Question

calcule cos 15 + sen 105​

Answer 1

Olá, siga a explicação:

$\boxed { \sf cos ~ 15^o + sen ~ 105^o }$

Inicialmente iremos calcular o cosseno:

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

$\sf cos ~ 15^o = cos(45^o-30^o) \\ \\\sf cos(45^o-30^o) = cos ~ 45^o \centerdot cos ~30^o + sen ~ 45^o \centerdot sen ~30^o \\ \\cos ~ 15^o = \dfrac{\sqrt{2} }{2} \cdot \dfrac{\sqrt{3} }{2} + \dfrac{\sqrt{2} }{2} \cdot \dfrac{1}{2} \\ \\cos ~ 15^o = \dfrac{\sqrt{6} }{4} + \dfrac{\sqrt{2} }{4} \\ \\\boxed { \sf cos ~ 15^o = \dfrac{\sqrt{6} + \sqrt{2} }{4}}$

Seno:

$\boxed { \sf sen (a+b) = sen ~ a \cdot cos ~ b + sen ~ b \cdot cos ~ a} \\ \\ \sf sen (60^o +45^o ) = sen ~ 60^o \cdot cos ~ 45^o + sen ~ 45^o \cdot cos ~ 60^o \\ \\sen (60^o+45^o) = \dfrac{\sqrt{3} }{2} \cdot \dfrac{\sqrt{2} }{2} + \dfrac{\sqrt{2} }{2 } \cdot \dfrac{1}{2} \\ \\sen (60^o+45^o) = \dfrac{\sqrt{6} }{4} + \dfrac{\sqrt{2} }{4} \\ \\\boxed { \sf sen (60^o+45^o) = \dfrac{\sqrt{6}+ \sqrt{2} }{4} }$

Logo:

$\sf \boxed { \sf cos ~ 15^o+ sen ~ 105^o} = \dfrac{\sqrt{6} + \sqrt{2} }{4} + \dfrac{\sqrt{6} + \sqrt{2} }{4} = \dfrac{\sqrt{6} + \sqrt{6} + \sqrt{2} + \sqrt{2} }{4} =\sf \dfrac{ \not {2}\sqrt{6} + \not {2} \sqrt{2} }{ \not {4}} = \boxed { \sf \bf\dfrac{\sqrt{6} + \sqrt{2} }{2} }$

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