Trigonometric Expressions and Identities: Very Difficult Problems with Solutions

Solution:
[tex]\sin x\sin (60^{\circ}-x)\sin (60^{\circ}+x)=\sin x\left(\frac{1}{2}\cos 2x+\frac{1}{4}\right)=\frac{1}{4}\sin 3x[/tex]
In the same way
[tex]\cos x\cos(60^{\circ}-x)\cos(60^{\circ}+x)=\frac{1}{4}\cos 3x[/tex]
So [tex]\tan x\tan (60^{\circ}-x)\tan (60^{\circ}+x)=\tan 3x[/tex]. For [tex]x=20^{\circ}[/tex]: [tex]\tan 20^{\circ}\tan 40^{\circ}\tan 80^{\circ}=\tan 60^{\circ}=\sqrt{3}[/tex]. The required product is [tex]\sqrt{3}\times \tan 60^{\circ}=3[/tex].

Solution:
[tex]\tan x+\tan (90^{\circ}-x)=\tan x+\cot x=\frac{2}{\sin 2x}[/tex]
So [tex]\tan 9^{\circ}+\tan 81^{\circ}=\frac{2}{\sin 18^{\circ}}[/tex] and [tex]\tan 27^{\circ}+\tan 63^{\circ}=\frac{2}{\sin 54^{\circ}}[/tex]. Since [tex]\sin 54^{\circ}-\sin 18^{\circ}=2\cos 36^{\circ}\sin 18^{\circ}[/tex] and [tex]\sin 54^{\circ}=\cos 36^{\circ}[/tex], the expression equals
[tex]\frac{2}{\sin 18^{\circ}}-\frac{2}{\sin 54^{\circ}}=\frac{2(\sin 54^{\circ}-\sin 18^{\circ})}{\sin 18^{\circ}\sin 54^{\circ}}=\frac{4\cos 36^{\circ}\sin 18^{\circ}}{\sin 18^{\circ}\cos 36^{\circ}}=4[/tex]

Solution:
Since [tex]-\cos\frac{2\pi}{7}=\cos\frac{5\pi}{7}[/tex], the sum is [tex]S=\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7}[/tex]. Multiply by [tex]2\sin \frac{\pi}{7}[/tex] and use [tex]2\sin x\cos y=\sin (x+y)+\sin (x-y)[/tex]:
[tex]2\sin \frac{\pi}{7}\times S=\sin \frac{2\pi}{7}+\left(\sin \frac{4\pi}{7}-\sin \frac{2\pi}{7}\right)+\left(\sin \frac{6\pi}{7}-\sin \frac{4\pi}{7}\right)=\sin \frac{6\pi}{7}[/tex]
Since [tex]\sin \frac{6\pi}{7}=\sin \frac{\pi}{7}\ne 0[/tex], we get [tex]S=\frac{1}{2}[/tex].

Solution:
From the triple-angle formulas:
[tex]\cos^3\alpha=\frac{\cos 3\alpha+3\cos\alpha}{4}, \quad\sin ^3\alpha=\frac{3\sin \alpha-\sin 3\alpha}{4}[/tex]
The left side becomes
[tex]\frac{1}{4}\left(\cos^2 3\alpha-\sin ^2 3\alpha+3(\cos 3\alpha\cos\alpha+\sin 3\alpha\sin \alpha)\right)=\frac{1}{4}(\cos 6\alpha+3\cos 2\alpha)[/tex]
With [tex]\cos 6\alpha=4\cos^3 2\alpha-3\cos 2\alpha[/tex] this equals [tex]\cos^3 2\alpha[/tex].

