진행시각 | 교재쪽 | 제목 | 설명 | | 14초 | 20쪽 | 삼각함수 사이의 관계 | 삼각함수 사이의 관계
① , ,![[tex]cot heta=frac{1}{ an heta}[/tex]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tWyKGgAi0hRDy3aFQ6_SGeYJFETvNtIRF_vVf9RfvEnU_c2sZsccHIplp2zR0XelDEzVCa184mfs9lQlwb7BgXSi-6QhR4ptaoq2mpGR0sTe2MQahmwxNzBT6Z4aLKsdKxOix07n-xB-mNLlQWHPEIvwu9JFD1hMR7tqjnOEc=s0-d)
[cscθ = 1/sinθ, secθ = 1/cosθ, cotθ = 1/tanθ]
② , ![[tex]cot heta=frac{cos heta}{sin heta}[/tex]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vEapytwo0iC0ZdvBGmhz0wE2we68-mC8Us2lAN60YZQbUwU_MKQ6oreuM4jHABK-sB_hD551mM07ZG8U4lnXhwWxRDrE8gY8BUZmKhlxFiQHb_IZpVtNw1DEiUIuDxP-4WGxbD9cDV4j1CcP0UmD7A4WqF1OtPWf7bWuU5x-X76o3t-DHyZOzUcbXKSw=s0-d)
[tanθ = sinθ/cosθ, cotθ = cosθ/sinθ]
③ ![[tex]sin^2 heta+cos^2 heta=1[/tex]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ujsrSlnHy-b6GvZsDY1qHJSASYdspmBDXVQd_PpCuYCge30crNy4wDurpGz0f9f6nYnyeM-rzbQ7Agf3WOMdvLcdjVs10yLMUyw6DeCPQZ5g2lccR-dZfOTH092y_uccGWfDggbsW1ZY5k3wvjsI9J7gZ9mVg=s0-d)
[sin^2θ + cos^2θ=1]
④ , ![[tex]1+cot^2 heta=csc^2 heta[/tex]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vDjqUrb3Tcu-SYCVTYeSKD3oDu1ccJH-GrzoGDgMsH2H8oiMIeMgGykQsIRFP6WzZmcY-PFlWHp6ZRW5n8MtjtNqBK3GmCcvRRalTVyGhaKwwtFRsNkq8AI8-s9Cf31lCnFr4lurfo6sQvUd-JC_M=s0-d)
[ tan^2θ+1 = sec^2θ, 1+cot^2θ = csc^2θ] | | 9분 20초 | 21쪽 | 예제1) | [sinθ+cosθ=1/√2]일 때, 다음 식의 값을 구하여라.
(1) [sinθcosθ]
(2) [cos^3θ + sin^3θ ]
(3) [tanθ + cotθ]
| | 17분 29초 | 22쪽 | 예제2) | 이차방정식 [2x^2+px-1=0]의 두 근이 sinθ, cosθ 일 때, 상수 p의 값을 구하여라. | | 20분 6초 | 23쪽 | 예제3) | [sinθ+cosθ = (1-√3)/2]일 때, sinθ, cosθ 를 두 근으로 하는 x에 대한 이차방정식을 구하여라.
| | 23분 50초 | 23쪽 | 여러 가지 각의 삼각함수 | (1) 2nπ+θ의 삼각함수(단, n은 정수)
①![[tex]sin(2npi+ heta)=sin heta[/tex]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vVURaCNg_FhmNU6WTIK6AbU9c4jGyJZeiZpuLHcZVQzH5zqYKb7tpFXj-iR1Jv3te8lNI2fkl83v7K_IZyeeXYBT2H81efqmheLQ0f7eCq1c2wQaWRS5XWSE7DdwU_fTdvhm_ywMpFIk53SWU4n_5frvlzNig=s0-d)
[sin(2nπ+θ)=sinθ]
② ![[tex]cos(2npi+ heta)=cos heta[/tex]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tIzUjGYU0VyX-gon1sxsmHxQvNF5pvRXBTeESmJk2_ppyeresvD5TjD9h0fV5DhkIWc3hw1v7EFXxHigC6bnN3syGuoChn2MTn0WMxRDYud1QLy8iHuCxXzFUKaTu6bGCGhk6sCqpgpR9ZgVgdUsWnhw=s0-d)
[cos(2nπ+θ)=cosθ]
③ ![[tex] an(2npi+ heta)= an heta[/tex]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u2LFWLb7PFscGyLLzEqv9ZS9xUP5CzFq1vX4U3cMmI8v4U0pHmVaDSsOqpu8YGethHhZcPFDUY8xvtCS0Oqx5bgCJthub0WWQz5Og7xDd1iw95hcrnhFOtgAP2BGTcZ-79YlTDVnYqNxzhccvU6_v6Olx4v40=s0-d)
[tan(2nπ+θ)=tanθ]
(2) -θ의 삼각함수
④ ![[tex]sin(- heta)=-sin heta[/tex]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tXkddq-sPA5rxWfQ9gcXP8Xpg7ZifWq2PDy9E_UJuIXfu8r8hMZLr38Qu4MQ-SYbZjMeyNXQcNwavpe7kB9kOwWHpUU-dHjpSXS7x7GBEJDkzJS2MNlcSEOI7l1Y46JbNWi5ehIbhkkklfdiNDPVQ=s0-d)
[sin(-θ)=-sinθ]
⑤ ![[tex]cos(- heta)=cos heta[/tex]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_use-wTQV0hp8sqMfQrOyg6Jp9dNLGiWRUgLXcpT_1LEMidMnCK3h64_UomA9kQPS9ZWBXRtN5Goje5AHLw1-MUE8ybPN5Qx6SmicHeNCfrZsNhB3a1-yke97XVREj46w7HE_ur-v1WQf2yBQ=s0-d)
[cos(-θ)=cosθ]
⑥ ![[tex] an(- heta)=- an heta[/tex]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sMpiiBsdUzogiA3WIjZf2yZmzx4KWz-ymQLLjnrng6jX3qO4zX_9toa-lXF0oReeiZztpH1kIFNBIPwsj47lFCt3SYbUmboPsNQ3R1XKYG-FA3G1wMAX6vd-wdJ9-rR_7AxrGDhBitTHlFvshrcw=s0-d)
[tan(-θ)=-tanθ] | | 51분 6초 | 29쪽 | 예제1) | 다음 삼각함수의 값을 구하여라.
(1) [cos 15/4 π ]
(2) [sin 3/4 π ]
(3) [cos 7/6 π ]
(4) tan 150° | | 55분 15초 | 30쪽 | 삼각함수 사이의 관계 보충설명 | | | 1시간 1분 | 30쪽 | 예제2) | 다음 식의 값을 구하여라.
(1) sin50° + tan110° + cos140° + cot200°
(2) (sin10° + cos10° )2 + (sin80° - cos80°)2
(3) tan(20° + θ) · tan(70° - θ) | |
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