Question

Trigonometrie
doar g) h) i)​

Answer 1

Răspuns:

g)

$sin^23x-sin^24x=(sin3x-sin4x)(sin3x+sin4x) =\\= (2sin\frac{3x-4x}{2}*cos(\frac{3x+4x}{2})(2sin\frac{3x+4x}{2}*cos\frac{3x-4x}{2})$

$(2sin\frac{-x}{2}*cos\frac{7x}{2}) (2sin\frac{7x}{2}*cos\frac{-x}{2})=-4sin\frac{x}{2}*cos\frac{x}{2} *sin\frac{7x}{2}*cos\frac{7x}{2}$

h)

$cos^2x-cos^23x=(cosx-cos3x)(cosx+cos3x) = \\= (-2sin\frac{x+3x}{2}*sin\frac{x-3x}{2}) (2cos\frac{x+3x}{2}*cos\frac{x-3x}{2} )\\=(-2sin2x*sin(-2x))(2cos2x*cos(-2x)\\=(-2sin2x*(-sin(x))(2cos2x*cos2x)\\=4sin2x*sinx*cos^22x$

i)

$cos^2x-sin^24x =cos^2x-(1-cos^24x) = cos^2x-1+cos^24x\\= cos^2x+cos^24x-1 <=> 1+cos(x+4x)*cos(x-4x)-1 =\\=cos(5x)*cos(-3x) <=> cos(5x)*cos(3x)$

La i) am folosit formula : $cos^2x+cos^2y=1+cos(x+y)*cos(x-y)$