$\bf \textit{Sum and Difference Identities}
\\ \quad \\
sin({{ \alpha}} + {{ \beta}})=sin({{ \alpha}})cos({{ \beta}}) + cos({{ \alpha}})sin({{ \beta}})
\\ \quad \\
sin({{ \alpha}} - {{ \beta}})=sin({{ \alpha}})cos({{ \beta}})- cos({{ \alpha}})sin({{ \beta}})\\\\
-------------------------------\\\\$

$\bf sin(165^o)\implies sin(120^o+45^o)
\\\\\\
sin(120^o)cos(45^o)~+~cos(120^o)sin(45^o)\implies \cfrac{\sqrt{3}}{2}\cdot \cfrac{\sqrt{2}}{2}~+~\cfrac{-1}{2}\cdot \cfrac{\sqrt{2}}{2}
\\\\\\
\cfrac{\sqrt{6}}{4}~-~\cfrac{\sqrt{2}}{4}\implies \cfrac{\sqrt{6}-\sqrt{2}}{4}$

6 0

3 years ago

Simplify the expression:<br> 3j–1+4

RUDIKE [14]

Answer:

3j/3—5/3

J=5/3

hope helps

5 0

2 years ago