Solution:
Since [tex]\cos 84^{\circ}=-\cos 96^{\circ}[/tex], multiply [tex]\cos 12^{\circ}\cos 24^{\circ}\cos 48^{\circ}\cos 96^{\circ}[/tex] by [tex]16\sin 12^{\circ}[/tex] and use [tex]2\sin x\cos x=\sin 2x[/tex] four times:
[tex]16\sin 12^{\circ}\cos 12^{\circ}\cos 24^{\circ}\cos 48^{\circ}\cos 96^{\circ}=\sin 192^{\circ}=-\sin 12^{\circ}[/tex]
So [tex]\cos 12^{\circ}\cos 24^{\circ}\cos 48^{\circ}\cos 84^{\circ}=\frac{1}{16}[/tex]. In the same way [tex]4\sin 36^{\circ}\cos 36^{\circ}\cos 72^{\circ}=\sin 144^{\circ}=\sin 36^{\circ}[/tex], so [tex]\cos 36^{\circ}\cos 72^{\circ}=\frac{1}{4}[/tex]. With [tex]\cos 60^{\circ}=\frac{1}{2}[/tex] the product is
[tex]\frac{1}{16}\times \frac{1}{4}\times \frac{1}{2}=\frac{1}{128}[/tex]

Solution:
Use [tex]\tan \frac{x}{2}=\frac{1-\cos x}{\sin x}[/tex] with [tex]x=15^{\circ}[/tex], where [tex]\cos 15^{\circ}=\frac{\sqrt{6}+\sqrt{2}}{4}[/tex] and [tex]\sin 15^{\circ}=\frac{\sqrt{6}-\sqrt{2}}{4}[/tex]:
[tex]\tan 7^{\circ}30'=\frac{4-\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}=\frac{\left(4-\sqrt{6}-\sqrt{2}\right)\left(\sqrt{6}+\sqrt{2}\right)}{4}=\frac{4\sqrt{6}+4\sqrt{2}-8-4\sqrt{3}}{4}=\sqrt{6}-\sqrt{3}+\sqrt{2}-2[/tex]
Therefore the number equals [tex]-2[/tex].

Solution:
Since [tex]\tan (x+120^{\circ})=\tan (x-60^{\circ})[/tex]:
[tex]\tan (x+60^{\circ})+\tan (x-60^{\circ})=\frac{\sin 2x}{\cos(x+60^{\circ})\cos(x-60^{\circ})}=\frac{\sin 2x}{\cos^2 x-\frac{3}{4}}=\frac{8\sin x\cos x}{4\cos^2 x-3}[/tex]
Adding [tex]\tan x[/tex]:
[tex]\frac{\sin x\left(4\cos^2 x-3\right)+8\sin x\cos^2 x}{\cos x\left(4\cos^2 x-3\right)}=\frac{3\sin x\left(4\cos^2 x-1\right)}{\cos 3x}=\frac{3\sin 3x}{\cos 3x}=3\tan 3x[/tex]
because [tex]\cos 3x=\cos x\left(4\cos^2 x-3\right)[/tex] and [tex]\sin 3x=\sin x\left(3-4\sin ^2 x\right)=\sin x\left(4\cos^2 x-1\right)[/tex].

Solution:
With [tex]t=\tan x[/tex] we have [tex]\tan 3x=\frac{3t-t^3}{1-3t^2}[/tex]. For [tex]x=10^{\circ}[/tex], [tex]x=-50^{\circ}[/tex] and [tex]x=70^{\circ}[/tex] we get [tex]\tan 3x=\frac{1}{\sqrt{3}}[/tex], so [tex]\tan 10^{\circ}[/tex], [tex]-\tan 50^{\circ}[/tex] and [tex]\tan 70^{\circ}[/tex] are the three distinct roots of
[tex]\sqrt{3}\left(3t-t^3\right)=1-3t^2\iff t^3-\sqrt{3}t^2-3t+\frac{1}{\sqrt{3}}=0[/tex]
By Vieta's formulas [tex]t_1+t_2+t_3=\sqrt{3}[/tex] and [tex]t_1t_2+t_2t_3+t_3t_1=-3[/tex], so
[tex]t_1^2+t_2^2+t_3^2=\left(\sqrt{3}\right)^2-2\times (-3)=9[/tex]

