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:
[tex]\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}[/tex]
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°)
[tex]\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}}}[/tex]
b)
cos
135° = cos (90° + 45°)
[tex]\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}}}[/tex]
c)
cos
195° = cos (135° + 60)
[tex]\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}}}[/tex]
d)
sen 165° = sen (120° + 45°)
[tex]\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}}}[/tex]
e)
sen 225° = sen (180° + 45)
[tex]\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}}}[/tex]
f)
cos 225° = cos (180° + 45°)
[tex]\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}}}[/tex]
g)
cos 300° = cos (180° + 120°)
[tex]\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}}}[/tex]
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.
[tex]\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}}}[/tex]
Agora,
vamos aos cálculos do seno apresentado de maneira prima.
[tex]\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}}}[/tex]
Quaisquer
dúvidas, deixe nos comentários.
Bons estudos[tex].[/tex]
