Keltirish formulalari: matematika darsi

Jadvalda faqat $0^\circ$ dan $90^\circ$ gacha qiymatlar bor, lekin imtihonda $\sin 240^\circ$ yoki $\operatorname{tg}1665^\circ$ so'raladi. Keltirish formulalari istalgan burchakni o'tkir burchakka keltiradi. Formulalar soni ko'p, ammo ularning hammasi bitta qisqa qoidaga bo'ysunadi.

Keltirish qoidasi

Nom: burchak $\pi\pm\alpha$ , $2\pi\pm\alpha$ ko'rinishida bo'lsa, funksiya nomi o'zgarmaydi; $\dfrac{\pi}{2}\pm\alpha$ , $\dfrac{3\pi}{2}\pm\alpha$ ko'rinishida bo'lsa, nom almashadi: $\sin\leftrightarrow\cos$ , $\operatorname{tg}\leftrightarrow\operatorname{ctg}$ .

Ishora: $\alpha$ ni shartli ravishda o'tkir burchak deb olamiz, berilgan burchak qaysi chorakka tushishini aniqlaymiz va boshlang'ich funksiyaning shu chorakdagi ishorasini natija oldiga qo'yamiz.

Misol: $\cos\left(\dfrac{3\pi}{2}+\alpha\right)$ . Burchakda $\dfrac{3\pi}{2}$ turibdi, nom almashadi: $\cos$ o'rniga $\sin$ . O'tkir $\alpha$ da burchak IV chorakda, boshlang'ich funksiya kosinus u yerda musbat. Natija $\sin\alpha$ . $\alpha$ ni o'tkir deb olish faqat ishorani topish usuli; tayyor formula har qanday $\alpha$ uchun o'rinli.

To'liq jadval

Burchak	$\sin$	$\cos$	$\operatorname{tg}$	$\operatorname{ctg}$
Burchak $-\alpha$	$\sin$ $-\sin\alpha$	$\cos$ $\cos\alpha$	$\operatorname{tg}$ $-\operatorname{tg}\alpha$	$\operatorname{ctg}$ $-\operatorname{ctg}\alpha$
Burchak $\dfrac{\pi}{2}-\alpha$	$\sin$ $\cos\alpha$	$\cos$ $\sin\alpha$	$\operatorname{tg}$ $\operatorname{ctg}\alpha$	$\operatorname{ctg}$ $\operatorname{tg}\alpha$
Burchak $\dfrac{\pi}{2}+\alpha$	$\sin$ $\cos\alpha$	$\cos$ $-\sin\alpha$	$\operatorname{tg}$ $-\operatorname{ctg}\alpha$	$\operatorname{ctg}$ $-\operatorname{tg}\alpha$
Burchak $\pi-\alpha$	$\sin$ $\sin\alpha$	$\cos$ $-\cos\alpha$	$\operatorname{tg}$ $-\operatorname{tg}\alpha$	$\operatorname{ctg}$ $-\operatorname{ctg}\alpha$
Burchak $\pi+\alpha$	$\sin$ $-\sin\alpha$	$\cos$ $-\cos\alpha$	$\operatorname{tg}$ $\operatorname{tg}\alpha$	$\operatorname{ctg}$ $\operatorname{ctg}\alpha$
Burchak $\dfrac{3\pi}{2}-\alpha$	$\sin$ $-\cos\alpha$	$\cos$ $-\sin\alpha$	$\operatorname{tg}$ $\operatorname{ctg}\alpha$	$\operatorname{ctg}$ $\operatorname{tg}\alpha$
Burchak $\dfrac{3\pi}{2}+\alpha$	$\sin$ $-\cos\alpha$	$\cos$ $\sin\alpha$	$\operatorname{tg}$ $-\operatorname{ctg}\alpha$	$\operatorname{ctg}$ $-\operatorname{tg}\alpha$
Burchak $2\pi-\alpha$	$\sin$ $-\sin\alpha$	$\cos$ $\cos\alpha$	$\operatorname{tg}$ $-\operatorname{tg}\alpha$	$\operatorname{ctg}$ $-\operatorname{ctg}\alpha$
Burchak $2\pi+\alpha$	$\sin$ $\sin\alpha$	$\cos$ $\cos\alpha$	$\operatorname{tg}$ $\operatorname{tg}\alpha$	$\operatorname{ctg}$ $\operatorname{ctg}\alpha$

Tangens va kotangens qatnashgan tengliklar ikkala tomon mavjud bo'lganda o'rinli: masalan, $\operatorname{tg}\left(\dfrac{\pi}{2}-\alpha\right)=\operatorname{ctg}\alpha$ uchun $\sin\alpha\ne 0$ kerak.

To'ldiruvchi burchaklar

Yig'indisi $90^\circ$ bo'lgan ikki burchakda sinus va kosinus, tangens va kotangens o'rin almashadi. Bu imtihonda eng ko'p ishlatiladigan juftlik:

\sin(90^\circ-\alpha)=\cos\alpha

\operatorname{tg}(90^\circ-\alpha)=\operatorname{ctg}\alpha

Masalan, $\sin 70^\circ=\cos 20^\circ$ , $\operatorname{tg}35^\circ=\operatorname{ctg}55^\circ$ . Natijada bunday burchaklar juftlab qisqaradi yoki qo'shiladi:

Ifoda	Sababi	Qiymati
Ifoda $\operatorname{tg}35^\circ\cdot\operatorname{tg}55^\circ$	Sababi $\operatorname{tg}55^\circ=\operatorname{ctg}35^\circ$	Qiymati $1$
Ifoda $\log_2\operatorname{ctg}35^\circ+\log_2\operatorname{ctg}55^\circ$	Sababilogarifmlar yig'indisi $\log_2 1$	Qiymati $0$
Ifoda $\sin^2 10^\circ+\sin^2 80^\circ$	Sababi $\sin 80^\circ=\cos 10^\circ$	Qiymati $1$
Ifoda $\sin\dfrac{\pi}{8}-\cos\dfrac{3\pi}{8}$	Sababi $\dfrac{\pi}{8}+\dfrac{3\pi}{8}=\dfrac{\pi}{2}$	Qiymati $0$

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