Question

$sin x + sin ^{2} x = 1 \\ prove \: that \: cos ^{12} x + 3cos ^{10}x + 3cos ^{8} + cos ^{6} x = 1$
​

Answer 1

Answer:

Given: sinx+sin

2

x=1

sinx=1−sin

2

x=cos

2

x

Now,

cos

12

x+3cos

10

x+3cos

8

x+cos

6

x

⇒(cos

4

x)

3

+3(cos

4

x)

2

cos

2

x+3cos

4

x(cos

2

x)

2

+(cos

2

x)

3

⇒(sin

2

x)

3

+3sin

4

xcos

2

x+3sin

2

xcos

4

x+(cos

2

x)

3

⇒[sin

2

x+cos

2

x]

3

⇒1

3

⇒1.