Question

Evaluate sin18°. show the working​

Answer 1

Step-by-step explanation:

what are you write I am not understand

Answer 2

Answer:

Let θ = 18°….(i)

We can write 90° as:

90° = 36° + 54°

90 = 2 × 18° + 3 × 18°

Substituting (i) in the above equation,

90° = 2θ + 3θ

3θ = 90° – 2θ

Taking “sin” on both sides,

sin 3θ = sin(90° – 2θ)

Using the identity sin 3A = 3 sin A – 4 sin3A,

3 sin θ – 4 sin3θ = cos 2θ

Now, using the identity cos 2θ = 1 – 2 sin2θ,

3 sin θ – 4 sin3θ = 1 – 2 sin2θ = 0

4 sin3θ – 2 sin2θ – 3 sin θ + 1 = 0

Let us assume sin θ = x.

Thus,

4x3 – 2x2 – 3x + 1 = 0

Using factor method, we can write the above equation as:

(x – 1)(4x2 + 2x – 1) = 0

x – 1 = 0, 4x2 + 2x – 1 = 0

x = 1, 4x2 + 2x – 1 = 0

We know that -1 ≤ sin ≤ 1, where sine is maximum at 90 degrees.

Thus, sin 18 value will be less than 1.

So, x = 1 cannot be accepted.

Now,

4x2 + 2x – 1 = 0

Using quadratic formula,

x = [-2 ± √{4 – 4(-4)}]/2(4)

= [-2 ± √(4 + 16)]/8

= [-2 ± √20]/8

= [-2 ± 2√5]/8

= 2[-1 ± √5]/8

= (-1 ± √5)/4

The value of sin θ is positive when θ lies between 0 and 90 degrees.

Therefore, x = sin θ = sin 18° = (-1 + √5)/4

Or

sin 18° = (√5 – 1)/4

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