Math, asked by sakshi4242, 9 months ago

find the value of sin 15

Answers

Answered by kailash511
0

Answer:

√3-1/2√2

Step-by-step explanation:

sin(45-30)= sin45cos30-cos45sin30

1/√2×√3/2-1/√2×1/2

√3-1/2√2

Answered by altaf143baig
1

Answer:

Sin 15 = (√3−1)/(2√2)

(Sin P/2 + Cos P/2)2 = Sin2 P/2 + Cos2 P/2 +2Sin P/2Cos P/2

= 1 + sinP

Sin P/2 + Cos P/2 = +– √ (1 + sin P)

If P = 300 so P/2 = 30/2 =150

Putting this value in the above equation:

Sin 150 + Cos 150 = +– √ (1 + sin 30) …(1)

Also, (Sin P/2 – Cos P/2)2 = Sin2 P/2 + Cos2 P/2 – 2Sin P/2Cos P/2

= 1 – sinP

Sin P/2 – Cos P/2 = +– √ (1 – sin P)

Putting this value in the above equation:

Sin 150 – Cos 150 = +– √ (1 – sin 30) …(2)

As seen, 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 got, from (2) sin 15° – cos 15°= -√(1 – sin 30°) … (4)

add (3) and (4) we get,

2 sin 15° = √ (1 + ½) – √ (1 – ½)

2 sin 15° = (√3−1)/√2

∴ sin 15° = (√3−1)/2√2

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