Trigonometry Problems - sin, cos, tan, cot: Problems with Solutions

Solution:
$\sin(A) = \frac{11}{61}$

Solution:
$tan(A) = \frac{11}{60}$

Solution:
$cot(A) = \frac{60}{11}$

Solution:
$\sin(30°)=\frac{1}{2}$

Solution:
cos(90°) = 0

Solution:
Since $\sin^2\alpha + \cos^2\alpha = 1$,

sin²(43°) + cos²(43°) = 1

Solution:
[tex]\alpha[/tex] belongs to the II quadrant => [tex]cos\alpha[/tex] < 0, and [tex]cos\alpha=-\sqrt{1-sin^2\alpha}=-\sqrt{1-{\frac{25}{169}}}=-{\frac{12}{13}}[/tex]
[tex]tan\alpha=-{\frac{5}{12}}[/tex]
[tex]cot\alpha=-{\frac{12}{5}}[/tex].

Solution:
$\cot(\pi + x)=cot(x)$

Solution:
sin(-585°)=-sin(585°)=-sin(2π+225°)=-sin225°=-sin(π+45°)=sin45°=[tex]{\frac{\sqrt{2}}{2}}[/tex]

Solution:
[tex]\tan(270^\circ+\alpha)=\tan({\frac{3\pi}{2}}+\alpha)=-\cot\alpha[/tex]

Solution:
[tex]cos{\frac{8\pi}{3}}=cos(3\pi-{\frac{\pi}{3}})=cos(\pi-\frac{\pi}{3})=-cos{\frac{\pi}{3}}=-{\frac{1}{2}}[/tex]

Solution:
[tex]\sin 945^{\circ}[/tex][tex]=[/tex][tex]\sin(720^{\circ}+ 225^{\circ})[/tex]= [tex]\sin(225^{\circ}+2*360^{\circ})[/tex]= [tex]\sin225^{\circ}[/tex]= [tex]\sin(225^{\circ}-360^{\circ})[/tex]= [tex]\sin(-135^{\circ})[/tex]=[tex]-\sin 135^{\circ}=- \frac{1}{\sqrt{2}}=-\frac{\sqrt{2}}{2}[/tex]

Solution:
cos(175°) = cos(180 - 5) = -cos(5)
cos(204°) =cos(180+24) = -cos(24)
cos(300°) = cos(360-60) = cos(60)
So result would be cos(60) that is 1/2

Solution:
Cot x=2

$\frac{(2+2\sin x)(1-\sin x)}{(1+\cos x)(2-2\cos x)} = \frac{2[(1+\sin x)(1-\sin x)]}{2[(1+\cos x)(1-\cos x)]}=$
$\frac{1-\sin^2x}{1-\cos^2x} = \frac{\cos^2x}{\sin^2x}=\big(\frac{\cos x}{\sin x}\big)^2 =\cot^2x =2^2 =4$

Solution:
cos 15°=cos(45°-30°)=cos45°cos30°+sin45°sin30°=[tex]{\frac{\sqrt{2}}{2}}.{\frac{\sqrt{3}}{2}}+{\frac{\sqrt{2}}{2}}.{\frac{1}{2}}={\frac{\sqrt{6}+\sqrt{2}}{4}}[/tex].

Solution:
sin75°sin15°=sin(90°-15°)=cos15°sin15°=[tex]{\frac{1}{2}}sin30^{\circ}={\frac{1}{4}}[/tex]

Solution:
sin15°=sin(45°-30°)=sin45°cos30°-cos45°sin30°=[tex]{\frac{\sqrt{6}-\sqrt{2}}{4}}[/tex].

Solution:
sin4α+cos4αcot2α=[tex]{\frac{2tan2\alpha}{1+tan^22\alpha}}+{\frac{1-tan^22\alpha}{1+tan^22\alpha}}.{\frac{1}{tan2\alpha}}={\frac{2 \cdot 4}{1+16}}+{\frac{1-16}{1+16}}{\frac{1}{4}}={\frac{1}{4}}[/tex].

Solution:
3+4cos2α+cos4α=2+4cos2α+(1+cos4α)=2+4cos2α+2cos²2α=2(1+2cos2α+cos²2α)=2(1+cos2α)²=2(2cos²α)²=8cos⁴α

Solution:
Simplify the left part
[tex]4sin({\frac{\pi}{3}}-\alpha)cos({\frac{\pi}{6}}-\alpha)=4.{\frac{1}{2}}(sin({\frac{\pi}{3}}-\alpha+{\frac{\pi}{6}}-\alpha)+sin({\frac{\pi}{3}}-\alpha-{\frac{\pi}{6}}+\alpha))=2(sin({\frac{\pi}{2}}-2\alpha)+sin{\frac{\pi}{6}})=2(cos2\alpha+{\frac{1}{2}})={\frac{2sin\alpha(cos2\alpha+{\frac{1}{2}})}{sin\alpha}}={\frac{2sin\alpha cos2\alpha+sin\alpha}{sin\alpha}}={\frac{sin3\alpha-sin\alpha+sin\alpha}{sin\alpha}}={\frac{sin3\alpha}{sin\alpha}}[/tex]

Solution:
[tex]{\frac{sin^2\alpha+(1+cos\alpha)^2}{sin\alpha(1+cos\alpha)}}={\frac{sin^2\alpha+1+2cos\alpha+cos^2\alpha}{sin\alpha(1+cos\alpha)}}={\frac{2(1+cos\alpha)}{sin\alpha(1+cos\alpha)}}={\frac{2}{sin\alpha}}[/tex]

Solution:
[tex](cos\alpha+cos7\alpha)+(cos2\alpha+cos6\alpha)=2cos4\alpha cos3\alpha +2cos4\alpha cos2\alpha =2cos4\alpha(cos3\alpha+cos2\alpha)=4cos4\alpha cos{\frac{5\alpha}{2}}cos{\frac{\alpha}{2}}[/tex]