Ikkilangan burchak formulalari

Ushbu darsda qo‘shish formulalarining muhim xususiy holi — ikkilangan burchak ($2\alpha$) formulalari haqida so‘z yuritamiz. Ular $\alpha=\beta$ deb olganda qo‘shish formulalaridan darrov kelib chiqadi va juda ko‘p masalalarda ishlatiladi.

§1. Sinus va kosinusning ikkilangan burchagi

Ta’rif. $2\alpha$ burchak funksiyasini $\alpha$ burchak funksiyalari orqali ifodalovchi tengliklar ikkilangan burchak formulalari deyiladi.

$\sin(\alpha+\beta)$ da $\beta=\alpha$ deb olsak:

Qoida. $$ \sin2\alpha=2\sin\alpha\cos\alpha. $$

Xuddi shunday $\cos(\alpha+\alpha)$ dan:

Qoida. $$ \cos2\alpha=\cos^2\alpha-\sin^2\alpha=2\cos^2\alpha-1=1-2\sin^2\alpha. $$

Eslatma. Kosinusning uchta ko‘rinishi bejiz emas. Masalada faqat $\cos\alpha$ berilgan bo‘lsa $2\cos^2\alpha-1$, faqat $\sin\alpha$ berilgan bo‘lsa $1-2\sin^2\alpha$ shaklini tanlang — hisob qisqaradi.

Savol. $2\cos^2\alpha-1$ va $1-2\sin^2\alpha$ o‘zaro teng bo‘lishi mumkinmi? Ha — chunki ularning ayirmasi $2\cos^2\alpha-1-(1-2\sin^2\alpha)=2(\cos^2\alpha+\sin^2\alpha)-2=2-2=0$.

Misol. $\sin\alpha=\frac{3}{5}$, $\cos\alpha=\frac{4}{5}$. $\sin2\alpha$ va $\cos2\alpha$ ni toping.
Yechish. Formulalarga qo‘yamiz:

$$\sin2\alpha=2\cdot\frac{3}{5}\cdot\frac{4}{5}=\frac{24}{25}.$$

$$\cos2\alpha=1-2\sin^2\alpha=1-2\cdot\frac{9}{25}=1-\frac{18}{25}=\frac{7}{25}.$$

Javob. $\sin2\alpha=\dfrac{24}{25}$, $\cos2\alpha=\dfrac{7}{25}$.

§2. Tangensning ikkilangan burchagi va darajani pasaytirish

Qoida. $$ \tan2\alpha=\frac{2\tan\alpha}{1-\tan^2\alpha}. $$

Kosinus formulasini teskari o‘girsak, darajani pasaytirish (kvadratni yechish) formulalari chiqadi:

Qoida. $$ \sin^2\alpha=\frac{1-\cos2\alpha}{2},\qquad \cos^2\alpha=\frac{1+\cos2\alpha}{2}. $$

Misol. $\cos^2 15^\circ$ ni hisoblang.
Yechish. Darajani pasaytirish formulasi bilan, $2\alpha=30^\circ$:

$$\cos^2 15^\circ=\frac{1+\cos30^\circ}{2}=\frac{1+\frac{\sqrt3}{2}}{2}=\frac{2+\sqrt3}{4}.$$

Javob. $\dfrac{2+\sqrt3}{4}$.

§3. Formulalarni qo‘llash

Ikkilangan burchak formulalari ifodalarni soddalashtirishda ayniqsa kuchli.

Misol. $2\sin15^\circ\cos15^\circ$ ni hisoblang.
Yechish. Bu $\sin2\alpha$ formulasining o‘ng tomoni, $\alpha=15^\circ$:

$$2\sin15^\circ\cos15^\circ=\sin(2\cdot15^\circ)=\sin30^\circ=\frac{1}{2}.$$

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

Misol. $\dfrac{\sin2\alpha}{1+\cos2\alpha}$ ni soddalashtiring.
Yechish. Suratni $\sin2\alpha=2\sin\alpha\cos\alpha$, maxrajni $1+\cos2\alpha=2\cos^2\alpha$ bilan almashtiramiz:

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

Javob. $\tan\alpha$.

§ Lug‘at

ikkilangan burchak — double angle
darajani pasaytirish — power reduction
xususiy hol — special case
soddalashtirish — simplification
maxraj — denominator
surat — numerator