Question

Sa se calculeze
a) sin(a+b+c), dacă a, b, c €
(pi/2,pi),sina=15/3,sin b=4/5,sin c=4/5
b) cos(a - b+c), dacă a, b, c €(pi/2,pi)
cos a = -4/5,cos b=-3/5,sin c =5/13​

Answer 1

Răspuns:

Folosesti formula fundamentala a trigonometriei pentru a afla cos(a),cos(b) si cos(c) la punctul a) si la fel ptr. punctul b) pentru a afla sin(a),sin(b) si sin(c), iar apoi introduci valorile in ecuatie si calculezi.

Formula fundamentala a trigonometriei: $sin^2x+cos^2x=1$, ∀x∈R

$a) sin(a+b+c)=sin((a+b)+c)=sin(a+b)*cos(c)+sin(c)*cos(a+b)$

$(sin(a)*cos(b)+sin(b)*cos(a))*cos(c)+sin(c)*(cos(a)*cos(b)-sin(a)*sin(b))$

$b) cos(a-b+c)=cos((a-b)+c)=cos(a-b)*cos(c)-sin(a-b)*sin(c)$

$=(cos(a)*cos(b)+sin(a)*sin(b))*cos(c)-(sin(a)*cos(b)-sin(b)*cos(a))*cos(c)$

$sin^2a+cos^2a=1=>sin^2a=1-cos^2a<=>1-\frac{16}{25}=>sin^2a= \frac{9}{25}=>\\=>sin(a)=\frac{3}{5}$

$sin^2b+cos^2b=1=>sin^2b=1-cos^2b<=>1-\frac{9}{25} =>sin^2b=\frac{16}{25}=>\\=>sin(b)=\frac{4}{5}$

$sin^2c+cos^2c=1=>sin^2c=1-cos^2c<=>1-\frac{25}{169}=> sin^2c=\frac{25}{169}=>\\=>sin(c)=\frac{5}{13}$