Question

Aratati ca sin3x+sin2x+sinx=radical din 3 , x apartine (0, pi/2) si sinx=radical din 3/2

Answer 1

Explicație pas cu pas:

cosx=√(1-sin²x)⇒cosx=√1-3/4=1/2

sin3x+sin2x+sinx

3sinx-4sin³x+2*sinx*cosx+sinx=

=3*√3/2-4*3√3/8+2*√3/2*1/2+√3/2

=3√3/2-3√3/2+2√3/2=

=0+√3=√3

Bafta!

Answer 2

x∈(0;π/2)=>x∈C1=>sinx>0;cosx>0 (1)

sinx=√3/2

sin²x+cos²x=1=>cos²x=1-sin²x=1-(√3/2)²=1-3/4=(4-3)4=1/4=>cosx=±√(1/4)=±1/2 (2)

Din (1) si (2)=>cosx=1/2

sin2x=2sinx*cosx=2*√3/2*1/2=√3/2

sin3x=3sinx-4sin³x=3√3/2-4(√3/2)³=3√3/2-12√3/8=3√3/2-3√3/2=0

sin3x+sin2x+sinx=0+√3/2+√3/2=√3