Trigonometry> If sinx + sin^2 x = 1 then value of cos^1...

If sinx + sin^2 x = 1 then value of cos^12 x + 3 cos^10 x + 3 cos^8 x + cos^6 x + 2 cos^4 x + cos^2 x - 2 is equal to

Kuldeepak , 7 Years ago

Grade 11

2 Answers

Shashank Singh

Last Activity: 7 Years ago

sinx= 1-sin^2 x= cos^2 x
=sin^6 x +3 sin^5 x + 3sin^4 x + sin^3 x + 2 sin^2 x + sinx – 2
=sin^4x( sin^2 x+ sinx) + 2sin^3 x (sin^2 x+ sinx) + sin^2 x( sin^2x + sinx)+ 2(sin^2 x + sinx) -sinx -2
as we know that sin^2 x+ sinx= 1
=sin^4x + sin^3x + sin^3 x + sin^2 x +2 – sinx- 2
=sin^2 x(sin^2 x+ sinx) + sinx(sin^2x+ sinx) -sinx
=sin^2 x+ sinx-sinx
=sin^2 x [ans]

Sripad Sambrani

Last Activity: 5 Years ago

Given : sinx+sin^2x = 1

RTF : cos^12x+3cos^10x+3cos^8x+cos^6x-2

Solution :

sinx=1-sin^2x [given] => sinx=cos^2x

Hence, cos^12x=sin^6x; cos^10x=sin^5x; cos^8x=sin^4x; cos^6x=sin^3x

Substitute the above values in required expression

sin^6x+3sin^5x+3sin^4x+sin^3x-2

= (sin^2x)^3+3*(sin^2x)^2*(sinx)+3*(sin^2x)*(sinx)^2+(sinx)^3-2

= (sin^2x+sin^x)^3-2 [since, a^3+3*(a^2)*(b)+3*(a)*(b^2)+b^3 = (a+b)^3]

= (1)^3-2 [given: sin^2x+sin^x=1]

= 1-2 = -1