Sinuslar va kosinuslar teoremalari: matematika darsi

Tomonlari $7$ va $8$ , orasidagi burchagi $120^\circ$ bo'lgan uchburchakning uchinchi tomoni butun son chiqadi: $13$ . To'g'ri burchak yo'q joyda Pifagor ishlamaydi, uning o'rnini kosinuslar teoremasi egallaydi. Sinuslar teoremasi esa tomonni qarshisidagi burchak va tashqi aylana radiusi bilan bog'laydi. Bu ikki teorema bilan istalgan uchburchak yechiladi.

Belgilashlar va burchaklar qiymatlari

$a$ , $b$ , $c$ tomonlar, $\alpha$ , $\beta$ , $\gamma$ mos ravishda ular qarshisidagi burchaklar, ya'ni $a=BC$ qarshisida $\angle A=\alpha$ , $R$ tashqi chizilgan aylana radiusi.

$\alpha$	$\sin\alpha$	$\cos\alpha$
$\alpha$ $30^\circ$	$\sin\alpha$ $\dfrac{1}{2}$	$\cos\alpha$ $\dfrac{\sqrt3}{2}$
$\alpha$ $45^\circ$	$\sin\alpha$ $\dfrac{\sqrt2}{2}$	$\cos\alpha$ $\dfrac{\sqrt2}{2}$
$\alpha$ $60^\circ$	$\sin\alpha$ $\dfrac{\sqrt3}{2}$	$\cos\alpha$ $\dfrac{1}{2}$
$\alpha$ $90^\circ$	$\sin\alpha$ $1$	$\cos\alpha$ $0$
$\alpha$ $120^\circ$	$\sin\alpha$ $\dfrac{\sqrt3}{2}$	$\cos\alpha$ $-\dfrac{1}{2}$
$\alpha$ $135^\circ$	$\sin\alpha$ $\dfrac{\sqrt2}{2}$	$\cos\alpha$ $-\dfrac{\sqrt2}{2}$
$\alpha$ $150^\circ$	$\sin\alpha$ $\dfrac{1}{2}$	$\cos\alpha$ $-\dfrac{\sqrt3}{2}$
$\alpha$ $15^\circ$	$\sin\alpha$ $\dfrac{\sqrt6-\sqrt2}{4}$	$\cos\alpha$ $\dfrac{\sqrt6+\sqrt2}{4}$
$\alpha$ $75^\circ$ , $105^\circ$	$\sin\alpha$ $\dfrac{\sqrt6+\sqrt2}{4}$	$\cos\alpha$ $75^\circ$ da $\dfrac{\sqrt6-\sqrt2}{4}$ , $105^\circ$ da $-\dfrac{\sqrt6-\sqrt2}{4}$

\sin 105^\circ=\sin(60^\circ+45^\circ)

=\frac{\sqrt3}{2}\cdot\frac{\sqrt2}{2}+\frac{1}{2}\cdot\frac{\sqrt2}{2}

=\frac{\sqrt6+\sqrt2}{4}

Jadvaldagi qiymat yig'indi sinusi formulasi bilan shu tarzda topiladi

\sqrt{2+\sqrt3}=\frac{\sqrt6+\sqrt2}{2}

Bu qiymat ildiz ostida ham uchraydi: o'ng tomon kvadrati

{\dfrac{8+2\sqrt{12}}{4}=2+\sqrt3}

Uchburchakda hamma burchak $0^\circ$ va $180^\circ$ orasida, shuning uchun sinus doim musbat, kosinus esa o'tkir burchakda musbat, o'tmasda manfiy. Sinus ma'lum bo'lsa, $\cos\gamma=\pm\sqrt{1-\sin^2\gamma}$ , ishora burchak turiga qarab tanlanadi.

Kosinuslar teoremasi

c^2=a^2+b^2-2ab\cos\gamma

Tomon kvadrati: qolgan ikki tomon kvadratlari yig'indisi minus ularning ikkilangan ko'paytmasi va orasidagi burchak kosinusi

Isbot uchun

CH

balandlikni tushiramiz

CH=b\sin\alpha

AH=b\cos\alpha

BH=c-b\cos\alpha

a^2=CH^2+BH^2

a^2=b^2+c^2-2bc\cos\alpha

CHB

uchburchakda Pifagor, keyin

{\sin^2\alpha+\cos^2\alpha=1}

Teorema har qanday burchak uchun o'rinli. $\gamma=90^\circ$ da $\cos\gamma=0$ va Pifagor qoladi. O'tmas burchakda minus ishora plyusga aylanadi: $120^\circ$ da $c^2=a^2+b^2+ab$ , $60^\circ$ da $c^2=a^2+b^2-ab$ .

\cos\gamma=\frac{a^2+b^2-c^2}{2ab}

Uch tomon ma'lum bo'lsa, burchak topiladi. Manfiy natija o'tmas burchakni bildiradi

\cos\gamma=\frac{9+25-49}{30}=-\frac{1}{2}

Tomonlar

3

5

7

da eng katta burchak

120^\circ

Teskari masala: $a^2=b^2+c^2-\sqrt2\,bc$ ko'rinish berilsa, uni teorema bilan solishtiramiz: $2\cos\alpha=\sqrt2$ , ya'ni $\alpha=45^\circ$ .

Eng katta tomon $c$ uchun	Uchburchak	Misol
Eng katta tomon $c$ uchun $c^2<a^2+b^2$	Uchburchako'tkir burchakli	Misol $5,\ 6,\ 7$ : $49<61$
Eng katta tomon $c$ uchun $c^2=a^2+b^2$	Uchburchakto'g'ri burchakli	Misol $5,\ 12,\ 13$ : $169=169$
Eng katta tomon $c$ uchun $c^2>a^2+b^2$	Uchburchako'tmas burchakli	Misol $4,\ 5,\ 7$ : $49>41$

c^2=(a+b)^2-2ab(1+\cos\gamma)

a+b

\gamma

berilganda

ab

topiladi, undan yuz:

S=\dfrac{1}{2}ab\sin\gamma

$\alpha=60^\circ$ bo'lsa, $a^2=b^2-bc+c^2$ , shuning uchun:

b^3+c^3=(b+c)(b^2-bc+c^2)

=(b+c)a^2

\frac{a^3+b^3+c^3}{a+b+c}=a^2

Tomonlar emas, ifoda so'ralganda shunday qisqarishlar qidiriladi

Tangens so'ralsa, avval kosinus, keyin sinus topiladi va ${\mathrm{tg}\,\gamma=\dfrac{\sin\gamma}{\cos\gamma}}$ . Tomonlar $5$ , $6$ , $7$ da $5$ qarshisidagi burchak uchun:

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