Тригонометрични функции - sin, cos, tg, cotg - трудни задачи с решения

Решение:
$\cos^4x-\sin^4x=(\cos^2x)^2-(\sin^2x)^2$
$=(\cos^2 x+\sin^2 x)(\cos^2 x-\sin^2 x)$             $/a^2-b^2=(a+b)(a-b)$

$=\cos^2x-\sin^2x$         $/\cos^2 x+\sin^2 x = 1$
$=\cos^2x-(1-\cos^2x)$         $/\sin^2 x = 1 - \cos^2 x$

$=2\cos^2x-1$

Решение:
Привеждаме лявата страна във вида: [tex]tg4\alpha-{\frac{1}{cos4\alpha}}={\frac{tg4\alpha cos4\alpha-1}{cos4\alpha}}={\frac{sin4\alpha-1}{cos4\alpha}}={\frac{2sin2\alpha cos2\alpha-(sin^22\alpha+cos^22\alpha)}{cos^22\alpha-sin^22\alpha}}={\frac{-(cos^22\alpha-sin^22\alpha)^2}{(cos2\alpha-sin2\alpha)(cos2\alpha+sin2\alpha)}}={\frac{sin2\alpha-cos2\alpha}{sin2\alpha+cos2\alpha}}[/tex]

Решение:
[tex]{\frac{1-cos2x}{2}}+cos2x={\frac{1}{4}}; 1+cos2x={\frac{1}{2}}; cos2x=-{\frac{1}{2}}; 2x=2\kappa \pi \pm{\frac{2\pi}{3}}; x=\kappa \pi \pm{\frac{\pi}{3}}[/tex].

Решение:
[tex]sin5x-cos7x=0; sin5x-sin({\frac{\pi}{2}}-7x)=0; 2sin(6x-{\frac{\pi}{4}})cos({\frac{\pi}{4}}-x)=0; sin(6x-{\frac{\pi}{4}})cos({\frac{\pi}{4}}-x)=0[/tex].
[tex]sin(6x-{\frac{\pi}{4}})=0;6x-{\frac{\pi}{4}}=\kappa \pi;x=\kappa{\frac{\pi}{6}}+{\frac{\pi}{24}}[/tex].
[tex]cos({\frac{\pi}{4}}-x)=0; x-{\frac{\pi}{4}}=2\kappa \pi \pm{\frac{\pi}{2}}, x=2\kappa \pi \pm{\frac{\pi}{2}}+{\frac{\pi}{4}}[/tex].

Решение:
Тъй като [tex]cos^2x = 1-sin^2x[/tex], получаваме квадратно уравнение [tex]sin x[/tex]:
[tex]1-\sin^2x+\sin x={\frac{5}{4}}[/tex]
[tex]4\sin^2x-4sin x+1=0[/tex]
[tex](2\sin x-1)^2=0[/tex]
[tex]2\sin x-1=0[/tex]
[tex]\sin x={\frac{1}{2}}[/tex]
[tex]x=n\pi+(-1)^n{\frac{\pi}{6}}[/tex]

Решение:
[tex]2sin^2x+3sin x cos x+7cos^2x=6(sin^2x+cos^2x), 4sin^2x-3sin x cos x-cos^2x=0[/tex] => [tex]cos x\ne0[/tex]. Разделяме двете страни на [tex]cos^2x[/tex]
[tex]{\frac{4sin^2x}{cos^2x}}-{\frac{3sin x cos x}{cos^2x}}-{\frac{cos^2x}{cos^2x}}=0; 4tg^2x-3tg x-1=0;tg x={\frac{3\pm\sqrt{9+16}}{8}}={\frac{3\pm5}{8}}[/tex].
а)[tex]tg x =1; x=\kappa \pi \pm+{\frac{\pi}{4}}[/tex].
б)[tex]tg x=-{\frac{1}{4}}; x\approx180^{\circ}\kappa-13^{\circ}30\prime[/tex].

Решение:
Използваме формула [tex]sin\alpha cos\beta={\frac{sin(\alpha+\beta)+sin(\alpha-\beta)}{2}}; {\frac{sin4x+sin2x}{2}}={\frac{sin12x+sin2x}{2}}; sin4x=sin12x; sin12x-sin4x=0; 2sin4x cos8x=0[/tex]. a) [tex]sin4x=0; 4x=\kappa\pi; x=\kappa{\frac{\pi}{4}}[/tex]. б)[tex]cos8x=0; 8x=2\kappa\pi\pm{\frac{\pi}{2}}; x=\kappa{\frac{\pi}{4}}\pm{\frac{x}{16}}[/tex].

