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Matemātika I 11 (2024./2025.m.g.)

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Vienkāršojiet izteiksmi   fraction numerator sin beta over denominator t g beta end fraction

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Vienkāršojiet izteiksmi  sinα + sinβ

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Pierādiet identitātes.

  1. space space space fraction numerator t g alpha over denominator sin alpha end fraction space minus space fraction numerator sin alpha over denominator c t g alpha end fraction space equals space cos alpha space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space 4 space p.

   2.    cos(60° - β) = sin (30° + β)                        4 p.

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Vispirms aplūkosim piemēru   ( sinα + cosα )2 - sin2α = 1  pacelsim iekavu kvadrātā, izmantojot formulu  

(a + b )2 = a2 + 2ab + b2

  sin2 α + 2sinα•cosα + cos2 α - 2sinα•cosα = 1,

sin2α tika pārveidots saskaņā ar divkāršā argumenta formulu, pēc saīsināšanas iegūstam     sin2 α + cos2 α = 1      vai saskaņā ar trigonom. pamatidentitāti  1 = 1,     kas bija jāpierāda.

Tagad pierādiet identitātes patstāvīgi.

  1. (2 p.)    1 - ( sinα - cosα )2 = sin2α
  2. (2 p.)    cos2α = 2cos2 α - 1
  3. (4 p.)    sin(∏ - α) + sin(∏ + α) - cos(3∏/2 + α) + sinα = 0
  4. (4 p.)    sinα - cosα = √2•cos(∏/4 - α)
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Vienādojuma sinβ = 0 atrisinājums ir ....

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Vienādojuma  sinx = 1 atrisinājums ir ...

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Nosaki vērtību izteiksmei   arc sinfraction numerator square root of 2 over denominator 2 end fraction

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Atrisiniet vienādojumu   sinx = 1 third

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Atrisiniet vienādojumu  sinx = -1.1

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Nosakiet vērtību izteiksmei  arc sinsquare root of 3

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