Question

c$frac{cos2b-cos6b}{sin6b+sin2b}, где b=[tex] 22^{0} 30^{'}$

Answer 1

$b=22.30'\ frac{cos(2*22.30')-cos(6*22.30')}{sin(6*22.30])+sin(2*22.30')}=frac{cos45-cos135}{sin135+sin45}= frac{sqrt{2}}{sqrt{2}}=1$