Basic Trigonometry Problems: Difficult Problems with Solutions

Solution:
Use [tex]\frac{\pi}{2}\approx 1. 57[/tex], [tex]\pi\approx 3. 14[/tex] and [tex]2\pi\approx 6. 28[/tex]. Since [tex]1. 57\lt 2\lt 3. 14[/tex] and [tex]1. 57\lt 3\lt 3. 14[/tex], the angles [tex]2[/tex] and [tex]3[/tex] are in the second quadrant. For [tex]-4[/tex] add a full turn: [tex]-4+2\pi\approx 2. 28[/tex], which is also between [tex]1. 57[/tex] and [tex]3. 14[/tex]. So all three angles are in the second quadrant.

Solution:
[tex]\sin \frac{7\pi}{6}=\sin \left(\pi+\frac{\pi}{6}\right)=-\frac{1}{2}, \quad\cos\left(-\frac{5\pi}{3}\right)=\cos\left(2\pi-\frac{\pi}{3}\right)=\frac{1}{2}[/tex]
Since [tex]\frac{\pi}{7}+\frac{5\pi}{14}=\frac{\pi}{2}[/tex], we have [tex]\tan \frac{5\pi}{14}=\cot \frac{\pi}{7}[/tex], so [tex]\tan \frac{\pi}{7}\times \tan \frac{5\pi}{14}=1[/tex]. The value is [tex]-\frac{1}{2}+\frac{1}{2}+1=1[/tex].

Solution:
By the reduction formulas:
[tex]\cos(270^{\circ}-x)=-\sin x, \ \sin (270^{\circ}+x)=-\cos x, \ \tan (180^{\circ}-x)=-\tan x, \ \cos(270^{\circ}+x)=\sin x, \ \sin (180^{\circ}-x)=\sin x[/tex]
The numerator is [tex]2(-\sin x)(-\cos x)(-\tan x)=-2\sin ^2 x[/tex] and the denominator is [tex]\sin ^2 x[/tex], so the value is [tex]-2[/tex] for every admissible [tex]x[/tex].

Solution:
[tex]\sin (\alpha-270^{\circ})=\cos\alpha, \quad\cos(\alpha-90^{\circ})=\sin \alpha, \quad\tan (\alpha+90^{\circ})=-\cot \alpha[/tex]
The product is
[tex]\cos\alpha\times \sin \alpha\times \left(-\frac{\cos\alpha}{\sin \alpha}\right)=-\cos^2\alpha=-\frac{1}{1+\tan ^2\alpha}=-\frac{1}{1+\frac{1}{4}}=-\frac{4}{5}[/tex]

Solution:
[tex]x-2\pi\in\left[\frac{\pi}{2}, \pi\right)[/tex], so the terminal side of [tex]x[/tex] is in the second quadrant: [tex]\sin x\gt 0[/tex] and [tex]\cos x\lt 0[/tex]. From [tex]\cot x=\frac{\cos x}{\sin x}=-2[/tex] we get [tex]\cos x=-2\sin x[/tex], so [tex]\sin ^2 x+4\sin ^2 x=1[/tex] and [tex]\sin ^2 x=\frac{1}{5}[/tex]. Hence
[tex]\sin x=\frac{\sqrt{5}}{5}, \quad\cos x=-\frac{2\sqrt{5}}{5}[/tex]

Solution:
In the second quadrant [tex]\cos\alpha\lt 0[/tex], so
[tex]\cos\alpha=-\sqrt{1-\frac{9}{16}}=-\frac{\sqrt{7}}{4}, \quad\tan \alpha=\frac{\sin \alpha}{\cos\alpha}=-\frac{3}{\sqrt{7}}=-\frac{3\sqrt{7}}{7}[/tex]
So [tex]7\tan \alpha=-3\sqrt{7}[/tex] and [tex]-8\cos\alpha=2\sqrt{7}[/tex], and the value is [tex]-3\sqrt{7}+2\sqrt{7}=-\sqrt{7}[/tex].

Solution:
[tex]\frac{3\sin \alpha+2\tan \alpha}{\cot \alpha}=\left(3\sin \alpha+\frac{2\sin \alpha}{\cos\alpha}\right)\times \frac{\sin \alpha}{\cos\alpha}=\frac{\sin ^2\alpha}{\cos\alpha}\left(3+\frac{2}{\cos\alpha}\right)[/tex]
Only [tex]\sin ^2\alpha[/tex] appears, so the sign of [tex]\sin \alpha[/tex] does not matter: [tex]\sin ^2\alpha=1-\frac{1}{9}=\frac{8}{9}[/tex] and [tex]3+\frac{2}{\cos\alpha}=9[/tex]. The value is
[tex]\frac{\frac{8}{9}}{\frac{1}{3}}\times 9=\frac{8}{3}\times 9=24[/tex]

Solution:
Replace [tex]\sin ^2\alpha=1-\cos^2\alpha[/tex]:
[tex]20\cos^2\alpha-21\cos\alpha+4=0[/tex]
so [tex]\cos\alpha=\frac{4}{5}[/tex] or [tex]\cos\alpha=\frac{1}{4}[/tex]. For [tex]\alpha\in\left(315^{\circ}, 360^{\circ}\right)[/tex] we have [tex]\cos\alpha\gt\cos 315^{\circ}=\frac{\sqrt{2}}{2}\approx 0. 71[/tex], so [tex]\cos\alpha=\frac{4}{5}[/tex]. In the fourth quadrant [tex]\sin \alpha\lt 0[/tex], so [tex]\sin \alpha=-\frac{3}{5}[/tex] and [tex]\tan \alpha=-\frac{3}{4}[/tex].

