²2A=(1/4)+(3/4)cos²2A

={(1+3cos²2A)/4}=RHS proved.