Question

sin2x+sin5x-sinx /cosx +cos2x +cos5x =tq2 доказать равенство

Answer 1

sinx+12sin2x=cosx+cos2x ⇔⇔ sinx+sinxcosx=cosx+2cos2x−1 ⇔⇔ sinx(1+cosx)=1+cosx+2(cosx−1)(cosx+1) ⇔⇔ 2sinxcos2x2=(1+cosx)(2cosx−1) ⇔⇔ sinxcos2x2=(2cosx−1)cos2x2 ⇔⇔ (sinx−2cosx+1)cos2x2=0.sin⁡x+12sin⁡2x=cos⁡x+cos⁡2x ⇔⇔ sin⁡x+sin⁡xcos⁡x=cos⁡x+2cos2⁡x−1 ⇔⇔ sin⁡x(1+cos⁡x)=1+cos⁡x+2(cos⁡x−1)(cos⁡x+1) ⇔⇔ 2sin⁡xcos2⁡x2=(1+cos⁡x)(2cos⁡x−1) ⇔⇔ sin⁡xcos2⁡x2=(2cos⁡x−1)cos2⁡x2 ⇔⇔ (sin⁡x−2cos⁡x+1)cos2⁡x2=0.