Question

me pueden ayudar con esto por favor? :/​

Answer 1

Respuesta:

SOLUCIÓN #1.

A) sen60°tan60°+sen45°cos45°

$\frac{\sqrt{3} }{2}*\sqrt{3}+\frac{\sqrt{2} }{2} *1=\\\\\\\frac{(\sqrt{3}) ^{2} }{2} +\frac{\sqrt{2} }{2} =\\\\\frac{3}{2} + \frac{\sqrt{2} }{2}=\\\\\frac{1}{2}(3+\sqrt{2} )$

B) sen45°+cos60°-tan30°

$\frac{\sqrt{2} }{2} +\frac{1}{2} -\frac{\sqrt{3} }{3}= \\\\\frac{3\sqrt{2}+3-2\sqrt{3} }{6}$

SOLUCIÓN #2.

A)  tan30°(cos45°-csc60°)=

$\frac{\sqrt{3} }{3} (\frac{\sqrt{2}}{2} -\frac{2\sqrt{3}}{3 } )=\\\\\\\frac{\sqrt{6} }{6} -\frac{2*(\sqrt{3}) ^{2} }{9 } =\\\\\frac{\sqrt{6} }{6} -\frac{2*3}{6}=\\\\\frac{\sqrt{6} }{6} -\frac{6}{6}=\\\\\frac{\sqrt{6} }{6} -1$

B)  2(cos240°-tan120°)+sen90°=

$2 (-\frac{1}{2} -\sqrt{3} )+1=\\\\-1-2\sqrt{3} +1=\\\\-2\sqrt{3}$

C)  sen²120°+cos²120°=

$(\frac{\sqrt{3} }{2} )^{2} +(-\frac{1}{2} )^{2} =\\\\\frac{3}{2} -\frac{1}{4}=\\\\\frac{12-2}{8} =\\\\\frac{10}{8} =\\\\\frac{5}{4}$

D)  sec²45°-tan²45°=

$(\sqrt{2} )^{2} -(1)^{2} =\\\\2-1=\\\\1$

SOLUCIÓN #3.

A)  1-2sen²30°=cos60°

$1-2*(\frac{1}{2} )^{2}=\frac{1}{2} \\\\1-2*\frac{1}{4}= \frac{1}{2}\\\\1-\frac{1}{2}=\frac{1}{2}\\\\\frac{1}{2}=\frac{1}{2}$

B)  sen²60°=1-cos²60°

$(\frac{\sqrt{3} }{2} )^{2} =1-(\frac{1}{2} )^{2}\\\\\frac{3}{4} =1-\frac{1}{4} \\\\\frac{3}{4} =\frac{3}{4}$

C) tan45°=sen45°/cos45°

$1=\frac{\frac{\sqrt{2} }{2} }{\frac{\sqrt{2} }{2} } \\\\1=\frac{2\sqrt{2} }{2\sqrt{2} } \\\\1=1$