Question

Usando as formulas da adiçao, determine:
a) Sen 105° 
b) Cos 135°
c) Cos 195°
d) Sen 165°
e) Sen 225°
f) Cos 225°
g) Cos 300°
h) Sen 345°

Answer 1

Olá

 

Para resolver essa questão, de soma de arcos, temos que usar as expressões para o cálculo em seno (sen) e cosseno (cos). 

Para os cálculos devemos usar ângulos conhecidos, como por exemplo os ângulos notáveis. Adiciono em anexo uma tabela com os ângulo notáveis até 180°, donde podemos nos basear para resolver essa questão.

 

Seguem as expressões:

 

$\mathsf{sen~(a+b)=sen~a\cdot cos~b+sen~b\cdot cos~a}\\\\ \mathsf{cos~(a+b)=cos~a\cdot cos~b-sen~a\cdot sen~b}$

 

No desenvolvimento de cada questão, irei pegar direto cada seno e cosseno, igualar a um equivalente em forma de soma e resolver, de forma direta. Agora, vamos aos cálculos.

 

a)

sen 105° = sen (60° + 45°)

 

$\mathsf{sen~(a+b)=sen~a\cdot cos~b+sen~b\cdot cos~a}\\\\ \mathsf{sen~(60^{\circ}+45^{\circ})=sen~60^{\circ}\cdot cos~45^{\circ}+sen~45^{\circ}\cdot cos~60^{\circ}}\\\\ \mathsf{sen~(60^{\circ}+45^{\circ})=\dfrac{\sqrt3}{2}\cdot\dfrac{\sqrt2}{2}+\dfrac{\sqrt2}{2}\cdot\dfrac{1}{2}}\\\\ \mathsf{sen~(60^{\circ}+45^{\circ})=\dfrac{\sqrt6}{4}+\dfrac{\sqrt2}{4}}\\\\ \boxed{\mathsf{sen~(60^{\circ}+45^{\circ})=\dfrac{\sqrt6+\sqrt2}{4}}}$

 

b)

cos 135° = cos (90° + 45°)

 

$\mathsf{cos~(a+b)=cos~a\cdot cos~b-sen~a\cdot sen~b}\\\\ \mathsf{cos~(90^{\circ}+45^{\circ})=cos~90^{\circ}\cdot cos~45^{\circ}-sen~90^{\circ}\cdot sen~45^{\circ}}\\\\ \mathsf{cos~(90^{\circ}+45^{\circ})=0\cdot\dfrac{\sqrt2}{2}-1\cdot \dfrac{\sqrt2}{2}}\\\\ \boxed{\mathsf{cos~(90^{\circ}+45^{\circ})=\dfrac{-\sqrt2}{2}}}$

 

c)

cos 195° = cos (135° + 60)

 

$\mathsf{cos~(a+b)=cos~a\cdot cos~b-sen~a\cdot sen~b}\\\\ \mathsf{cos~(135^{\circ}+60^{\circ})=cos~135^{\circ}\cdot cos~60^{\circ}-sen~135^{\circ}\cdot sen~60^{\circ}}\\\\ \mathsf{cos~(135^{\circ}+60^{\circ})=\dfrac{-\sqrt2}{2}\cdot\dfrac{1}{2}-\dfrac{\sqrt2}{2}\cdot\dfrac{\sqrt3}{2}}\\\\ \mathsf{cos~(135^{\circ}+60^{\circ})=\dfrac{-\sqrt2}{4}-\dfrac{\sqrt6}{4}}\\\\ \boxed{\mathsf{cos~(135^{\circ}+60^{\circ})=\dfrac{-\sqrt2-\sqrt6}{4}}}$

 

d)

sen 165° = sen (120° + 45°)

 

$\mathsf{sen~(a+b)=sen~a\cdot cos~b+sen~b\cdot cos~a}\\\\ \mathsf{sen~(120^{\circ}+45^{\circ})=sen~120^{\circ}\cdot cos~45^{\circ}+sen~45^{\circ}\cdot cos~120^{\circ}}\\\\ \mathsf{sen~(120^{\circ}+45^{\circ})=\dfrac{\sqrt3}{2}\cdot\dfrac{\sqrt2}{2}+\dfrac{\sqrt2}{2}\cdot\dfrac{-1}{2}}\\\\ \mathsf{sen~(120^{\circ}+45^{\circ})=\dfrac{\sqrt6}{4}+\dfrac{-\sqrt2}{4}}\\\\ \boxed{\mathsf{sen~(120^{\circ}+45^{\circ})=\dfrac{\sqrt6-\sqrt2}{4}}}$

