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provar sin(60)=2sin(30)cos(30)

Solução

provar

sin (6 0^{\circ}) = 2 sin (3 0^{\circ}) cos (3 0^{\circ})

Verdadeiro

Passos da solução

sin (6 0^{\circ}) = 2 sin (3 0^{\circ}) cos (3 0^{\circ})

Manipular o lado direito

sin (6 0^{\circ})

Simplificar

= \frac{3}{2}

Manipular o lado esquerdo

2 sin (3 0^{\circ}) cos (3 0^{\circ})

Simplificar

2 cos (3 0^{\circ}) sin (3 0^{\circ}) : \frac{3}{2}

2 cos (3 0^{\circ}) sin (3 0^{\circ})

Simplificar

cos (3 0^{\circ}) : \frac{3}{2}

cos (3 0^{\circ})

Utilizar a seguinte identidade trivial:

cos (3 0^{\circ}) = \frac{3}{2}

cos (x)

tabela de periodicidade com ciclo de

36 0^{\circ} n

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{3}{2}

= 2 \cdot \frac{3}{2} sin (3 0^{\circ})

Simplificar

sin (3 0^{\circ}) : \frac{1}{2}

sin (3 0^{\circ})

Utilizar a seguinte identidade trivial:

sin (3 0^{\circ}) = \frac{1}{2}

sin (x)

tabela de periodicidade com ciclo de

36 0^{\circ} n

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{1}{2}

= 2 \cdot \frac{1}{2} \cdot \frac{3}{2}

Simplificar

2 \cdot \frac{3}{2} \cdot \frac{1}{2}

Multiplicar frações:

a \cdot \frac{b}{c} \cdot \frac{d}{e} = \frac{a \cdot b \cdot d}{c \cdot e}

= \frac{3 \cdot 2}{2 \cdot 2}

Eliminar o fator comum:

2

= \frac{3}{2}

= \frac{3}{2}

= \frac{3}{2}

Demonstramos que os dois lados podem adquirir a mesma forma

\Rightarrow Verdadeiro

p ro v e \frac{sin ( 2 θ )}{sin ( θ )} - \frac{cos ( 2 θ )}{cos ( θ )} = sec (θ)

p ro v e sec (2 θ) = \frac{sec ^{2} ( θ )}{1 - tan ^{2} ( θ )}

p ro v e cos (2 a) = cos^{2} (a) - sin^{2} (a)

p ro v e tan (θ) + cot (θ) = sec (θ) \cdot csc (θ)

p ro v e \frac{cos ( x )}{csc ( x ) + 1} + \frac{cos ( x )}{csc ( x ) - 1} = 2 tan (x)