OpenStax Algebra and Trigonometry · 9.4절 · 59번우리말 번역 다음 문제에서 항등식을 증명하세요. cos(2y)−cos(4y)sin(2y)+sin(4y)=tan y\frac{\cos (2y)-\cos (4y)}{\sin (2y)+\sin (4y)}=\tan \,ysin(2y)+sin(4y)cos(2y)−cos(4y)=tany 영어 원문 보기 For the following exercises, prove the identity. cos(2y)−cos(4y)sin(2y)+sin(4y)=tan y\frac{\cos (2y)-\cos (4y)}{\sin (2y)+\sin (4y)}=\tan \,ysin(2y)+sin(4y)cos(2y)−cos(4y)=tany 풀이 공책 펴기한 줄씩 식 확인정답·풀이 보기cos(2y)−cos(4y)sin(2y)+sin(4y)=−2 sin(3y)sin(−y)2 sin(3y)cos y=2 sin(3y)sin(y)2 sin(3y)cos y=tan y\begin{array}{ccc}\frac{\cos (2y)-\cos (4y)}{\sin (2y)+\sin (4y)} & = & \frac{-2\,\sin (3y)\sin (-y)}{2\,\sin (3y)\cos \,y}= \\ \frac{2\,\sin (3y)\sin (y)}{2\,\sin (3y)\cos \,y} & = & \tan \,y\end{array}sin(2y)+sin(4y)cos(2y)−cos(4y)2sin(3y)cosy2sin(3y)sin(y)==2sin(3y)cosy−2sin(3y)sin(−y)=tany 원본 교재 풀이(영어)cos(2y)−cos(4y)sin(2y)+sin(4y)=−2 sin(3y)sin(−y)2 sin(3y)cos y=2 sin(3y)sin(y)2 sin(3y)cos y=tan y\begin{array}{ccc}\frac{\cos (2y)-\cos (4y)}{\sin (2y)+\sin (4y)} & = & \frac{-2\,\sin (3y)\sin (-y)}{2\,\sin (3y)\cos \,y}= \\ \frac{2\,\sin (3y)\sin (y)}{2\,\sin (3y)\cos \,y} & = & \tan \,y\end{array}sin(2y)+sin(4y)cos(2y)−cos(4y)2sin(3y)cosy2sin(3y)sin(y)==2sin(3y)cosy−2sin(3y)sin(−y)=tany 한 번 더 — 비슷한 문제OpenStax Algebra and Trigonometry · 9.1절 · 21번우리말 번역 다음 문제에서 첫 번째 삼각식을 간단히 한 꼴을 두 번째 식으로 나타내세요. (sec x+csc x)(sin x+cos x)−2−cot x; tan x(\sec \,x+\csc \,x)(\sin \,x+\cos \,x)-2-\cot \,x;\,\tan \,x(secx+cscx)(sinx+cosx)−2−cotx;tanx 풀이 공책 펴기한 줄씩 식 확인답 쓰고 채점채점다 풀었으면 정답 확인tan x\tan \,xtanx 이 문제로 이어서 풀기 # 삼각함수의 뜻과 관계증명·논술자동 분류