Question

if sinx+sin^2x=1 then find cos^12x + 3cos^10x + 3cos^8x + cos^6x -1

Answer 1

Explainations

1. Find value of cos²x, we get sinx.

2. Remember formula (a + b)³?

→ (a + b)³ = a³ + b³ + 3a²b + 3ab².

3. Bring equation as equal to above identity.

4. Bring that into formula.

5. Substitute values accordingly.

6. Since sinx + sin²x = 1.

7. Hence whole cube of 1 is 1.

8. By doing 1 - 1 we get answer 0.

Hence required answer is 0.

Answer 2

Answer:

sinx + sin^2x =1

sinx =1 - sin^2x =cos^2x

now,

cos^12x+3cos^10x +3cos^8x +cos^6x-1

=> (cos^4x)^3+3. (cos^4x)^2.cos^2x +3cos^4x.(cos^2x)^2+(cos^2x)^3 - 1

=>(sin^2x)^3+3sin^4x.cos^2x+3sin^2x.cos^4x +(cos^2x)^3 -1

=>{sin^2x+ cos^2x}^3 = 1^3 =1-1

=> 0