# Find The Numerical Value Of (sin12x + 3sin10x + 3sin8x + sin6x – 1)

Cos x + cos^{2}x = 1, then the numerical value of (sin^{12}x + 3sin^{10}x + 3sin^{8}x + sin^{6}x – 1) is

- 0
- 1
- -1
- 2

**Answer: (1) 0
**

**Explanation:-**

Cos X + cos

^{2}x = 1

cos x = 1 – cos

^{2}x

cos x = sin

^{2}x ……………………………. (1)

Again, cos x + cos

^{2}x = 1

cube on both side,

(cos x + cos

^{2}x)

^{3}= 1

^{3 }cos

^{3}x + (cos

^{2}x)

^{3}+ 3 cos x X cos

^{2}x(cos x + cos

^{2}x) = 1

cos

^{3}x + cos

^{6}x + 3cos

^{3}x (cos x + cos

^{2}x) = 1

cos

^{3}x + cos

^{6}x + 3cos

^{4}x + 3cos

^{5}– 1 = 0

Put the value of cos x = sin

^{2}x [from equation (1)]

(sin

^{2}x)

^{3}+ (sin

^{2}x)

^{6}+ 3(sin

^{2}x)

^{4}+ 3(sin

^{2}x)

^{5}– 1 = 0

sin

^{6}x + sin

^{12}x + 3sin

^{8}x + 3sin

^{10}x – 1 = 0

sin

^{12}x + 3sin

^{10}x + 3sin

^{8}x + 3sin

^{6}x – 1 = 0

Hence, the answer is (1) 0.