Решение:
Свеждаме уравнението до квадратно относно [tex]tg{\frac{x}{2}}[/tex]: [tex]3{\frac{1-tg^2{\frac{x}{2}}}{1+tg^2{\frac{x}{2}}}}+2\sqrt{3}\cdot{\frac{2tg{\frac{x}{2}}}{1+tg^2{\frac{x}{2}}}}={\frac{9}{2}}; 15tg^2{\frac{x}{2}}-8\sqrt{3}tg{\frac{x}{2}}+3=0[/tex]. a)[tex]tg{\frac{\sqrt{3}}{3}}; {\frac{x}{2}}=\kappa\pi+{\frac{\pi}{6}}; x=2\kappa\pi+{\frac{\pi}{3}}[/tex]. б)[tex]tg{\frac{x}{2}}={\frac{\sqrt{3}}{5}}; {\frac{x}{2}}\approx180^{\circ}\kappa+13^{\circ}; x\approx360^{\circ}\kappa+26^{\circ}[/tex].

Решение:
Използваме формулата [tex]sin\alpha cos\alpha={\frac{sin2\alpha}{2}}; {\frac{sin2x}{2}}cos2x={\frac{1}{8}}; sin2x cos2x={\frac{1}{4}}[/tex]. Още един път използваме формулата [tex]sin\alpha cos\alpha={\frac{sin2\alpha}{2}}[/tex] и получаваме: [tex]{\frac{sin4x}{2}}={\frac{1}{4}}; sin4x={\frac{1}{2}}; 4x=n\pi+(-1)^n{\frac{\pi}{24}}[/tex].

Решение:
Към лявата страна прибавяме 2 взаимно унищожяеми се члена [tex]2sin^2x cos^2x[/tex] и [tex]-2sin^2x cos^2x[/tex]:[tex]sin^4x+2sin^2cos^2x+2cos^4x-2sin^2x cos^2x={\frac{5}{8}}; 1-2\cdot({\frac{sin2x}{2}})^2={\frac{5}{8}}; {\frac{sin^22x}{2}}={\frac{3}{8}}; sin^22x={\frac{3}{4}}; sin2x=\pm{\frac{\sqrt{3}}{2}}[/tex].
[tex]sin2x={\frac{\sqrt{3}}{2}}; 2x=n\pi+(-1)^2{\frac{\pi}{3}}; x=n{\frac{\pi}{2}}+(-1)^n{\frac{\pi}{6}}[/tex].

Решение:
Ако [tex]cos x=0[/tex], за да имаме решените трябва и [tex]sin x =0[/tex]. Но [tex]sin x[/tex] и [tex]cos x[/tex] не могат да са 0 едновременно. Затова можем да разделим двете страни на [tex]cos x[/tex]:[tex]{\frac{sin x}{cos x}}-{\frac{cos x}{sin x}}=0;tg x=1; x=\kappa \pi+{\frac{\pi}{4}}[/tex]

Решение:
Разделяме 2-те страни на уравнението на [tex]\sqrt{2}[/tex]: [tex]{\frac{1}{\sqrt{2}}}sin x+{\frac{1}{\sqrt{2}}}cos x=sin5x; cos{\frac{\pi}{4}}sin x+sin{\frac{\pi}{4}}cos x=sin5x; sin(x+{\frac{\pi}{4}})=sin5x; sin5x-sin(x+{\frac{\pi}{4}})=0; 2sin(2x-{\frac{\pi}{8}})cos(3x+{\frac{\pi}{8}})=0[/tex].
[tex]sin(2x-{\frac{\pi}{8}})=0; 2x-{\frac{\pi}{8}}=n\pi; x=n{\frac{\pi}{2}}+{\frac{\pi}{16}}[/tex].

Решение:
Извършваме следното преобразование: [tex]sin3x=sin({\frac{\pi}{2}}-2x)[/tex] => [tex]3x-({\frac{\pi}{2}}-2x)=2\kappa\pi[/tex] и [tex]3x+({\frac{\pi}{2}}-2x)=(2\kappa+1)\pi[/tex]. От тук [tex]x={\frac{2\kappa\pi}{5}}+{\frac{\pi}{10}}[/tex] и [tex]x=(2\kappa+1)\pi-{\frac{\pi}{2}}[/tex].

Решение:
[tex]tg3x=tg({\frac{\pi}{2}}-2x). 3x-({\frac{\pi}{2}}-2x)=\kappa\pi; 5x=\kappa\pi+{\frac{\pi}{2}}; x=\kappa{\frac{\pi}{5}}+{\frac{\pi}{10}}[/tex].