Question

pls answer this question​

Answer 1

Answer:

Step-by-step explanation:

Given relation

sin

x

+

sin

2

x

+

sin

3

x

=

1

⇒

sin

x

+

sin

3

x

=

1

−

sin

2

x

⇒

(

sin

x

+

sin

3

x

)

2

=

(

1

−

sin

2

x

)

2

⇒

sin

2

x

+

sin

6

x

+

2

sin

4

x

=

cos

4

x

⇒

1

−

cos

2

x

+

(

1

−

cos

2

x

)

3

+

2

(

1

−

cos

2

x

)

2

=

cos

4

x

⇒

1

−

cos

2

x

+

1

−

3

cos

2

x

+

3

cos

4

x

−

cos

6

x

+

2

−

4

cos

2

x

+

2

cos

4

x

=

cos

4

x

⇒

cos

6

x

−

4

cos

4

x

+

8

cos

2

x

=

4

Answer 2

Answer:

Step-by-step explanation:

Sin theta + sin^3 theta = 1- sin^2 theta

Sin theta (1+ sin^2 theta) = cos^2theta

Squaring on both sides

Sin^2 theta(1+ sin^2theta)^2 = cos^4 theta

(1- cos^2theta)(2-cos^2 theta )^2= cos^4 theta

S(1-cos^2 theta) (4 + cos^4 theta-4cos^theta)

4 + cos^4theta - 4cos^2 theta-4cos^2 theta - cos^6 theta + 4 cos^4theta =cos^4 theta

4-8cos^2 theta - cos^6 theta + cos^4 theta =0

8cos^2 theta + cos^6 theta - 4cos ^4 theta = 4

Hence, proved