Yarim burchak formulalari

Ushbu darsda burchakni ikkiga bo‘lganda hosil bo‘ladigan yarim burchak formulalari haqida so‘z yuritamiz. Ular ikkilangan burchak formulalarini "teskari o‘qish" natijasi bo‘lib, jadvalda bo‘lmagan burchaklarni aniq hisoblashga yordam beradi.

§1. Yarim burchakning sinusi va kosinusi

Ta’rif. $\frac{\alpha}{2}$ burchak funksiyasini butun $\alpha$ burchak funksiyalari orqali ifodalovchi tengliklar yarim burchak formulalari deyiladi.

Darajani pasaytirish formulalarida $\alpha$ o‘rniga $\frac{\alpha}{2}$ qo‘yamiz. $\cos^2\frac{\alpha}{2}=\frac{1+\cos\alpha}{2}$ dan ildiz olib:

Qoida. $$ \sin\frac{\alpha}{2}=\pm\sqrt{\frac{1-\cos\alpha}{2}},\qquad \cos\frac{\alpha}{2}=\pm\sqrt{\frac{1+\cos\alpha}{2}}. $$

Eslatma. Ishora ($\pm$) $\frac{\alpha}{2}$ qaysi chorakda yotishiga qarab bittasi tanlanadi — ikkalasi emas. Masalan, $\frac{\alpha}{2}$ I chorakda bo‘lsa, ikkalasi ham musbat.

Savol. Nima uchun ildiz ostida $1-\cos\alpha$ turadi? Chunki $\cos\alpha=1-2\sin^2\frac{\alpha}{2}$, bundan $\sin^2\frac{\alpha}{2}=\frac{1-\cos\alpha}{2}$ — biz shunchaki ikkilangan burchak formulasini teskari yozdik.

Misol. $\sin15^\circ$ ning aniq qiymatini toping.
Yechish. $15^\circ=\frac{30^\circ}{2}$ deb olamiz, $\cos30^\circ=\frac{\sqrt3}{2}$. $15^\circ$ I chorakda, sinus musbat:

$$\sin15^\circ=\sqrt{\frac{1-\cos30^\circ}{2}}=\sqrt{\frac{1-\frac{\sqrt3}{2}}{2}}=\sqrt{\frac{2-\sqrt3}{4}}=\frac{\sqrt{2-\sqrt3}}{2}.$$

Javob. $\dfrac{\sqrt{2-\sqrt3}}{2}$.

§2. Yarim burchakning tangensi

Tangens uchun ildizsiz, ancha qulay formulalar mavjud:

Qoida. $$ \tan\frac{\alpha}{2}=\frac{1-\cos\alpha}{\sin\alpha}=\frac{\sin\alpha}{1+\cos\alpha}. $$

Bu ko‘rinishlar $\pm$ ishorasidan xoli — shuning uchun amalda ular ildizli formuladan qulayroq.

Misol. $\cos\alpha=\frac{3}{5}$, $\sin\alpha=\frac{4}{5}$. $\tan\frac{\alpha}{2}$ ni toping.
Yechish. Ikkinchi ko‘rinishni olamiz:

$$\tan\frac{\alpha}{2}=\frac{\sin\alpha}{1+\cos\alpha}=\frac{\frac{4}{5}}{1+\frac{3}{5}}=\frac{\frac{4}{5}}{\frac{8}{5}}=\frac{4}{8}=\frac{1}{2}.$$

Javob. $\dfrac{1}{2}$.

§3. Universal almashtirish

Yarim burchak tangensi $t=\tan\frac{\alpha}{2}$ orqali barcha funksiyalarni ratsional ifodalash mumkin — bu "universal trigonometrik almashtirish".

Qoida. $t=\tan\frac{\alpha}{2}$ bo‘lsa: $$ \sin\alpha=\frac{2t}{1+t^2},\qquad \cos\alpha=\frac{1-t^2}{1+t^2},\qquad \tan\alpha=\frac{2t}{1-t^2}. $$

Misol. $\tan\frac{\alpha}{2}=\frac{1}{3}$ bo‘lsa, $\sin\alpha$ ni toping.
Yechish. $t=\frac{1}{3}$, demak $t^2=\frac{1}{9}$. Universal formulani qo‘llaymiz:

$$\sin\alpha=\frac{2t}{1+t^2}=\frac{2\cdot\frac{1}{3}}{1+\frac{1}{9}}=\frac{\frac{2}{3}}{\frac{10}{9}}=\frac{2}{3}\cdot\frac{9}{10}=\frac{18}{30}=\frac{3}{5}.$$

Javob. $\dfrac{3}{5}$.

Misol. $\frac{1-\cos\alpha}{\sin\alpha}$ ni soddalashtiring.
Yechish. Bu aynan yarim burchak tangensining birinchi ko‘rinishi:

$$\frac{1-\cos\alpha}{\sin\alpha}=\tan\frac{\alpha}{2}.$$

Javob. $\tan\dfrac{\alpha}{2}$.

§ Lug‘at

yarim burchak — half angle
universal almashtirish — universal substitution
ratsional ifoda — rational expression
ildiz — root
ishora — sign
qulaylik — convenience