Question

ДОКАЖИТЕ ИДЕНТИЧНОСТЬ! 1.) sin3α + sinα / cos3α+ cosα = tg2α2.) ( 2tgα/1-tg²α + tg2α) : sin2α/ cos2α = 2

Answer 1

sin2α+cos2α =11+tg2α=1/cos2α1+ctg2α=1/sin2αtgα *ctgα=1sin(α+β)=sinαcosβ +cosαsinβsin(α-β)=sinαcosβ-cosαsinβcos(α+β)=cosαcosβ-sinαsinβcos(α-β)=cosαcosβ+sinαsinβtg(α+β)=(tgα+tgβ)/(1-tgαtgβ)tg(α-β)=(tgα-tgβ)/(1+tgαtgβ)sin2α=2sinαcosβcos2α=cos2α-sin2α= 2cos2α-1=1-2 sin2αtg2α=2tg/(1-tg2α)sin2α=(1-cos2α)/2cos2α=(1+cos2α)/2tg(α/2)=sinα/(1+cos α)=(1-cos α)/sin αsin3α=3sinα-4sin3αcos3α=4cos3α-3cosαtg3α=(3tgα-tg3α)/(1-3tg2α)ctg3α=(ctg3α-3ctgα)/(3ctg2α-1)sin α+sin β=2sin[(α+β)/2]cos[(α-β)/2]sin α-sin β=2sin[(α-β)/2]cos[(α+β)/2]cosα+cosβ=2cos[(α+β)/2]cos[(α-β)/2]cosα-cosβ=-2sin[(α+β)/2]sin[(α-β)/2]tgα±tgβ=sin(α±β)/cosαcosβctgα±ctgβ=sin(β±α)/sinαsinβsinαcosβ=1/2*(sin(α+β)+sin(α-β))cosαcosβ=1/2*(cos(α+β)+cos(α-β))sinαsinβ=1/2*(cos(α-β)+cos(α+β))

Answer 2

Identity 1: sin(3α) + sin(α) / cos(3α) + cos(α) = tan(2α)

To prove the identity, we can start by using the trigonometric identity for the sum of two sines:

sin(A) + sin(B) = 2 * sin((A + B) / 2) * cos((A - B) / 2)

Applying this identity to the numerator of the left-hand side (LHS) of the equation, we have:

sin(3α) + sin(α) = 2 * sin((3α + α) / 2) * cos((3α - α) / 2)

Simplifying the numerator:

sin(3α) + sin(α) = 2 * sin(2α) * cos(α)

Now, let's apply the trigonometric identity for the sum of two cosines:

cos(A) + cos(B) = 2 * cos((A + B) / 2) * cos((A - B) / 2)

Applying this identity to the denominator of the LHS of the equation, we have:

cos(3α) + cos(α) = 2 * cos((3α + α) / 2) * cos((3α - α) / 2)

Simplifying the denominator:

cos(3α) + cos(α) = 2 * cos(2α) * cos(α)

Now, substituting the simplified numerator and denominator back into the original equation, we have:

(2 * sin(2α) * cos(α)) / (2 * cos(2α) * cos(α))

The 2's in the numerator and denominator cancel out, leaving us with:

sin(2α) / cos(2α)

And finally, using the trigonometric identity for the tangent:

tan(2α)

Therefore, we have proven that:

sin(3α) + sin(α) / cos(3α) + cos(α) = tan(2α)

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