Question

Довести тотожність:
(sin4α/cosα+cos4α/sinα)*(1/cos3α-1/cosα)=4tgα

Answer 1

(sin4α/cosα+cos4α/sinα)*(1/cos3α-1/cosα)=

((sin4α*sinα+cos4α*cosα)/(sinα*cosα))*((cosα-cos3α)/(cos3α*cosα))=

cos3α*2sin2α*sinα/(sinα*cosα*cos3α*cosα)=2sin2α/(cos²α)=

4sinα*cosα/cos²α=4sinα/cosα=4tgα

Доказано.

использовал формулы sin2α=2sinα*сosα

cosα*cosβ+sinα*sinβ=cos(α-β)

cosα-cosβ=2(sin(β-α)/2)sin((β+α)/2)

Answer 2

$(\frac{Sin4\alpha }{Cos\alpha}+\frac{Cos4\alpha }{Sin\alpha})*(\frac{1}{Cos3\alpha}-\frac{1}{Cos\alpha}})=\frac{Sin4\alpha Sin\alpha+Cos4\alpha Cos\alpha}{Sin\alpha Cos\alpha}*\frac{Cos\alpha-Cos3\alpha}{Cos\alpha Cos3\alpha}= \\\\\ =\frac{Cos(4\alpha-\alpha)}{Sin\alpha Cos\alpha} *\frac{-2Sin\frac{\alpha+3\alpha}{2}*Sin\frac{\alpha-3\alpha}{2}}{Cos\alpha Cos3\alpha} =\frac{Cos3\alpha }{Sin\alpha Cos\alpha}*\frac{2Sin2\alpha Sin\alpha}{Cos\alpha Cos3\alpha}=$

$=\frac{2Sin2\alpha }{Cos^{2}\alpha}=\frac{2*2Sin\alpha Cos\alpha}{Cos^{2}\alpha}=\frac{4Sin\alpha }{Cos\alpha }=4tg\alpha\\\\4tg\alpha=4tg\alpha$

Тождество доказано