ขอความช่วยเหลือหน่อยครับ เรื่องตรีโกณครับ

[Toy City]

ขอบคุณมากๆนะครับ

[กิตติ]

1.$\cos^3 10^\circ +\sin^3 20^\circ=(\cos 10^\circ +\sin 20^\circ)(\cos^2 10^\circ +\sin^2 20^\circ-\cos 10^\circ \sin 20^\circ) $
$\frac{\cos^3 10^\circ +\sin^3 20^\circ}{\cos 10^\circ +\sin 20^\circ} =(\cos^2 10^\circ +\sin^2 20^\circ-2\cos 10^\circ \sin 20^\circ)+\cos 10^\circ \sin 20^\circ$
$=(\cos 10^\circ -\sin 20^\circ )^2+\cos 10^\circ \sin 20^\circ $
$=(\sin 80^\circ -\sin 20^\circ )^2+\cos 10^\circ \sin 20^\circ$
$=(2\cos 50^\circ \sin 30^\circ )^2+\cos 10^\circ \sin 20^\circ$
$=\cos^2 50^\circ+\frac{1}{2} (2\cos 10^\circ \sin 20^\circ)$
$=\sin^2 40^\circ+\frac{1}{2} (\sin 30^\circ +\sin 10^\circ)$
$=\sin^2 40^\circ+\frac{1}{2} (\sin 30^\circ +\cos 80^\circ)$
$=\sin^2 40^\circ+\frac{1}{4}+\frac{1}{2} (1-2\sin^2 40^\circ)$
$=\frac{3}{4} $

[กิตติ]

ข้อ 2 น่าจะให้พิสูจน์มากกว่าจะแก้สมการ
2. $\cos^2 \theta \sin^3 \theta=\frac{1}{16} (2\sin \theta+\sin 3\theta-\sin 5\theta)$

$2\sin \theta+\sin 3\theta-\sin 5\theta$
$=(\sin \theta+\sin 3\theta)+(\sin \theta-\sin 5\theta) $
$=2\sin 2\theta \cos \theta +2\cos 3\theta \sin (-2\theta)$
$=2\sin 2\theta \cos \theta -2\cos 3\theta \sin (2\theta)$
$=2\sin 2\theta (\cos \theta -\cos 3\theta )$
$=4 \sin \theta \cos \theta(-2 \sin 2\theta \sin (-\theta)$
$=8 \sin \theta \cos \theta(\sin 2\theta \sin (\theta))$
$=16 \sin^3 \theta \cos^2 \theta$

$\sin^3 \theta \cos^2 \theta=\frac{1}{16} (2\sin \theta+\sin 3\theta-\sin 5\theta)$

[I am Me.]

ข้อความเดิมเขียนโดยคุณ กิตติ

1.$\cos^3 10^\circ +\sin^3 20^\circ=(\cos 10^\circ +\sin 20^\circ)$($\cos^2 10^\circ +\sin^2 20^\circ$$-\cos 10^\circ \sin 20^\circ) $

$\frac{\cos^3 10^\circ +\sin^3 20^\circ}{\cos 10^\circ +\sin 20^\circ} =$$1$$-\cos 10^\circ \sin 20^\circ$

$=1-\frac{1}{2}(2\cos 10^\circ \sin 20^\circ) $

$=1-\frac{1}{2}(\sin (20^\circ+10^\circ )+\sin (20^\circ-10^\circ)) $

$=1-\frac{1}{2}(\sin (30^\circ)+\sin (10^\circ))$

$=\frac{3}{4} -\frac{1}{2}(\sin (10^\circ))$

มันเป็น 10 องศากับ 20 องศา นะครับ หรือว่าผมเข้าใจผิด

[กิตติ]

ใช่ครับ เดี๋ยวแก้ใหม่ครับ
3.$\cos^2 \theta \sin^4 \theta=\frac{1}{32}(2-\cos 2\theta-2\cos 4 \theta+\cos 6\theta) $

