• Предмет: Математика
  • Автор: Мортиgg
  • Вопрос задан 2 года назад

любые 6 Математика тригонометрия

Приложения:

Ответы

Ответ дал: QDominus
0

1)

3 \sin( \frac{\pi}{6} )  \tg( \frac{\pi}{4} )  = 3 \sin( \frac{\pi}{6} )  \times 1 = 3 \times  \frac{ 1 }{2}  =  \frac{3}{2}  = 1.5

2)

3 \sin( \frac{\pi}{6} )  -  \tg( \frac{\pi}{4} )  + 2 \sin( \frac{\pi}{2} )  -  \cos(\pi)  = 3 \times  \frac{1}{2}  - 1 + 2 \times 1 - ( - 1) =  \frac{3}{2}  - 1 + 2 + 1 = 1.5 + 2 = 3.5

3)

 \sin(0)  +  \cos( \frac{\pi}{2} )  +  \sin( \frac{\pi}{4} )  = 0 + 1 +  \frac{ \sqrt{2} }{2}  = 1 +  \frac{ \sqrt{2} }{2}  =  \frac{2 +  \sqrt{2} }{2}

4)

 \tg( \frac{23\pi}{4} )  \cos( \frac{23\pi}{3} )  \sin( \frac{23\pi}{6} )  =  \tg( \frac{24\pi}{4}  -  \frac{\pi}{4} )  \cos( \frac{24\pi}{3} -  \frac{\pi}{3}  )  \sin( \frac{24\pi}{6}  -  \frac{\pi}{6} )  =  \tg(6\pi -  \frac{\pi}{4} )  \cos(8\pi - \frac{\pi}{3}  )  \sin(4\pi -  \frac{\pi}{6} )  =  \tg( -  \frac{\pi}{4} ) \cos( -  \frac{\pi}{3} )  \sin( -  \frac{\pi}{6} )  =  -  \tg( \frac{\pi}{4} )  \cos( \frac{\pi}{3} ) ( -  \sin( \frac{\pi}{6} ) ) = 1 \times  \frac{1}{2}  \times  \frac{1}{2}  =  \frac{1}{4}  = 0.25

5)

 \frac{12 \sin(11°) \cos(11°)  }{ \sin(22°) }  =  \frac{6 \times 2 \sin(11°)  \cos(11°) }{ \sin(22°) }  =  \frac{6 \sin(2 \times 11°) }{ \sin(22°) }  =  \frac{6 \sin(22°) }{ \sin(22°) }  = 6

6)

 \tg(45°)  \ctg(45°)  =  \frac{ \sin(45°) }{ \cos(45°) }  \times  \frac{ \cos(45°) }{ \sin(45°) }  = 1

7)

3 \tg( \frac{\pi}{4} )  -  \sin {}^{2} ( \frac{\pi}{3} )  +  \cos {}^{2} ( \frac{\pi}{6} )  = 3 \times 1 - ( \frac{ \sqrt{3} }{2} ) {}^{2}  + ( \frac{ \sqrt{3} }{2} ) {}^{2}  = 3

8)

3 \sin( \frac{\pi}{6} )  + 2 \cos(\pi)  +  \ctg {}^{2} ( \frac{\pi}{6} ) = 3 \times  \frac{1}{2}  + 2 ( - 1) + ( \sqrt{3} ) {}^{2}  =  \frac{3}{2}  - 2 + 3 = 1.5  + 1 = 2.5

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