Question

cos⁶A-sin⁶A= 1-3/4 sin²2A​

Answer 1

Answer:

sin^6A = (sin^2A)^3

cos^6A = (cos^2A)^3

We know that;

x^3+y^3 = (x+y)(x^2-xy+y^2)

sin^6A+cos^6A = (sin^2A+cos^A)(sin^4A-sin^2Acos^2A+cos^4A)

We can write;

x^2+y^2 = (x+y)^2-2xy

sin^4A+cos^4A = (sin^2A+cos^2A)^2-2sin^2Acos^2A

We know that sin^2A+cos^A = 1

sin^6A+cos^6A

= (sin^2A+cos^A)(sin^4A-sin^2Acos^2A+cos^4A)

= 1xx((sin^2A+cos^2A)^2-2sin^2Acos^2A-sin^2Acos^2A)

= 1xx(1-3sin^2Acos^2A)

sin2A = 2sinAcosA

(sin2A)^2 = 4sin^2Acos^2A

3/4sin^2(2A) = 3sin^2Acos^2A

So this gives the answer as;

sin^6A+cos^6A

=(1-3sin^2Acos^2A)

= 1-3/4sin^2(2A)

Answer 2

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