Question

Bantuin aku nomor 20 dan 21 yaa, makasih banyak

Answer 1

20. Nilai limx->π/3[(cosx-sinπ/6)/(π/6 - x/2)]=√3

21. Nilai dari limx->π/4[(cos2x/(cosx-sinx)]=√2

Pembahasan:

beberapa identitas yang diperlukan untuk kasus ini adalah

sin(π/2-x)=cosx

sina-sinb=2cos1/2(a+b) sin1/2(a-b)

cos2x=cos²x-sin²x

Kembali kesoal

limx->π/3[(cosx-sinπ/6):(π/6 - x/2)]

=limx->π/3[sin(π/2-x)-sinπ/6:(π/6 - x/2)]

=limx->π/3[2cos1/2(π/2-x+π/6) sin1/2(π/2-x-π/6):(π/6 - x/2)]

=limx->π/3[2cos1/2(4π/6 - x) sin1/2(2π/6 - x):(π/6-x/2)]

=limx->π/3[2cos(2π/6-x/2) sin[(π/6-x/2):(π/6-x/2)]

=2 cos(2π/6-π/6) lim(x-π/3)->0 [sin1/2(π/3-x):1/2(π/3-x)]

=2 cosπ/6

=2(1/2√3)

=√3

note:lim(x-π/3)->0 [sin1/2(π/3-x):1/2(π/3-x)]=1

        limx->π/3=lim(x-π/3)->0

20.limx->π/4[(cos2x/(cosx-sinx)]

=limx->π/4[cos²x-sin²x/(cosx-sinx)]

=limx->π/4[((cosx+sinx)(cosx-sinx)/(cosx-sinx)]

=limx->π/4[cosx+sinx]

=cosπ/4+sinπ/4

=1/2√2+1/2√2

=√2

note:cos²x-sin²x=(cosx+sinx)(cosx-sinx)

Pelajari juga:

brainly.co.id/tugas/5036976

Detai jawaban:

Mapel:Matematika

Kelas:11

Kode soal:2

Bab:8

Kode kategori:11.2.8

Kata kunci:Limit trigonometri

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