Question

Wiadomo że sin alfa =2/5.Zatem wartość wyrażenia tgalfax(1/tg2alfa -cos^alfa) jest równa

Answer 1

a --- to inaczej alfa ;)

 

sina = 2/5

 

sin²a + cos²a = 1 ---- wzór na 1 trygonometryczną

(2/5)² + cos²a = 1

4/25 + cos²a = 1

cos²a = 25/25 - 4/25

cos²a = 21/25

cosa = √(21/25)  lub  cosa = -√(21/25)

cosa = √21/5      lub  cosa = -√21/5

 

tga = sina / cosa

tga = 2/5 * √21/5    lub tga = 2/5 / (-√21/5)

tga = 2/5 * 5/√21    lub tga = 2/5 * (-5/√21)

tga = 2/√21             lub  tga = -2/√21

tga = 2√21/21         lub tga = -2√21/21

 

dla tga = 2√21/21 mamy:

tgα x (1/tg²α - cos²α) = 2√21/21 x (1/(2√21/21)² - (√21/5)² =

2√21/21 x [(21*21)/(4 * 21) - 21/25] = 2√21/21 x (21/4 - 21/25) =

√21/2 - 2√21/25 = 25√21/50 - 4√21/50 = 21√21/50

 

dla tga = -2√21/21 mamy:

tgα x (1/tg²α - cos²α) = -2√21/21 x (1/(-2√21/21)² - (√21/5)² =

-2√21/21 x [(21*21)/(4 * 21) - 21/25] = -2√21/21 x (21/4 - 21/25) =

-√21/2 - 2√21/25 = -25√21/50 - 4√21/50 = -29√21/50