 

e)

sen 225° = sen (180° + 45)

 

$\mathsf{sen~(a+b)=sen~a\cdot cos~b+sen~b\cdot cos~a}\\\\ \mathsf{sen~(180^{\circ}+45^{\circ})=sen~180^{\circ}\cdot cos~45^{\circ}+sen~45^{\circ}\cdot cos~180^{\circ}}\\\\ \mathsf{sen~(180^{\circ}+45^{\circ})=0\cdot\dfrac{\sqrt2}{2}+\dfrac{\sqrt2}{2}\cdot-1}\\\\ \boxed{\mathsf{sen~(180^{\circ}+45^{\circ})=\dfrac{-\sqrt2}{2}}}$

 

f)

cos 225° = cos (180° + 45°)

 

$\mathsf{cos~(a+b)=cos~a\cdot cos~b-sen~a\cdot sen~b}\\\\ \mathsf{cos~(180^{\circ}+45^{\circ})=cos~180^{\circ}\cdot cos~45^{\circ}-sen~180^{\circ}\cdot sen~45^{\circ}}\\\\ \mathsf{cos~(180^{\circ}+45^{\circ})=1\cdot\dfrac{\sqrt2}{2}-0\cdot\dfrac{\sqrt2}{2}}\\\\ \boxed{\mathsf{cos~(180^{\circ}+45^{\circ})=\dfrac{\sqrt2}{2}}}$

 

g)

cos 300° = cos (180° + 120°)

 

$\mathsf{cos~(a+b)=cos~a\cdot cos~b-sen~a\cdot sen~b}\\\\ \mathsf{cos~(180^{\circ}+120^{\circ})=cos~180^{\circ}\cdot cos~120^{\circ}-sen~180^{\circ}\cdot sen~120^{\circ}}\\\\ \mathsf{cos~(180^{\circ}+120^{\circ})=1\cdot\dfrac{-1}{2}-0\cdot\dfrac{\sqrt3}{2}}\\\\ \boxed{\mathsf{cos~(180^{\circ}+120^{\circ})=\dfrac{-1}{2}}}$

 

h)

sen 345° = sen (180° + sen 165°)

 

Nesse caso, temos o seno de 165° (questão d) mas não o cosseno. Primeiro vou descobrir o cosseno de 165° e depois continuar esse cálculo.

 

$\mathsf{cos~(a+b)=cos~a\cdot cos~b-sen~a\cdot sen~b}\\\\ \mathsf{cos~(120^{\circ}+45^{\circ})=cos~120^{\circ}\cdot cos~45^{\circ}-sen~120^{\circ}\cdot sen~45^{\circ}}\\\\ \mathsf{cos~(120^{\circ}+45^{\circ})=\dfrac{-1}{2}\cdot \dfrac{\sqrt2}{2}-\dfrac{\sqrt3}{2}\cdot\dfrac{\sqrt2}{2}}\\\\ \mathsf{cos~(120^{\circ}+45^{\circ})=\dfrac{-\sqrt2}{4}-\dfrac{\sqrt6}{4}}\\\\ \boxed{\mathsf{cos~(120^{\circ}+45^{\circ})=\dfrac{-\sqrt2+\sqrt6}{4}}}$

 

Agora, vamos aos cálculos do seno apresentado de maneira prima.

 

$\mathsf{sen~(a+b)=sen~a\cdot cos~b+sen~b\cdot cos~a}\\\\ \mathsf{sen~(180^{\circ}+165^{\circ})=sen~180^{\circ}\cdot cos~165^{\circ}+sen~165^{\circ}\cdot cos~180^{\circ}}\\\\ \mathsf{sen~(180^{\circ}+165^{\circ})=0\cdot\dfrac{-\sqrt2+\sqrt6}{4}+\dfrac{\sqrt6-\sqrt2}{4}\cdot-1}\\\\ \boxed{\mathsf{sen~(180^{\circ}+165^{\circ})=\dfrac{-\sqrt6+\sqrt2}{4}}}$

 

Quaisquer dúvidas, deixe nos comentários.

Bons estudos$.$