Question

O valor de 105° é igual a:
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Answer 1

Resposta:

1) Alternativa d)

2) sen(165°)=(√6-√2)/4

3) cos(15°)=(√6+√2)/4

Explicação passo-a-passo:

1)

sen(105°)=sen(60°+45°)

sen(60°+45°)=sen60°cos45°+cos60°sen45°=

√3/2.√2/2+1/2.√2/2=√2/2(√3/2+1/2)= √2/2(√3+1)/2=(√6+√2)/4

2)

sen(165°)=sen(120°+45°)

sen(120°+45°)=sen120°cos45°+cos120°.sen45°

sen120°=sen60°=√3/2

cos120°= -cos60°= -1/2

sen120°cos45°+cos120°.sen45°=√3/2.√2/2-1/2.√2/2=√2/2(√3/2-1/2)=√2/2(√3-1).1/2=√2/4(√3-1)=(√6-√2)/4

3) cos(15°)=cos(45°-30°)

cos(45°-30°)=cos45°cos30°+sen45°sen30°= √2/2.√3/2+√2/2.1/2=√2/2(√3/2+1/2)=

√2/2(√3+1)1/2=√2/4(√3+1)=(√6+√2)/4

Answer 2

Explicação passo-a-passo:

1)

$\sf sen~(a+b)=sen~a\cdot cos b +sen~b\cdot cos~a$

$\sf sen~(45^{\circ}+60^{\circ})=sen~45^{\circ}\cdot cos~60^{\circ}+sen~60^{\circ}\cdot cos~45^{\circ}$

$\sf sen~105^{\circ}=\dfrac{\sqrt{2}}{2}\cdot\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt{2}}{2}$

$\sf sen~105^{\circ}=\dfrac{\sqrt{2}}{4}+\dfrac{\sqrt{6}}{4}$

$\sf sen~105^{\circ}=\dfrac{\sqrt{2}+\sqrt{6}}{4}$

Letra D

2)

$\sf sen~(a+b)=sen~a\cdot cos b +sen~b\cdot cos~a$

$\sf sen~(45^{\circ}+120^{\circ})=sen~45^{\circ}\cdot cos~120^{\circ}+sen~120^{\circ}\cdot cos~45^{\circ}$

• $\sf sen~120^{\circ}=sen~60^{\circ}=\dfrac{\sqrt{3}}{2}$

• $\sf cos~120^{\circ}=-cos~60^{\circ}=\dfrac{-1}{2}$

$\sf sen~165^{\circ}=\dfrac{\sqrt{2}}{2}\cdot\left(\dfrac{-1}{2}\right)+\dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt{2}}{2}$

$\sf sen~105^{\circ}=\dfrac{-\sqrt{2}}{4}+\dfrac{\sqrt{6}}{4}$

$\sf sen~105^{\circ}=\dfrac{\sqrt{6}-\sqrt{2}}{4}$

3)

$\sf cos~(a-b)=cos~a\cdot cos~b+sen~a\cdot sen~b$

$\sf cos~(60^{\circ}-45^{\circ})=cos~60^{\circ}\cdot cos~45^{\circ}+sen~60^{\circ}\cdot sen~45^{\circ}$

$\sf cos~15^{\circ}=\dfrac{1}{2}\cdot\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt{2}}{2}$

$\sf cos~15^{\circ}=\dfrac{\sqrt{6}}{4}+\dfrac{\sqrt{2}}{4}$

$\sf cos~15^{\circ}=\dfrac{\sqrt{6}+\sqrt{2}}{4}$