Maktab dasturi· 35 daqiqa· 21 ta savol
Avval o‘rganing: Trigonometrik ayniyatlar
Bu mavzudagi formulalar ikki burchak funksiyalarini bog‘laydi. Ular yordamida jadvalda yo‘q burchaklarni (15°, 75°, 105°) hisoblash, ifodalarni soddalashtirish, trigonometrik tenglamalarni yechish va funksiyaning eng katta hamda eng kichik qiymatini topish mumkin. Barcha formulalar bitta — cos(α − β) formulasidan kelib chiqadi, shuning uchun mantiqni tushunish yodlashdan muhimroq: biror formulani unutsangiz, uni bir-ikki qatorda tiklay olasiz.
Birlik aylanada A(cos α; sin α) va B(cos β; sin β) nuqtalarni olamiz. OA va OB vektorlar orasidagi burchak α − β, ularning uzunligi 1. Skalyar ko‘paytmani ikki usulda yozamiz:
cos(α − β) = cos α cos β + sin α sin β — barcha formulalarning manbai
Qolganlari shundan almashtirishlar bilan olinadi: β ni −β ga almashtirsak — cos(α + β); keltirish formulasi sin x = cos(π/2 − x) bilan — sin(α ± β); sin ni cos ga bo‘lsak — tg(α ± β); β = α desak — qo‘sh burchak formulalari.
sin(α ± β) = sin α cos β ± cos α sin β
cos(α ± β) = cos α cos β ∓ sin α sin β
tg(α ± β) = (tg α ± tg β)/(1 ∓ tg α tg β)
Diqqat: cos formulasida ishora teskari — yig‘indida minus, ayirmada plyus. sin formulasida esa "aralash" ko‘paytmalar (sin · cos), cos formulasida — "bir xil" ko‘paytmalar (cos · cos, sin · sin).
Bu formulalar ikki tomonga ham o‘qiladi: sin 40° cos 20° + cos 40° sin 20° = sin 60° = √3/2.
sin 75° ni hisoblang.
Xuddi shunday cos 15° = cos(45° − 30°) = (√6 + √2)/4, tg 15° = 2 − √3.
tg α = 2, tg β = 3. tg(α + β) ni toping.
β = α deb olsak:
sin 2α = 2 sin α cos α
cos 2α = cos²α − sin²α = 2cos²α − 1 = 1 − 2sin²α
tg 2α = 2tg α/(1 − tg²α)
sin 3α = 3sin α − 4sin³α, cos 3α = 4cos³α − 3cos α
cos 2α ning uch ko‘rinishi asosiy ayniyat sin²α + cos²α = 1 yordamida bir-biridan olinadi — masalaga qaysi biri qulay bo‘lsa, shuni tanlang.
Darajani pasaytirish (integral va tenglamalarda juda kerak) — cos 2α formulasidan sin²α va cos²α ni ifodalash:
sin²α = (1 − cos 2α)/2, cos²α = (1 + cos 2α)/2
Yarim burchak — shu formulalarda α ni α/2 ga almashtirish: sin²(α/2) = (1 − cos α)/2. Masalan, sin 15° = √((1 − cos 30°)/2) = (√6 − √2)/4 (15° I chorakda, ishora +).
Tayyor qoliplarni taniy bilish: 2 sin 15° cos 15° = sin 30° = 1/2; cos²15° − sin²15° = cos 30° = √3/2; 1 − 2sin²22,5° = cos 45° = √2/2.
sin α = 3/5, α I chorakda. sin 2α va cos 2α ni toping.
cos 20° · cos 40° · cos 80° ni hisoblang.
sin α + sin β = 2 sin((α + β)/2) cos((α − β)/2)
sin α − sin β = 2 cos((α + β)/2) sin((α − β)/2)
cos α + cos β = 2 cos((α + β)/2) cos((α − β)/2)
cos α − cos β = −2 sin((α + β)/2) sin((α − β)/2)
Ular §2 formulalarini qo‘shishdan kelib chiqadi: sin(x + y) + sin(x − y) = 2 sin x cos y, so‘ng x + y = α, x − y = β deymiz. Asosiy foydasi — tenglamada yig‘indini ko‘paytmaga aylantirib, har bir ko‘paytuvchini nolga tenglash.
(cos 3α + cos α)/(sin 3α − sin α) ni soddalashtiring.
sin α cos β = ½(sin(α + β) + sin(α − β))
cos α cos β = ½(cos(α − β) + cos(α + β))
sin α sin β = ½(cos(α − β) − cos(α + β))
Tenglamalarda ko‘paytma bo‘lsa, uni yig‘indiga aylantirib, ikki tomonni tenglashtirish qulay; integralda esa ko‘paytmani integrallash oson emas, yig‘indini — oson.
Masalan, sin 15° sin 75° = ½(cos 60° − cos 90°) = ½ · ½ = 1/4.
a sin x + b cos x ko‘rinishidagi ifodani bitta sinusga keltirish mumkin:
a sin x + b cos x = √(a² + b²) · sin(x + φ), bu yerda cos φ = a/√(a² + b²), sin φ = b/√(a² + b²)
Isboti — sin(x + φ) ni §2 formulasi bilan ochish. Natija: ifodaning eng katta qiymati √(a² + b²), eng kichigi −√(a² + b²). Masalan, √3 sin x + cos x = 2 sin(x + π/6), chunki √(3 + 1) = 2, cos φ = √3/2, sin φ = 1/2.
y = 3 sin x + 4 cos x ning eng katta va eng kichik qiymatlarini toping.