Question

can anyone solve this??​

Answer 1

Step-by-step explanation:

Solution :-

Given that cos x + cos² x = 1

=> cos x = 1 - cos² x

=> cos x = sin² x

since, sin² A + cos² A = 1

Therefore, sin² x = cos x

On squaring both sides

=> ( sin² x )² = ( cos x )²

=> sin⁴ x = cos² x

=> sin⁴ x = 1-sin² x

=> sin⁴ x + sin² x = 1 ---------------(1)

Since , sin² A + cos² A = 1

The value of sin¹²x+3 sin¹⁰x+3 sin⁸x+sin⁶x

=> sin¹²x+sin¹⁰x+2 sin¹⁰x+2sin⁸x+sin⁸x +sin⁶x

=>(sin¹²x+sin¹⁰x)+(2 sin¹⁰x+2 sin⁸x)+(sin⁸x+sin⁶x )

=> sin⁸x( sin⁴x+sin²x)+2sin⁶x(sin⁴x+sin²x)+sin⁴x(sin⁴x+sin²x)

=> sin⁸x(1)+2sin⁶x(1)+sin⁴x(1)

Since ,sin⁴ x + sin² x = 1 from (1)

=> sin⁸x+2sin⁶x+sin⁴x

=> sin⁸x+sin⁶x+sin⁶x+sin⁴x

=> ( sin⁸x+sin⁶x)+(sin⁶x+sin⁴x)

=> sin⁴x(sin⁴x+sin²x)+sin²x(sin⁴x+sin²x)

=> sin⁴x(1)+sin²x(1)

Since, sin⁴ x + sin² x = 1 from (1)

=> sin⁴x + sin² x

=> 1

since , sin⁴ x + sin² x = 1 from (1)

The required value = 1

Answer :-

The value of sin¹²x+3sin¹⁰x+3sin⁸x+sin⁶x is 1

Used formulae:-

→ sin² A + cos² A = 1

Answer 2

Step-by-step explanation:

Solution:

Given,

cos x + cos²x = 1

cos x = 1 – cos²x

cos x = sin²x….(i)

Let us take the given equation again.

cos x + cos²x = 1

Cubing on both sides,

(cos x + cos²x)³ = (1)³

cos³x + $\cos^6x$ + 3(cos x)(cos²x)(cos x + cos²x) = 1

cos³x + $\cos^6x$ + 3 cos⁴x + 3 $\cos^5x$ = 1….(ii)

Substituting (i) in (ii),

(sin²x)³ + $(\sin²x)^6$ + 3(sin²x)⁴ + $3(\sin²x)^5$ = 1

$\sin^6x$ + sin¹²x + 3 $\sin^8x$ + 3 $\sin^{10}x$ = 1

sin¹²x + 3$\sin^{10}x$ + 3 $\sin8x$ +$\sin^6x$ – 1 = 0