Asosiy trigonometrik ayniyatlar

Ushbu darsda trigonometriyaning tayanch tenglamalari — asosiy ayniyatlar haqida so‘z yuritamiz. Ayniyat deganda o‘zgaruvchining barcha qiymatlarida to‘g‘ri bo‘lib qoladigan tenglik tushuniladi. Bu formulalar bitta trigonometrik funksiya orqali qolganlarini topish imkonini beradi.

§1. Pifagor ayniyati

Eng muhim ayniyat birlik aylanadan bevosita kelib chiqadi. Aylanadagi $P(\cos\alpha;\sin\alpha)$ nuqta uchun $x^2+y^2=1$ bo‘lgani sababli:

Qoida. $$ \sin^2\alpha+\cos^2\alpha=1. $$

Savol. Nima uchun bu "Pifagor" ayniyati? Chunki birlik aylanadagi radius — gipotenuzasi $1$, katetlari $\sin\alpha$ va $\cos\alpha$ bo‘lgan to‘g‘ri burchakli uchburchakning aynan Pifagor teoremasi.

Misol. $\sin\alpha=\frac{3}{5}$ va $\alpha$ o‘tkir burchak. $\cos\alpha$ ni toping.
Yechish. Pifagor ayniyatidan $\cos^2\alpha=1-\sin^2\alpha$:

$$\cos^2\alpha=1-\left(\frac{3}{5}\right)^2=1-\frac{9}{25}=\frac{16}{25}.$$

O‘tkir burchakda kosinus musbat, shuning uchun:

$$\cos\alpha=\sqrt{\frac{16}{25}}=\frac{4}{5}.$$

Javob. $\dfrac{4}{5}$.

§2. Tangens va kotangens bilan bog‘liq ayniyatlar

Tangens va kotangens sinus hamda kosinus orqali ta’riflanadi.

Ta’rif. $$ \tan\alpha=\frac{\sin\alpha}{\cos\alpha}\ (\cos\alpha\ne0),\qquad \cot\alpha=\frac{\cos\alpha}{\sin\alpha}\ (\sin\alpha\ne0). $$

Bulardan darrov ikkita foydali natija chiqadi:

Qoida. $$ \tan\alpha\cdot\cot\alpha=1,\qquad 1+\tan^2\alpha=\frac{1}{\cos^2\alpha},\qquad 1+\cot^2\alpha=\frac{1}{\sin^2\alpha}. $$

$1+\tan^2\alpha=\frac{1}{\cos^2\alpha}$ formulasi Pifagor ayniyatini $\cos^2\alpha$ ga bo‘lishdan kelib chiqadi:

Misol. $\tan\alpha=2$ bo‘lsa, $\frac{1}{\cos^2\alpha}$ ni toping.
Yechish. To‘g‘ridan-to‘g‘ri formulaga qo‘yamiz:

$$\frac{1}{\cos^2\alpha}=1+\tan^2\alpha=1+2^2=1+4=5.$$

Javob. $5$.

§3. Ayniyatlarni isbotlashda qo‘llash

Ayniyatlar ifodalarni soddalashtirishda ham juda foydali.

Misol. $\sin^4\alpha-\cos^4\alpha$ ifodani soddalashtiring.
Yechish. Kvadratlar ayirmasi sifatida yoyamiz, so‘ng Pifagor ayniyatini qo‘llaymiz:

$$\sin^4\alpha-\cos^4\alpha=(\sin^2\alpha-\cos^2\alpha)(\sin^2\alpha+\cos^2\alpha).$$

Ikkinchi qavs $1$ ga teng, demak:

$$=(\sin^2\alpha-\cos^2\alpha)\cdot1=\sin^2\alpha-\cos^2\alpha.$$

Javob. $\sin^2\alpha-\cos^2\alpha$.

Eslatma. Ildizdan chiqarganda ishorani unutmang. $\cos\alpha=\pm\sqrt{1-\sin^2\alpha}$ da qaysi belgini olishni burchak qaysi chorakda ekaniga qarab tanlaymiz.

Misol. $\cos\alpha=-\frac{5}{13}$ va $\alpha$ II chorakda. $\sin\alpha$ va $\tan\alpha$ ni toping.
Yechish. $\sin^2\alpha=1-\cos^2\alpha=1-\frac{25}{169}=\frac{144}{169}$. II chorakda sinus musbat:

$$\sin\alpha=\frac{12}{13},\qquad \tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{12/13}{-5/13}=-\frac{12}{5}.$$

Javob. $\sin\alpha=\dfrac{12}{13}$, $\tan\alpha=-\dfrac{12}{5}$.

§ Lug‘at

ayniyat — identity
tangens — tangent
kotangens — cotangent
soddalashtirish — simplification
isbotlash — proof
ifoda — expression