Ответ:
Найти значение выражения [tex]\bf cos\Big(2arctg\dfrac{1}{4}+arccos\dfrac{3}{5}\Big)[/tex] .
Применим формулу косинуса суммы углов .
[tex]\bf cos\Big(2arctg\dfrac{1}{4}+arccos\dfrac{3}{5}\Big)=\\\\=\bf cos\Big(2arctg\dfrac{1}{4}\Big)\cdot cos\Big(arccos\dfrac{3}{5}\Big)-sin\Big(2arctg\dfrac{1}{4}\Big)\cdot sin\Big(arccos\dfrac{3}{5}\Big)=[/tex]
Теперь распишем косинус синус двойных углов .
[tex]\bf =\left (2cos^2\Big(arctg\dfrac{1}{4}\Big)-1\right)\cdot cos\Big(arccos\dfrac{3}{5}\Big)-\\\\-2\cdot sin\Big(arctg\dfrac{1}{4}\Big)\cdot cos\Big(arctg\dfrac{1}{4}\Big)\cdot sin\Big(arccos\dfrac{3}{5}\Big)\ \ .[/tex]
Вычислим значения тригонометрических функций .
[tex]\bf 1)\ \ cos\Big(arctg\dfrac{1}{4}\Big)=cos\alpha \ \ \ \Rightarrow \ \ \ tg\alpha =\dfrac{1}{4}\ \ ,\\\\1+tg^2\alpha =\dfrac{1}{cos^2\alpha }\ \ \ \Rightarrow \ \ \ cos^2\alpha =\dfrac{1}{1+tg^2\alpha }=\dfrac{1}{1+\frac{1}{16}}=\dfrac{16}{17}\ ,\\\\cos\alpha=\dfrac{4}{\sqrt{17}}\ \ ,\ \ \ cos\Big(arctg\dfrac{1}{4}\Big)=\dfrac{4}{\sqrt{17}}[/tex]
[tex]\bf 2)\ \ sin\Big(arctg\dfrac{1}{4}\Big)=sin\alpha \ \ \ \Rightarrow \ \ tg\alpha =\dfrac{1}{4}\\\\sin\alpha =tg\alpha \cdot cos\alpha =\dfrac{1}{4}\cdot \dfrac{4}{\sqrt{17}}=\dfrac{1}{\sqrt{17}}[/tex]
[tex]\bf 3)\ \ cos\Big(arccos\dfrac{3}{5}\Big)=\dfrac{3}{5}[/tex]
[tex]\bf 4)\ \ sin\Big(arccos\dfrac{3}{5}\Big)=sin\beta \ \ \ \Rightarrow \ \ \ cos\beta =\dfrac{3}{5}\ \ ,\\\\sin^2\beta +cos^2\beta =1\ \ \ \Rightarrow \ \ \ sin^2\beta =1-cos^2\beta =1-\dfrac{9}{25}=\dfrac{16}{25}\ ,\\\\sin\beta =\dfrac{4}{5}\ \ ,\ \ \ sin\Big(arccos\dfrac{3}{5}\Big)=\dfrac{4}{5}[/tex]
Подставим значения тригонометрических функций в формулу.
[tex]\displaystyle \bf =\Big(2\cdot \dfrac{16}{17}-1\Big)\cdot \frac{3}{5}-2\cdot \frac{1}{\sqrt{17}}\cdot \frac{4}{\sqrt{17}\cdot}\frac{4}{5}=\frac{32-17}{17}\cdot \frac{3}{5}-\frac{32}{17\cdot 5}=\\\\\\=\frac{45-32}{17\cdot 5}=\frac{13}{85}[/tex]
[tex]\bf cos\Big(2arctg\dfrac{1}{4}+arccos\dfrac{3}{5}\Big)=\dfrac{13}{85}[/tex]
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Ответ:
Найти значение выражения [tex]\bf cos\Big(2arctg\dfrac{1}{4}+arccos\dfrac{3}{5}\Big)[/tex] .
Применим формулу косинуса суммы углов .
[tex]\bf cos\Big(2arctg\dfrac{1}{4}+arccos\dfrac{3}{5}\Big)=\\\\=\bf cos\Big(2arctg\dfrac{1}{4}\Big)\cdot cos\Big(arccos\dfrac{3}{5}\Big)-sin\Big(2arctg\dfrac{1}{4}\Big)\cdot sin\Big(arccos\dfrac{3}{5}\Big)=[/tex]
Теперь распишем косинус синус двойных углов .
[tex]\bf =\left (2cos^2\Big(arctg\dfrac{1}{4}\Big)-1\right)\cdot cos\Big(arccos\dfrac{3}{5}\Big)-\\\\-2\cdot sin\Big(arctg\dfrac{1}{4}\Big)\cdot cos\Big(arctg\dfrac{1}{4}\Big)\cdot sin\Big(arccos\dfrac{3}{5}\Big)\ \ .[/tex]
Вычислим значения тригонометрических функций .
[tex]\bf 1)\ \ cos\Big(arctg\dfrac{1}{4}\Big)=cos\alpha \ \ \ \Rightarrow \ \ \ tg\alpha =\dfrac{1}{4}\ \ ,\\\\1+tg^2\alpha =\dfrac{1}{cos^2\alpha }\ \ \ \Rightarrow \ \ \ cos^2\alpha =\dfrac{1}{1+tg^2\alpha }=\dfrac{1}{1+\frac{1}{16}}=\dfrac{16}{17}\ ,\\\\cos\alpha=\dfrac{4}{\sqrt{17}}\ \ ,\ \ \ cos\Big(arctg\dfrac{1}{4}\Big)=\dfrac{4}{\sqrt{17}}[/tex]
[tex]\bf 2)\ \ sin\Big(arctg\dfrac{1}{4}\Big)=sin\alpha \ \ \ \Rightarrow \ \ tg\alpha =\dfrac{1}{4}\\\\sin\alpha =tg\alpha \cdot cos\alpha =\dfrac{1}{4}\cdot \dfrac{4}{\sqrt{17}}=\dfrac{1}{\sqrt{17}}[/tex]
[tex]\bf 3)\ \ cos\Big(arccos\dfrac{3}{5}\Big)=\dfrac{3}{5}[/tex]
[tex]\bf 4)\ \ sin\Big(arccos\dfrac{3}{5}\Big)=sin\beta \ \ \ \Rightarrow \ \ \ cos\beta =\dfrac{3}{5}\ \ ,\\\\sin^2\beta +cos^2\beta =1\ \ \ \Rightarrow \ \ \ sin^2\beta =1-cos^2\beta =1-\dfrac{9}{25}=\dfrac{16}{25}\ ,\\\\sin\beta =\dfrac{4}{5}\ \ ,\ \ \ sin\Big(arccos\dfrac{3}{5}\Big)=\dfrac{4}{5}[/tex]
Подставим значения тригонометрических функций в формулу.
[tex]\displaystyle \bf =\Big(2\cdot \dfrac{16}{17}-1\Big)\cdot \frac{3}{5}-2\cdot \frac{1}{\sqrt{17}}\cdot \frac{4}{\sqrt{17}\cdot}\frac{4}{5}=\frac{32-17}{17}\cdot \frac{3}{5}-\frac{32}{17\cdot 5}=\\\\\\=\frac{45-32}{17\cdot 5}=\frac{13}{85}[/tex]
[tex]\bf cos\Big(2arctg\dfrac{1}{4}+arccos\dfrac{3}{5}\Big)=\dfrac{13}{85}[/tex]