Question

prove that sin75°=(√6+√2)/4​

Answer 1

Answer:

The 21st term of an AP whose first two terms are – 3 and 4, is *

1 point

(a) 17

(b) 137

(c) 143

(d)-143

Answer 2

Step-by-step explanation:

sin75° = sin(90° - 15°)

= cos15°

we know, sin30° = 1/2

=> sin30° = 1/2 → sin(2 x 15°) = 1/2

=> 2sin15°cos15° = 1/2 → sin15°cos15° = 1/4

=> sin²15°.cos²15° = 1/16

=> (1 - cos²15°) cos²15° = 1/16

=> cos²15° - cos⁴15° = 1/16

=> cos⁴15° - cos²15° + 1/16 = 0

=> cos²15° = (2 ± √3)/4

=> cos15 = (6 + 2 ± 4√3)/16 = (√6 ± √2)/4

Hence we can say,

sin75° = (√6 + √2)/4