$2-\cos 2\theta-2\cos 4 \theta+\cos 6\theta$
$=2-2\cos 4 \theta+(\cos 6\theta-\cos 2 \theta)$
$=2-2\cos 4 \theta-2\sin 4\theta \sin 2\theta$
$=2-2(1-2\sin^2 2\theta)-4\sin^2 2\theta \cos 2\theta$
$=4\sin^2 2\theta-4\sin^2 2\theta \cos 2\theta$
$=4\sin^2 2\theta (1-\cos 2\theta)$
$=16\sin^2 \theta \cos^2 \theta(2\sin^2 \theta)$
$=32\sin^4 \theta \cos^2 \theta$

[กิตติ]

4.$\sin A+\sin B=1$.......(1)
$\cos A+\cos B=\frac{1}{2} $........(2)
$\sin 2A+\sin 2B=2\sin (A+B) \cos(A-B)$

$\sin A+\sin B=1=2\sin(\frac{A+B)}{2})\cos (\frac{A-B)}{2}) $.....(3)
$\cos A+\cos B=\frac{1}{2} =2\cos(\frac{A+B)}{2})\cos (\frac{A-B)}{2})$...(4)
(3)หารด้วย(4)
$\tan (\frac{A+B)}{2}) =2$
$\sin 2A=\frac{2\tan A}{1+\tan^2 A} $
$\sin(A+B)=\frac{4}{5} $

(1)ยกกำลังสอง+(2)ยกกำลังสอง
$2+2\sin A \sin B+2\cos A \cos B=\frac{5}{4} $
$2\cos (A-B)=-\frac{3}{4} $
$\cos (A-B)=-\frac{3}{8}$

$\sin 2A+\sin 2B=2\sin (A+B) \cos(A-B)=2\times \frac{4}{5}\times (-\frac{3}{8})= -\frac{3}{5}$

[กิตติ]

7.$\dfrac{1}{\sin 10^\circ } -\dfrac{\sqrt{3} }{\cos 10^\circ } $
$=\dfrac{\cos 10^\circ -\sqrt{3}\sin 10^\circ }{\sin 10^\circ \cos 10^\circ } $
$=2\left(\,\dfrac{\dfrac{\cos 10^\circ}{2}-\dfrac{\sqrt{3}\sin 10^\circ}{2} }{\sin 10^\circ \cos 10^\circ} \right) $

$=4(\frac{\sin 30^\circ \cos 10^\circ-\cos 30^\circ\sin 10^\circ}{\sin 20^\circ} )$

$=4(\frac{\sin (30^\circ- 10^\circ)}{\sin 20^\circ} )$

$=4$

[กิตติ]

5.$\dfrac{\sin 23A-\sin 3A}{\sin 16A+\sin 4A} $

$=\dfrac{\cos 13A \sin 10A}{\sin 10A \cos 6A} $

$=\dfrac{\cos 13A }{ \cos 6A}$

แทนค่ามุม Aลงไปจะได้

$=\dfrac{\cos \frac{13\pi}{19} }{ \cos \frac{6\pi}{19} }$

$=\dfrac{\cos (\pi-\frac{6\pi}{19}) }{ \cos \frac{6\pi}{19} }$

$=-1$

[กิตติ]

6.$\cos(A+B)=\frac{1}{2} =1-2\sin^2 (\frac{A+B}{2}) $
$\sin^2 (\frac{A+B}{2})=\frac{1}{4} \rightarrow \sin (\frac{A+B}{2})=\frac{1}{2}$
$\sin A+\sin B=\frac{1}{4}=2\sin (\frac{A+B}{2}) \cos (\frac{A-B}{2}) $
จะได้ว่า $\cos (\frac{A-B}{2})=\frac{1}{4} \rightarrow \tan (\frac{A-B}{2})=\frac{\sqrt{15} }{4} $