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# Find the value of sin 15 using sin 30.

We can find the value of sin 15 using the value of sin 30.

## Answer: The value of sin 15 using sin 30 is (√3−1)/2√2

Let see, how we can find the value of sin 15° using sin 30°.

**Explanation:**

(sin P/2 + cos P/2)^{2} = sin^{2 }P/2 + cos^{2 }P/2 + 2 sin P/2 cos P/2 = 1 + sin P [Using sin 2A = 2 sin A cos A]

sin P/2 + cos P/2 = ± √ (1 + sin P)

(sin P/2 - cos P/2)^{2} = sin^{2 }P/2 + cos^{2 }P/2 - 2 sin P/2 cos P/2 = 1 - sin P [Using sin 2A = 2 sin A cos A]

sin P/2 - cos P/2 = ± √ (1 - sin P)

If P = 30°, then P/2 = 30°/2 = 15°

Substituting P/2 = 15° in the sin P/2 + cos P/2 = ± √ (1 + sin P) equation, we get

sin 15° + cos 15° = ±√ (1 + sin 30°) …(1)

Similarly, we have

sin 15° – cos 15° = ±√ (1 – sin 30°) …(2)

Since sin 15° > 0 and cos 15˚ > 0 ⇒ Hence, sin 15° + cos 15° > 0

From (1) we will get,

sin 15° + cos 15° = √ (1 + sin 30°) …(3)

Also, sin 15° – cos 15° = √2 (1/√2 sin 15˚ – 1/√2 cos 15˚)

or, sin 15° – cos 15° = √ 2 (cos 45° sin 15˚ – sin 45° cos 15°)

or, sin 15° – cos 15° = √ 2 sin (15˚ – 45˚)

or, sin 15° – cos 15° = √ 2 sin (- 30˚)

or, sin 15° – cos 15° = -√ 2 sin 30°

or, sin 15° – cos 15° = -√ 2 x 1/2

or, sin 15° – cos 15° = – √2/2

So, sin 15° – cos 15° < 0

Now we get, from (2), sin 15° – cos 15°= -√(1 – sin 30°) … (4)

Adding equations (3) and (4) we get,

2 sin 15° = √(1 + ½) – √(1 – ½)

2 sin 15° = (√3−1)/√2

∴ sin 15° = (√3−1)/2√2

### Thus, sin 15° = (√3−1)/2√2

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