sin²2α+sin²β+cos(2α+β)cos(2α-β). Created by Rahmann. algebra-ru

sin²2α+sin²β+cos(2α+β)cos(2α-β)...

Nullpo201

Я сначала произведение косинусов перевёл в сумму косинусов, а потом понизил степень у синусов в квадрате

1 votes Thanks 1

resenans

$\sin^22\alpha+\sin^2\beta+\cos(2\alpha+\beta)\cos(2\alpha-\beta)=\\\\=\sin^22\alpha+\sin^2\beta +\dfrac{1}{2}\bigg(\cos\big((2\alpha+\beta)+(2\alpha-\beta)\big)+\cos\big((2\alpha +\beta )-(2\alpha -\beta )\big)\bigg)=\\\\=\sin^22\alpha +\sin^2\beta +\dfrac{1}{2}\big(\cos4\alpha +\cos2\beta\big)=\sin^22\alpha +\sin^2\beta +\dfrac{1}{2}\cos4\alpha +\dfrac{1}{2}\cos2\beta =\\\\=\sin^22\alpha +\sin^2\beta +\dfrac{1}{2}\big(\cos^22\alpha -\sin^22\alpha \big)+\dfrac{1}{2}(\cos^2\beta -\sin^2\beta \big)=$

$=\sin^22\alpha +\sin^2\beta +\dfrac{1}{2}\cos^22\alpha -\dfrac{1}{2}\sin^22\alpha +\dfrac{1}{2}\cos^2\beta -\dfrac{1}{2}\sin^2\beta =\\\\=\dfrac{1}{2}\sin^22\alpha +\dfrac{1}{2}\sin^2\beta+\dfrac{1}{2}\cos^22\alpha+\dfrac{1}{2}\cos^2\beta=\\\\=\dfrac{1}{2}\bigg(\sin^22\alpha+\sin^2\beta+\cos^22\alpha+\cos^2\beta\bigg)=\dfrac{1}{2}\bigg(1+1\bigg)=\dfrac{1}{2}\cdot2=1$

4 votes Thanks 3

Answers & Comments

srochno pomogite pozhalujsta sostavit formulu225 trimetil 4 vtorbutiloktantr

sostavte uravneniya himicheskih reakcij soglasno sheme i nazovite poluchennye veshe

reshite sovokupnost neravenstv xlt2 2x 17

ochen srochno sochinenie rassuzhdenie po povesti posle bala l n tolstoj t

50b geometriyazadacha na podobie treugolnikovzadacha vo vlozhenii proshu ne spam

30b himiyazadacha 2 dano i reshenie podrobno pishite tolko esli uvereny proshu

put the correct prepositions9f11c835b37f36c446ee79879d9b5edb 60225

has been studying in the libraryvyberite vernyj variant otveta

complete the sentences with does our second conditionals

biologiyakakie priznaki harakterny dlya bolshinstva podrazdelenij carstva monera

рекомендуемые вопросы

Helpful Links

Helpful Social