Math, asked by annuchauchan1024, 1 day ago

1


17. If 2 sin 0 = x +1/x
prove that sin 3 theta + 1/2( x^3 + x^3) = 0




Answers

Answered by stualvinfelix0063
0

Step-by-step explanation:

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Answered by nathan008
0

Answer:

✪ᴀɴsᴡᴇʀ✪

☆ɢɪᴠᴇɴ☆

2sin(\theta) = x + \frac{1}{x} 2sin(θ)=x+

x

1

☆ᴛᴏ ᴘʀᴏᴠᴇ☆

sin(3\theta) + \frac{1}{2}(x^3 + \frac{1}{x^3}) = 0sin(3θ)+

2

1

(x

3

+

x

3

1

)=0

☆ᴘʀᴏᴏғ☆

\:\:\:\:sin(3\theta)sin(3θ)

= 3sin(\theta) -4sin^3(\theta) =3sin(θ)−4sin

3

(θ)

= \frac{3}{2}(x + \frac{1}{x}) -4[\frac{1}{2}(x + \frac{1}{x})]^3=

2

3

(x+

x

1

)−4[

2

1

(x+

x

1

)]

3

= \frac{3}{2}(x + \frac{1}{x}) -\frac{4}{8}(x + \frac{1}{x})^3=

2

3

(x+

x

1

)−

8

4

(x+

x

1

)

3

= \frac{3}{2}(x + \frac{1}{x})=

2

3

(x+

x

1

)

\:\:\:\: -\frac{1}{2}[x^3 + \frac{1}{x^3} + 3x.\frac{1}{x}(x + \frac{1}{x})]−

2

1

[x

3

+

x

3

1

+3x.

x

1

(x+

x

1

)]

= \frac{3}{2}(x + \frac{1}{x})-\frac{1}{2}(x^3 + \frac{1}{x^3}) -\frac{3}{2}(x + \frac{1}{x})=

2

3

(x+

x

1

)−

2

1

(x

3

+

x

3

1

)−

2

3

(x+

x

1

)

\implies sin(3\theta) = -\frac{1}{2}(x^3 + \frac{1}{x^3})⟹sin(3θ)=−

2

1

(x

3

+

x

3

1

)

\implies sin(3\theta) + \frac{1}{2}(x^3 + \frac{1}{x^3}) = 0⟹sin(3θ)+

2

1

(x

3

+

x

3

1

)=0

Hence Proved

Step-by-step explanation:

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