Question

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PROVE :✨

sin 18° = (√5 - 1)/4



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Answer 1

Heya!!

so,we have to find the value of sin18° .

Let, y = 18° .

Now,

5y = 18 × 5° = 90°

We know,

sin90° = 1 (rule)

Sin5y =1.

now, 2y + 3y = 90°

=> 2y = 90° - 3y

Taking sin on both sides

sin2y = sin(90-3y)

=> sin2y = cos3y. [ sin(90-y)= cosy]. ...........(I)

Now,

sin2y = 2siny cos y. (rule)

cos3y = 4cos³y - 3 cosy. (rule)

so,equation (I) becomes

2 siny cosy = 4cos³y - 3 cosy

4cos³y - 3 cosy - 2 siny cosy = 0

=>cos y [4cos²y - 3 - 2 siny ] = 0

=>4cos²y - 3 - 2 siny = 0

=>4(1 - sin²y] - 2 siny - 3 = 0

=>4 - 4sin²y - 2 siny - 3 = 0

=>4 sin²y + 2 siny - 1 = 0

by solving the equation by sreedharacharya's rule,we get

siny = √5 - 1 /4 Ans.

Regards

@EVERkshitijEST

Answer 2

Heya brainliac!

Thnx for asking the question...

♠ TRIGONOMETRIC INDUCTIONS ♠


>>> Let x = 18°

5x = 90°

(3x + 2x) = 90°

2x = 90° - 3x

Applying sin both sides,

sin 2x = sin ( 90° - 3x)

sin 2x = cos 3x

2 sin x cos x = 4 cos³x - 3 cos x

2 sin x cos x - 4 cos³x + 3 cos x = 0

cos x ( 2 sin x - 4 cos²x + 3) = 0

2 sin x - 4 cos²x + 3 = 0

2 sin x - 4(1 - sin² x) + 3 = 0

2 sin x - 4 + 4 sin² x + 3 = 0

4 sin² x + 2 sin x - 1 = 0

Applying quadratic formula,

sin x = -2 ±√(2)² - 4 × 4 × (-1)/ 2×4

sin x = - 2 ± √20 /8

sin x = -2 ± 2√5 /8

sin x = -1 ± √5 /4

sin x = √5 -1/ 4 ( Since x is an acute angle)

Therefore, sin 18° = √5 - 1 /4

Hope this would help you
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