Sin75°+sin45°/sin285°

Question

Артём Хохлов

Математика

2 января, 13:47

Sin75°+sin45°/sin285°

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Никита Павловский · Answer 1

Sin 75° = sin (45 + 30) = sin45*cos30 + sin30*cos45 = (√2/2) * (√3/2) + (1/2) * (√2/2) = (√6/4) + (√2/4) = √2 * (√3+1) / 4.

sin285° = (2*120 + 45) = sin (2*120) * cos45 + sin45*cos (2*120) = sin (2*120) * (√2/2) + (√2/2) * cos (2*120) = (√2/2) * (sin (2*120) + cos (2*120)) = (√2/2) * (2*sin120*cos120 + (cos120) ^2 - (sin120) ^2) = (√2/2) * (4*sin60*cos60 * ((cos60) ^2 - (sin60) ^2) + (cos120) ^2 - (sin120) ^2) =

= (√2/2) * (4 * (√3/2) * (1/2) * ((1/2) ^2 - (√3/2) ^2) + ((cos60) ^2 - (sin (60) ^2) ^2 - (2*sin60*cos60) ^2) =

= (√2/2) * (√3 * (1/4 - ¾) + (¼ - ¾) ^2 - (2 * (√3/2) * (1/2)) ^2) = (√2/2) * (-2*√3/4 + 4/16 - ¾) =

= (√2/2) * (-2*√3/4 + 4/16 - ¾) = - (√2/2) * ((2*4*√3+8) / 16) = - √2 * (√3+1) / 4.

Sin75°+sin45°/sin285° = √2 * (√3+1) / 4 + (√2/2) / (-√2 * (√3+1) / 4) = √2 * (√3+1) / 4 + 4*√2/2 * (-√2 * (√3+1) = √2 * (√3+1) / 4 + 2*√2 / (-√2 * (√3+1) = (-2 * (√3+1) * (√3+1) + 2√2) / (-√2 * (√3+1) =

= (-2 * (3+2√3 + 1 + 2√2) / (-√2 * (√3+1) = ( - 8 - 4√3 + 4√2) / (-√2 * (√3+1).