Solution:
Let [tex]ABC[/tex] have [tex]\angle C=90^{\circ}[/tex] and altitude [tex]CH[/tex]. Let [tex]AH=x[/tex] be the projection of [tex]AC[/tex]; then [tex]AB=5x[/tex] and [tex]HB=4x[/tex]. In a right triangle [tex]CH^2=AH\times HB=4x^2[/tex], so [tex]CH=2x[/tex]. The acute angle adjacent to [tex]AC[/tex] is [tex]\angle A[/tex], and in the right triangle [tex]AHC[/tex]:
[tex]\tan A=\frac{CH}{AH}=\frac{2x}{x}=2[/tex]

Solution:
Let [tex]CM[/tex] be the median to the hypotenuse; then [tex]CM=MA=MB=\frac{AB}{2}=2CH[/tex]. In the right triangle [tex]CHM[/tex]: [tex]\sin \angle CMH=\frac{CH}{CM}=\frac{1}{2}[/tex], so the acute angle between [tex]CM[/tex] and [tex]AB[/tex] is [tex]30^{\circ}[/tex]. Let [tex]\angle A[/tex] be the smallest angle. The triangle [tex]AMC[/tex] is isosceles ([tex]MA=MC[/tex]), so its exterior angle [tex]\angle CMB=2\angle A\lt 90^{\circ}[/tex]. Hence [tex]2\angle A=30^{\circ}[/tex] and [tex]\angle A=15^{\circ}[/tex].

Solution:
Let [tex]AH\perp BC[/tex], [tex]H\in BC[/tex], so [tex]AH=h_a[/tex]. Since [tex]\beta[/tex] and [tex]\gamma[/tex] are acute, [tex]H[/tex] lies between [tex]B[/tex] and [tex]C[/tex], so [tex]a=BH+HC[/tex]. In the right triangle [tex]ABH[/tex]: [tex]\cot \beta=\frac{BH}{AH}[/tex], so [tex]BH=h_a\cot \beta[/tex]. In the same way [tex]HC=h_a\cot \gamma[/tex]. Then
[tex]a=h_a(\cot \beta+\cot \gamma)\iff\frac{a}{h_a}=\cot \beta+\cot \gamma[/tex]

Solution:
Let [tex]CH\perp AB[/tex]; [tex]H[/tex] is the midpoint of [tex]AB[/tex]. In the right triangle [tex]AHC[/tex]: [tex]AH=AC\times \cos\angle BAC=\frac{3}{5}AC[/tex], so [tex]AB=2AH=\frac{6}{5}AC[/tex]. By the angle bisector theorem
[tex]\frac{CL}{LB}=\frac{AC}{AB}=\frac{AC}{\frac{6}{5}AC}=\frac{5}{6}[/tex]

Solution:
The diagonals bisect each other at right angles, so the side is [tex]a=\sqrt{3^2+4^2}=5[/tex]. The area is [tex]S=\frac{6\times 8}{2}=24[/tex], so the height is [tex]h=\frac{S}{a}=\frac{24}{5}[/tex]. In the right triangle formed by the height and a side:
[tex]\sin \alpha=\frac{h}{a}=\frac{24}{25}, \quad\cos\alpha=\sqrt{1-\frac{576}{625}}=\frac{7}{25}, \quad\tan \alpha=\frac{24}{7}[/tex]

Solution:
The height equals the diameter: [tex]h=2r=4[/tex]. Let [tex]DD_1\perp AB[/tex]. In the right triangle [tex]AD_1D[/tex]: [tex]\tan A=\frac{DD_1}{AD_1}=4[/tex], so [tex]AD_1=1[/tex] and [tex]AD=\sqrt{1+16}=\sqrt{17}[/tex]. A circle is inscribed in the trapezoid, so [tex]AB+CD=AD+BC=2\sqrt{17}[/tex]. Also [tex]AB-CD=2AD_1=2[/tex]. Hence
[tex]AB=\sqrt{17}+1, \quad CD=\sqrt{17}-1, \quad S=\frac{AB+CD}{2}\times h=\sqrt{17}\times 4=4\sqrt{17}[/tex]

Solution:
Let [tex]BM[/tex] be the median, [tex]M[/tex] the midpoint of [tex]AC[/tex]. The point [tex]C[/tex] lies on [tex]BC[/tex], so the distance from [tex]M[/tex] to [tex]BC[/tex] is half the distance from [tex]A[/tex] to [tex]BC[/tex], i.e. [tex]\frac{h_a}{2}[/tex]. Let [tex]MP\perp BC[/tex], [tex]P\in BC[/tex]. In the right triangle [tex]BPM[/tex]:
[tex]\sin \angle MBP=\frac{MP}{BM}=\frac{\frac{h_a}{2}}{m_b}=\frac{1}{2}\;\Rightarrow\;\angle MBP=30^{\circ}[/tex]

Solution:
Let the base angle be [tex]\alpha[/tex]; the vertex angle is [tex]180^{\circ}-2\alpha[/tex]. Then
[tex]\sin \alpha=3\cos(180^{\circ}-2\alpha)=-3\cos 2\alpha=-3\left(1-2\sin ^2\alpha\right)\iff 6\sin ^2\alpha-\sin \alpha-3=0[/tex]
So [tex]\sin \alpha=\frac{1\pm\sqrt{73}}{12}[/tex]. Since [tex]\sin \alpha\gt 0[/tex], [tex]\sin \alpha=\frac{1+\sqrt{73}}{12}\approx 0. 795[/tex] (then [tex]\alpha\approx 52. 7^{\circ}[/tex], which is possible).