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WorksheetsҚосу, айырма, қос бұрыш, дәреже төмендету
Total questions: 33
Worksheet time: 17mins
sin(α + β) = ?
sinα·cosβ + cosα·sinβ
sinα·cosβ − cosα·sinβ
cosα·cosβ − sinα·sinβ
cosα·cosβ + sinα·sinβ
sin(α − β) = ?
sinα·cosβ − cosα·sinβ
sinα·cosβ + cosα·sinβ
cosα·cosβ − sinα·sinβ
cosα·cosβ + sinα·sinβ
cos(α + β) = ?
cosα·cosβ − sinα·sinβ
cosα·cosβ + sinα·sinβ
sinα·cosβ + cosα·sinβ
sinα·cosβ − cosα·sinβ
cos(α − β) = ?
cosα·cosβ + sinα·sinβ
cosα·cosβ − sinα·sinβ
sinα·cosβ − cosα·sinβ
sinα·cosβ + cosα·sinβ
tg(α + β) = ?
(tgα + tgβ) / (1 − tgα·tgβ)
(tgα − tgβ) / (1 + tgα·tgβ)
(1 − tgα·tgβ) / (tgα + tgβ)
(1 + tgα·tgβ) / (tgα − tgβ)
tg(α − β) = ?
(tgα − tgβ) / (1 + tgα·tgβ)
(tgα + tgβ) / (1 − tgα·tgβ)
(1 + tgα·tgβ) / (tgα − tgβ)
(1 − tgα·tgβ) / (tgα + tgβ)
ctg(α + β) = ?
(ctgα·ctgβ − 1) / (ctgβ + ctgα)
(1 − ctgα·ctgβ) / (ctgβ + ctgα)
(ctgα·ctgβ + 1) / (ctgβ − ctgα)
(1 + ctgα·ctgβ) / (ctgβ − ctgα)
ctg(α − β) = ?
(ctgα·ctgβ + 1) / (ctgβ − ctgα)
(ctgα·ctgβ − 1) / (ctgβ + ctgα)
(1 + ctgα·ctgβ) / (ctgβ + ctgα)
(1 − ctgα·ctgβ) / (ctgβ − ctgα)
Егер sinα = 3/5 және cosβ = 4/5 болса, онда sin(α + β) = ?
24/25
7/25
12/25
18/25
Егер cosα = 4/5, cosβ = 3/5 болса, онда cos(α − β) = ?
24/25
7/25
12/25
18/25
sin(2α) = ?
2sinα·cosα
sin²α + cos²α
2cosα·cosα
sinα + cosα
cos(2α) = ?
cos²α − sin²α
2cosα·sinα
1 − 2sin²α
sin²α + cos²α
cos(2α) формуласын басқа түрде жаз:
1 − 2sin²α
2cos²α + 1
sin²α − cos²α
1 + 2sin²α
cos(2α) формуласын тағы бір түрде жаз:
2cos²α − 1
1 − 2cos²α
2sin²α − 1
sin²α − cos²α
tg(2α) = ?
2tgα / (1 − tg²α)
1 − 2tg²α
2tgα / (1 + tg²α)
tg²α / (1 − tgα)
ctg(2α) = ?
(ctg²α − 1) / (2ctgα)
(1 − ctg²α) / (2ctgα)
(2ctgα) / (1 − ctg²α)
2ctgα / (ctg²α − 1)
sin(α/2) = ?
±√((1 − cosα)/2)
±√((1 + cosα)/2)
±√((1 + sinα)/2)
±√((1 − sinα)/2)
cos(α/2) = ?
±√((1 + cosα)/2)
±√((1 − cosα)/2)
±√((1 + sinα)/2)
±√((1 − sinα)/2)
tg(α/2) = ?
±√((1 − cosα)/(1 + cosα))
±√((1 + cosα)/(1 − cosα))
±√((1 + sinα)/(1 − sinα))
±√((1 − sinα)/(1 + sinα))
tg(α/2)-нің тағы бір түрі:
sinα / (1 + cosα)
sinα / (1 − cosα)
cosα / (1 + sinα)
cosα / (1 − sinα)
tg(α/2) формуласын басқа түрде жазуға болады:
(1 − cosα) / sinα
(1 + cosα) / sinα
sinα / (1 − cosα)
cosα / sinα
ctg(α/2) = ?
(1 + cosα) / sinα
(1 − cosα) / sinα
sinα / (1 + cosα)
sinα / (1 − cosα)
sin²α = ?
(1 − cos2α)/2
(1 + cos2α)/2
(1 − sin2α)/2
(1 + sin2α)/2
cos²α = ?
(1 + cos2α)/2
(1 − cos2α)/2
(1 + sin2α)/2
(1
sin²α + cos²α = ?
1
0
sin2α
cos2α
sinA + sinB = ?
2sin((A + B)/2)·cos((A − B)/2)
2cos((A + B)/2)·sin((A − B)/2)
sin(A + B)
cos(A + B)
sinA − sinB = ?
2cos((A + B)/2)·sin((A − B)/2)
2sin((A − B)/2)·cos((A + B)/2)
2cos((A + B)/2)·sin((A − B)/2)
sin(A − B)
cosA + cosB = ?
2cos((A + B)/2)·cos((A − B)/2)
2sin((A + B)/2)·sin((A − B)/2)
cos(A + B) + cos(A − B)
2cosA·cosB
cosA − cosB = ?
−2sin((A + B)/2)·sin((A − B)/2)
2cos((A + B)/2)·cos((A − B)/2)
2sin((A + B)/2)·sin((A − B)/2)
cos(A − B)
sinA·sinB = ?
½[cos(A − B) − cos(A + B)]
½[cos(A + B) − cos(A − B)]
½[sin(A − B) + sin(A + B)]
½[sin(A + B) − sin(A − B)]
cosA·cosB = ?
½[cos(A − B) + cos(A + B)]
½[cos(A + B) − cos(A − B)]
½[sin(A − B) + sin(A + B)]
½[sin(A + B) − sin(A − B)]
sinA·cosB = ?
½[sin(A + B) + sin(A − B)]
½[sin(A + B) − sin(A − B)]
½[cos(A − B) − cos(A + B)]
½[cos(A + B) + cos(A − B)]
cosA·sinB = ?
½[sin(A + B) − sin(A − B)]
½[sin(A + B) + sin(A − B)]
½[cos(A − B) − cos(A + B)]
½[cos(A + B) + cos(A − B)]
