Question

Помогите решить,пожалуйста)

Дано:а)sin60*cos135*tg120 б)cos60-2sin^2 135+cos^2 150 упростить эти выражения. 

 

Answer 1

$sin60^0* cos135^0*tan120^0=frac{sqrt{3}}{2} cos135^0*tan120^0$

 

$frac{sqrt{3}}{2} cos135^0*tan120^0=frac{sqrt{3}}{2} cos(90^0+45^0)*tan120^0$

 

$frac{sqrt{3}}{2} cos(90^0+45^0)*tan120^0=$

 

$=frac{sqrt{3}}{2}( cos90^0cos45^0-sin90^0*sin45^0)*tan120^0$

 

$frac{sqrt{3}}{2}left(-frac{sqrt{2}}{2}right)*tan120^0=-frac{sqrt{6}}{4}*tan120^0$

 

$-frac{sqrt{6}}{4}*tan120^0=-frac{sqrt{6}}{4}*frac{sin120^0}{cos 120^0}$

 

$-frac{sqrt{6}}{4}*frac{sin(90^0+30^0)}{cos (90^0+30^0)}=$

 

$=-frac{sqrt{6}}{4}*frac{sin90^0*cos 30^0+cos90^0sin30^0}{cos (90^0+30^0)}$

 

$-frac{sqrt{6}}{4}*frac{sin90^0*cos 30^0+cos90^0sin30^0}{cos (90^0+30^0)}=-frac{sqrt{6}}{4}*frac{frac{sqrt{3}}{2}}{cos (90^0+30^0)}$

 

$-frac{3sqrt{2}}{8}*frac{1}{cos (90^0+30^0)}=-frac{3sqrt{2}}{8}*frac{1}{cos 90^0cos30^0-sin90^0sin30^0}$

 

$-frac{3sqrt{2}}{8}*frac{1}{cos 90^0cos30^0-sin90^0sin30^0}=-frac{3sqrt{2}}{8}*frac{1}{-frac{1}{2}}$

 

$-frac{3sqrt{2}}{8}*frac{1}{-frac{1}{2}}=2*frac{3sqrt{2}}{8}$

 

$2*frac{3sqrt{2}}{8}=frac{3sqrt{2}}{4}$

 

$cos60^0-2sin^2 135^0+cos^2 150^0=frac{1}{2}-2sin^2 135^0+cos^2 150^0$

 

По формуле двойного угла

 

$frac{1}{2}-2sin^2 135^0+cos^2 150^0=frac{1}{2}-(1-cos(2*135^0))+cos^2150^0$

 

$frac{1}{2}-(1-cos(2*135^0))+cos^2150^0=frac{1}{2}-1+cos 270^0+cos^2150^0$

 

$frac{1}{2}-1+cos 270^0+cos^2150^0=-frac{1}{2}+cos 270^0+cos^2150^0$

 

Снова по формуле двойного угла

 

$-frac{1}{2}+cos 270^0+cos^2150^0=-frac{1}{2}+cos 270^0+frac{1}{2}(1+cos(2*150^0))$

 

$-frac{1}{2}+cos 270^0+frac{1}{2}(1+cos(2*150^0))=cos 270^0+<var>frac{1}{2}</var>cos300^0$

 

$cos 270^0+<var>frac{1}{2}</var>cos300^0=0+<var>frac{1}{2}</var>cos300^0$

 

$frac{1}{2}</var><var>cos300^0=<var>frac{1}{2}</var>cos(360^0-60^0)$

 

$<var>frac{1}{2}</var>cos(360^0-60^0)=<var>frac{1}{2}</var>(cos 360^0cos 60^0+sin 360^0sin60^0)=frac{1}{4}$