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_se3PhMcIp0uhmsjr-T34RAxcJ88kING6veAzABlca5aiJairmxr7XvNL0T42gZCYeBw0CbE8_mIAhSMUiB55Np800zbsxDJL0aA_eu-uvvneEnc_SRqB-WkTtbIfe7Wex0ugArTBk5ykEBcN_srjNHUB5TFStZDqolZ_GFFNc=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_sAYO0KKQe0El2BlxLDjXoXrnPQte7fEqbmUDwgFjqwV96F4JCDYyFoCS1MWYUKo6c3uHK8OIlnZO4xYSVUBkY_kShN8WKGkG27PoQ54xK7xIHZwqCNU_vb3Luu80DI98SKwJClNdbFoy5Rcoyr46rJUt3HmwacJJ_sWTXhEO9Qx7L2bgLfuJtUTs5Ytw=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_senV833W01puxjgPG4wnYFgi-IkzOrJY4N_C2CVJEb6qicNW9lrGtUvx6kFBezdKY6S3WDzvlRZAhs-8LGV7qFxbPr16ij55mDTzQQPWBiLAnYc5W_DCT_nVPmuNcw_mRVG0eG-xd_Ka7SUAo0PxuXebPQ7Qg=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_uH_N4XZih_zxbferSlfYmRhsTJTyvNe7UMXtMGvgSmcet6leboeq-IahtyBLSGT5iIhwNZjqkMw_b701jhTO1IICKTDEma0TtMijbaLK467U09D955Px6HvD-N8nO-kDB7CoV_UrSoQxg2kFJ2Fao=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_tFu0e83bOQi11TRVT-HArW_VJlJbXhcUWueZfJcEm3xFKxlNtreanUQeTGZ8Eeayv9BqGsBp6m2Sx1W3wqDnqtW_DVDRjo4HuffpbhMhAY7XKYqbu2Ty034t4HM0Hbi7QUB08YhOuZgHP1laqXtTItKZRVP9U=s0-d)
[sin(2nπ+θ)=sinθ]
② ![[tex]cos(2npi+ heta)=cos heta[/tex]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sNl01ojofPYDdcId6zOJUpSZ1bhzzbGd0zD_AOC6nyh7zt_BDdqz_bNvW-9TpXBQ4AV3ScgwSvu2IEdTJ5V7vMi011uS2dI-yVZUwC42I4rA2SoDp-Wh3GZVBPY6gE0ZlYZ2IPP8_MWOQ79Z9NH2dhlw=s0-d)
[cos(2nπ+θ)=cosθ]
③ ![[tex] an(2npi+ heta)= an heta[/tex]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uK-8phRWYMulD2baFW5RmZfdINWy97QHZ_d1eCtaRI57_4e7sdUNclLS4-SFklw1lO-d91o2BP7pSN4mfNVWt_Eedx8rtiQtOxfwAQyd29rowV3ohh5hHsJdYCPQLSXtYDKp3iOscJlizwM_dmcXU-WNkYm5o=s0-d)
[tan(2nπ+θ)=tanθ]
(2) -θ의 삼각함수
④ ![[tex]sin(- heta)=-sin heta[/tex]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s3TBsAr7YoLvfnJvjxOxc1nnyb1SKaug3-3x5rqpp9Rvyz4TO51AH3YlQTsQwhO8shvgSYIqF1ZDfoZ5sXiFTcgHTk6W7MrBH1kNCC6c5AkbpVWij0qE5gmHXh3BJcuQncembeEcUpbmRUmcFj4O4=s0-d)
[sin(-θ)=-sinθ]
⑤ ![[tex]cos(- heta)=cos heta[/tex]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vbYmmP1j6GITezCs02GyA1Xa7wjU7wyqtfb5QUQHrF4SdEzUXqZ8muKXvQYbu72IJSO8HC8nvYubzmzd7aC-JV3AbqUuR8PRDLvP38lNCNjQVaG-Ea3xXxgRmey8nfGXU3-iGwQTkoyg1-1Q=s0-d)
[cos(-θ)=cosθ]
⑥ ![[tex] an(- heta)=- an heta[/tex]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uxiosap8FPJFD9PC7Ab0_dBk6PrDHiEauT3RsvuOGzMSgVXqL9Y3EzJUU0Heyc_GarrSd9mnvRHpZc8icoON8Yr2Tlb8HOlrDDxMfZRma7tZvNaocSvmeN_-MIvUXvwpS7l3Kb9OWalxba1aDifw=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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