Question

prove that 4 sin 27 is equal to 5 + root 5 whole power 1 by 2 minus 3 minus root 5 whole power 1 by 2​

Answer 1

Answer:

proved

Step-by-step explanation:

Given  

prove that 4 sin 27 is equal to 5 + root 5 whole power 1 by 2 minus 3 minus root 5 whole power 1 by 2

We can write as

(sin 27 + cos 27)^2 = sin^2 27 + cos^2 27 + 2 sin 27 cos 27

(sin 27 + cos 27)^2 = 1 + sin 2. 27

(sin 27 + cos 27)^2 = 1 + sin 54

(sin 27 + cos 27)^2 = 1 + sin (90 – 36)

(sin 27 + cos 27)^2 = 1 + cos 36

(sin 27 + cos 27)^2 = 1 + √5 + 1/4

(sin 27 + cos 27)^2 = 1/4 (5 + √5)

Sin 27 + cos 27 = 1/2 (√5 + √5) -------------1

Similarly we get

(sin 27 – cos 27)^2 = 1 – cos 36

(sin 27 - cos 27)^2 = 1 - √5 + 1/4

(sin 27 - cos 27)^2 = 1/4 (3 - √5)

sin 27 - cos 27 = + - 1/2 √3 - √5-------------2

sin 27 - cos 27 = -1/2 √3 - √5 --------------3

Adding 1 and 3 we get,

2 sin 27 = 1/2 √5 + √5 – 1/2 √3 - √5

             = 1/2 (√5 + √5 –  √3 - √5)

  4 sin 27 =  (√5 + √5)^1/2 – (√3 - √5)^1/2