Question

43 балла
ПОМОГИТЕ ПОЖАЛУЙСТА​

Answer 1

Ответ:

Объяснение:

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Answer 2

$(frac{sin4beta+2sin2beta}{2(cosbeta+cos3beta)}=frac{2sin2beta cos2beta+2sin2beta}{4cos2betacosbeta}=frac{2sin2beta(cos2beta+1)}{4cos2betacosbeta}=\=frac{tg2beta(cos2beta+1)}{2cosbeta}=frac{tg2beta cos^2beta}{cosbeta}=cosbetacdot tg2beta)$

$(frac{2cospsi+cos3psi + cos5psi}{cos3psi+sinpsisin2psi}=frac{2cospsi+2cos4psi cospsi}{cos3psi+0,5cospsi-0,5cos3psi}=frac{2cospsi(1+cos4psi)}{0,5cos3psi+0,5cospsi}=frac{2cospsi(1+cos4psi)}{0,5(cos3psi+cospsi)}=\=frac{2cospsi(1+cos4psi)}{0,5cdot2cos2psi cospsi}=frac{2cospsi(1+cos4psi)}{cos2psi cospsi}=frac{2(1+cos4psi)}{cos2psi}=frac{4(1+cos4psi)}{2cos2psi}=frac{4cos^22psi}{cos2psi}=4cos2psi)$