Question

In fractions, how do you find the value of sin 27°?

Answer 1

Step-by-step explanation:

⇒ (sin 27° + cos 27°)22 = 1+ sin 2 ∙ 27°

⇒ (sin 27° + cos 27°)22 = 1 + sin 54° 

⇒ (sin 27° + cos 27°)22 = 1 + sin (90° - 36°)

⇒ (sin 27° + cos 27°)22 = 1 + cos 36° 

⇒ (sin 27° + cos 27°)22 = 1+ √5+14√5+14

⇒ (sin 27° + cos 27°)22 = 1414 ( 5 + √ 5)

Therefore,  sin 27° + cos 27° = 125+5–√−−−−−−√125+5 …………….….(i) [Since, sin 27° > 0 and cos 27° > 0)

Similarly, we have, (sin 27° - cos 27°)22 = 1 - cos 36°

⇒ (sin 27° - cos 27°)22 = 1 - √5+14√5+14

⇒ (sin 27° - cos 27°)22 = 1414 (3 - √5  )

Therefore, sin 27° - cos 27° = ± 123−5–√−−−−−−√123−5 …………….….(ii)

Now, sin 27° - cos 27° = √2 (1√21√2 sin 27˚ - 1√21√2 cos 27°)

                               = √2 (cos 45° sin 27° - sin 45° cos 27°)

                               = √2 sin (27° - 45°)

                               = -√2 sin 18° < 0

Therefore, from (ii) we get,

sin 27° - cos 27° = -123−5–√−−−−−−√123−5 …………….….(iii)

Now, adding (i) and (iii) we get,

2 sin 27° = 125+5–√−−−−−−√125+5 - 123−5–√−−−−−−√123−5

⇒ sin 27° = 14(5+5–√−−−−−−√−3−5–√−−−−−−√)14(5+5−3−5)

Therefore, sin 27° = 14(5+5–√−−−−−−√−3−5–√−−−−−−√)14(5+5−3−5)