11 клас тригонометрия

[alexm]

Моля, помогнете ми с оградените задачи. Благодаря предварително.

[math10.com]

[tex]\frac{sin^2{2\alpha}- 4sin^2\alpha}{sin^2{2\alpha}+ 4sin^2\alpha-4}=[/tex]

[tex]=\frac{(2sin\alpha.cos \alpha )^2- 4sin^2\alpha}{(2sin\alpha.cos \alpha )^2- 4(1-sin^2\alpha)}=[/tex]

[tex]=\frac{4sin^2\alpha.cos^2 \alpha- 4sin^2\alpha}{4sin^2\alpha.cos ^2\alpha- 4cos ^2\alpha}=[/tex]

[tex]=\frac{4sin^2\alpha(cos^2 \alpha-1)}{4cos ^2\alpha(sin^2\alpha-1)}=[/tex]

[tex]=\frac{-4sin^2\alpha(1-cos^2 \alpha)}{-4cos ^2\alpha(1-sin^2\alpha)}=[/tex]

[tex]=\frac{-4sin^2\alpha.sin^2\alpha}{-4cos ^2\alpha.cos ^2\alpha}=[/tex]

[tex]=tg^4 \alpha[/tex]

[alexm]

[tex]\frac{sin^2{2\alpha}- 4sin^2\alpha}{sin^2{2\alpha}+ 4sin^2\alpha-4}=[/tex]

[tex]=\frac{(2sin\alpha.cos \alpha )^2- 4sin^2\alpha}{(2sin\alpha.cos \alpha )^2- 4(1-sin^2\alpha)}=[/tex]

[tex]=\frac{4sin^2\alpha.cos^2 \alpha- 4sin^2\alpha}{4sin^2\alpha.cos ^2\alpha- 4cos ^2\alpha}=[/tex]

[tex]=\frac{4sin^2\alpha(cos^2 \alpha-1)}{4cos ^2\alpha(sin^2\alpha-1)}=[/tex]

[tex]=\frac{-4sin^2\alpha(1-cos^2 \alpha)}{-4cos ^2\alpha(1-sin^2\alpha)}=[/tex]

[tex]=\frac{-4sin^2\alpha.sin^2\alpha}{-4cos ^2\alpha.cos ^2\alpha}=[/tex]

[tex]=tg^4 \alpha[/tex]

Мерси