Trigonometry

Answer :

in ΔABC, sinA + cosB = 0

                   ∴ sinA = – cos B ⇒ sin A = – sin[(π/2) – B]

                   ∴ A = B – (π/2) ⇒ B = (π/2) + A

                   in ΔABC, A + B + C = π

                   ∴ C = π – [(π/2) + A + A]

                   ∴ C = (π/2) – 2A

                   ∴ A = (π/4) – (C/2)

                   ∴ range of A is [0, (π/4)].

Answer :

√[2 + √{2 + √(2[1 + cos(4π / 3)])}]

                   Let (4π / 3) = θ

                   ∴ given:  √[2 + √{2 + √(2 × 2cos²[θ/2])}] = √[2 + √{2 + 2cos(θ/2)}]

                                                                             = √[2 + 2cos(θ/4)]

                                                                             = √[4 cos²(θ/8)]

                                                                             = 2cos(θ/8)

                                                                             = 2cos[(4π / 3) × (1/8)]

                                                                             = 2cos(π/6)

                                                                             = √3.

Answer :

cot[52(1/2)]° = [1 / {tan[52(1/2)]°}]

                                      = [1 / {tan[45 + 7(1/2)]°}]

                                      = [{1 – [tan45 ∙ tan7(1/2)°]} / {tan45 + tan7(1/2)°}]

                                      = [{1 – tan7(1/2)°} / {1 + tan7(1/2)°}]

                                      = [{cos7(1/2)° – sin7(1/2)°} / {cos7(1/2)° + sin7(1/2)°}]

                                      = [{cos7(1/2)° – sin7(1/2)°}² / {cos²7(1/2)° – sin7(1/2)°}]

                                      = [(1 – sin15) / (cos15)]

                                      = [{1 – sin(45 – 30)} / {cos(45 – 30)}]

                                      = [{1 – [(√3 – 1) / (2√2)]} / {(√3 + 1) / (2√2)}]

                                      = [(2√2 – √3 + 1) / (√3 + 1)]

                                      = [{(2√2 – √3 + 1)(√3 – 1)} / (3 – 1)]

                                      = [{2√2(√3 – 1) – (√3 – 1)²} / 2]

                                      = [√2(√3 – 1) – (2 – √3)]

                                      = √6 – √2 – 2 + √3

                                      = √6 + √3 – √2 – 2.

Answer :

∵ cosx + cos2x + cos3x + cos4x = 0

                   ∴ cosx + cos3x + cos2x + cos4x = 0

                   ∴ 2cos2x ∙ cosx + 2cos3x ∙ cosx = 0

                   ∴ 2cosx(cos2x + cos3x) = 0

                   ∴ 4cosx ∙ cos(5x / 2) ∙ cos(x/2) = 0

                   ∴ x = (π/2), (3π / 2), (π/5), (3π / 5), (7π / 5), (9π / 5) & π

                   ∴ Total solution = 7.

Answer :

K[sin18° + cos36°] = 5

                   ∴ sin18 = [(√5 – 1) / 4] and cos36 = [(√5 + 1) / 4]

                   ∴ K[{(√5 – 1) / 4} + {(√5 + 1) / 4}] = 5

                   ∴ K(√5 / 2) = 5

                   ∴ K = 2√5.