Опростете cos(A+B)cos(A–B)+sin(A+B).sin(A-B)

[doomaster]

опростете израза
cos(A+B)cos(A–B)+sin(A+B).sin(A-B)
4-sinA.sin(60°-A)cos(30°-A)

докажете тъждеството
4sinA.cosA.(1-tg²A)=sin4A
¯¯¯¯1+tg²A¯¯¯¯¯¯¯

cos2A =tg(45°-A)
¯¯¯¯1+sin2A¯¯¯
представете като произведения
3-4cos²A
1-2cosA+cos2A

[ева]

Ще докажем,че [tex]\frac{4sin\alpha.cos\alpha(1-tg^{2}\alpha)}{1+tg^{2}\alpha}[/tex]=sin4[tex]\alpha[/tex]

2sin2[tex]\alpha[/tex].[tex]\frac{1-tg^{2}\alpha}{1+tg^{2}\alpha}[/tex]=

=2sin2[tex]\alpha[/tex].cos2[tex]\alpha[/tex]=

=sin4[tex]\alpha[/tex]

[ева]

[tex]\frac{cos2\alpha}{1+sin2\alpha}[/tex]=tg(45[tex]^\circ[/tex]-[tex]\alpha[/tex])

[tex]\frac{cos2\alpha}{1+sin2\alpha}[/tex]=[tex]\frac{-(1-cos2\alpha)+1}{2cos^{2}(45^\circ-\alpha)}[/tex]=[tex]\frac{-2sin^{2}\alpha+1}{2cos^{2}(45^\circ-\alpha)}[/tex]=[tex]\frac{cos^{2}\alpha-sin^{2}\alpha}{2(\frac{\sqrt{2}cos\alpha}{2}+\frac{\sqrt{2}sin\alpha}{2})^{2}}[/tex]=

=[tex]\frac{(cos\alpha-sin\alpha)(cos\alpha+sin\alpha)}{(cos\alpha+sin\alpha)^{2}}[/tex]=[tex]\frac{cos\alpha-sin\alpha}{cos\alpha+sin\alpha}[/tex]=[tex]\frac{\sqrt{2}sin(45^\circ-\alpha)}{\sqrt{2}cos(45^\circ-\alpha)}[/tex]=tg(45[tex]^\circ[/tex]-[tex]\alpha[/tex])