Question

СРОЧНО! $\frac{sin4b+2sin2b}{2(cosb+cos3b)}$

Answer 1

(sin4β+2sin2β)/(2(cosβ+cos3β)) =(2sin2βcos2β+2sin2β)/
(2*2cos(β+3β)/2*cos(β-3β)/2) =2sin2β(cos2β+1)/4cos2βcos(-β)=4sin2β*cos²β/(4cos2β*cosβ) =tq2β*cosβ.
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cos(-β)=cosβ