Oblicz wartość pozostałych funkcji trygonometrycznych kąta ostrego alfa jeśli: cos alfa = 0,8 cos alfa= 3/5 sin alfa=1/3 sin alfa=0,4 cos alfa=√13/7 tg alfa=3/2 tg alfa= 2 całości i 2/5 ctg alfa= 4/5 Gdybym mogł prosic to bardziej rozpisane

Oblicz wartość pozostałych funkcji trygonometrycznych kąta ostrego alfa jeśli: cos alfa = 0,8 cos alfa= 3/5 sin alfa=1/3 sin alfa=0,4 cos alfa=√13/7 tg alfa=3/2 tg alfa= 2 całości i 2/5 ctg alfa= 4/5 Gdybym mogł prosic to bardziej rozpisane
Odpowiedź

Korzystamy z trzech podstawowych tożsamości trygonometrycznych: [latex]sin^2alpha+cos^2alpha=1 ; qquadquad tgalpha=frac{sinalpha}{cosalpha} ;qquadquad ctgalpha=frac1{tgalpha}[/latex] [latex]1.\ cosalpha=0,8\\sin^2alpha+(0,8)^2=1\sin^2alpha+0,64=1\sin^2alpha=0,36\underline{sinalpha=0,6}\\ tgalpha=frac{0,6}{0,8}=frac34\\ ctg=dfrac1{frac34}=frac43\\\ 2.\ cosalpha=frac35=0,6 \\sin^2alpha+(0,6)^2=1\sin^2alpha+0,36=1\sin^2alpha=0,64\underline{sinalpha=0,8}\\ tgalpha=frac{0,8}{0,6}=frac43\\ ctg=dfrac1{frac43}=frac34[/latex] [latex]3.\ sinalpha=frac13\\(frac13)^2+cos^2alpha=1\frac19+cos^2alpha=1\cos^2alpha=frac89\underline{cosalpha=frac{sqrt8}3=frac{2sqrt2}3}\\ tgalpha=dfrac{frac13}{frac{2sqrt2}3}=frac13cdotfrac3{2sqrt2}=frac1{2sqrt2}=frac{sqrt2}4\\ ctg=dfrac1{frac1{2sqrt2}}=frac{2sqrt2}1=2sqrt2[/latex] [latex]4.\sinalpha=0,4=frac25\\(0,4)^2+cos^2alpha=1\0,16+cos^2alpha=1\cos^2alpha=0,84=frac{84}{100}=frac{21}{25}\underline{cosalpha=frac{sqrt{21}}5}\\ tgalpha=dfrac{frac25}{frac{sqrt{21}}5}=frac{ 2}{sqrt{21}}=frac{2sqrt{21}}{21}\\ ctgalpha=dfrac1{frac{ 2}{sqrt{21}}}=frac{sqrt{21}}{ 2}[/latex] [latex]5.\ cosalpha=frac{sqrt{13}}7\\sin^2alpha+(frac{sqrt{13}}7)^2=1\sin^2alpha+frac{13}{49}=1\~ sin^2alpha=frac{36}{49}\underline{sinalpha=frac{6}{7}}\\ tgalpha=dfrac{frac{6}{7}}{frac{sqrt{13}}7}=frac{6}{7}cdotfrac7{sqrt{13}}=frac6{sqrt{13}}=frac{6sqrt{13}}{13}\\ ctg=dfrac1{frac6{sqrt{13}}}=frac{sqrt{13}}6[/latex] [latex]6.\ tgalpha=frac32\\ctgalpha=dfrac{1}{frac32}=frac23\\\frac{sinalpha}{cosalpha}=frac32qquad/cdotcosalpha\~ sinalpha=frac32cosalpha implies cosalpha=frac23sinalpha\\sin^2alpha+(frac23sinalpha)^2=1\sin^2alpha+frac49sin^2alpha=1\frac{13}9sin^2alpha=1quad/:frac{13}9\sin^2alpha=frac9{13}\underline{sinalpha=frac3{sqrt{13}}=frac{3sqrt{13}}{13}}[/latex] [latex](frac3{sqrt{13}})^2+cos^2alpha=1\frac9{13}+cos^2alpha=1\cos^2alpha=frac4{13}\underline{ cosalpha=frac2{sqrt{13}}=frac{2sqrt{13}}{13} }[/latex] [latex]7.\tgalpha=2frac25=frac{12}5\\ctgalpha=dfrac{1}{frac{12}5}=frac5{12}\\\frac{sinalpha}{cosalpha}=frac{12}5qquad/cdotcosalpha\~ sinalpha=frac{12}5cosalpha implies cosalpha=frac5{12}sinalpha\\sin^2alpha+(frac5{12}sinalpha)^2=1\sin^2alpha+frac{25}{144}sin^2alpha=1\frac{169}{144}sin^2alpha=1qquad/:frac{169}{144}\sin^2alpha=frac{144}{169}\underline{sinalpha=frac{12}{13}}\\frac{144}{169}+cos^2alpha=1\cos^2alpha=frac{25}{169}\cosalpha=frac{5}{13}[/latex] [latex]8.\ctgalpha=frac45\\frac45=dfrac{1}{tgalpha}quadimpliesquad tgalpha=frac5{4}\\\frac{sinalpha}{cosalpha}=frac54qquad/cdotcosalpha\~ sinalpha=frac54cosalpha implies cosalpha=frac45sinalpha\\sin^2alpha+(frac45sinalpha)^2=1\sin^2alpha+frac{16}{25}sin^2alpha=1\frac{41}{25}sin^2alpha=1qquad/:frac{41}{25}\sin^2alpha=frac{25}{41}\underline{sinalpha=frac5{sqrt{41}}=frac{5sqrt{41}}{41}}[/latex] [latex](frac5{sqrt{41}})^2+cos^2alpha=1\frac{25}{41}+cos^2alpha=1\cos^2alpha=frac{16}{41}\underline{ cosalpha=frac4{sqrt{41}}=frac{4sqrt{41}}{41} }[/latex]

Dodaj swoją odpowiedź