Solution:
Let [tex]u=\cos(x+y)[/tex] and [tex]v=\cos(x-y)[/tex]. Then
[tex]\sin ^2 x+\sin ^2 y=1-\frac{\cos 2x+\cos 2y}{2}=1-uv=\frac{3}{2}[/tex]
so [tex]uv=-\frac{1}{2}[/tex]. The expression equals
[tex]\left(1-u^2\right)v+\left(1-v^2\right)u=(u+v)(1-uv)=\frac{3}{2}(u+v)[/tex]
From [tex]uv=-\frac{1}{2}[/tex], [tex]|u|\le 1[/tex] and [tex]|v|\le 1[/tex] we get [tex]\frac{1}{2}\le|u|\le 1[/tex] and [tex]u+v=u-\frac{1}{2u}[/tex]. This function increases on [tex]\left[-1, -\frac{1}{2}\right][/tex] and on [tex]\left[\frac{1}{2}, 1\right][/tex], so [tex]u+v\le\frac{1}{2}[/tex]. The maximum [tex]\frac{3}{2}\times \frac{1}{2}=\frac{3}{4}[/tex] is reached, for example, at [tex]x=60^{\circ}[/tex], [tex]y=-60^{\circ}[/tex].

Solution:
[tex]\frac{\sin 1^{\circ}}{\sin k^{\circ}\sin (k+1)^{\circ}}=\frac{\sin \left((k+1)^{\circ}-k^{\circ}\right)}{\sin k^{\circ}\sin (k+1)^{\circ}}=\cot k^{\circ}-\cot (k+1)^{\circ}[/tex]
For [tex]k=91, 93, \dots, 133[/tex] use [tex]\sin x=\sin (180^{\circ}-x)[/tex]: the term equals [tex]\frac{1}{\sin (179-k)^{\circ}\sin (180-k)^{\circ}}[/tex], i.e. these are the terms with [tex]m=46, 48, \dots, 88[/tex]. So the sum contains [tex]\frac{1}{\sin m^{\circ}\sin (m+1)^{\circ}}[/tex] for every [tex]m[/tex] from [tex]45[/tex] to [tex]89[/tex], and it telescopes:
[tex]\frac{1}{\sin 1^{\circ}}\left(\cot 45^{\circ}-\cot 90^{\circ}\right)=\frac{1}{\sin 1^{\circ}}[/tex]
So [tex]\sin n^{\circ}=\sin 1^{\circ}[/tex], and the smallest such [tex]n[/tex] is [tex]1[/tex].

Solution:
All tangents exist, so there is no right angle, and for [tex]\alpha+\beta+\gamma=180^{\circ}[/tex] we have
[tex]\tan \alpha+\tan \beta+\tan \gamma=\tan \alpha\tan \beta\tan \gamma[/tex]
Let the tangents be the nonzero integers [tex]a[/tex], [tex]b[/tex], [tex]c[/tex]. At most one angle is obtuse. If [tex]c\lt 0[/tex], then [tex]\alpha+\beta\lt 90^{\circ}[/tex] and [tex]\tan (\alpha+\beta)=\frac{a+b}{1-ab}\gt 0[/tex] gives [tex]ab\lt 1[/tex], which is impossible for positive integers. So [tex]1\le a\le b\le c[/tex] and [tex]abc=a+b+c\le 3c[/tex], hence [tex]ab\le 3[/tex]. The case [tex]a=b=1[/tex] gives [tex]c+2=c[/tex] (impossible); [tex]a=1[/tex], [tex]b=2[/tex] gives [tex]c=3[/tex]; [tex]a=1[/tex], [tex]b=3[/tex] gives [tex]c=2\lt b[/tex]. So the tangents are [tex]1, 2, 3[/tex] and their sum is [tex]6[/tex].

Solution:
Put [tex]a=\tan \alpha[/tex], [tex]b=\tan \beta[/tex], [tex]c=\tan \gamma[/tex] with [tex]\alpha, \beta, \gamma\in\left(0, \frac{\pi}{2}\right)[/tex]. Then
[tex]\tan (\alpha+\beta+\gamma)=\frac{a+b+c-abc}{1-ab-bc-ca}=0[/tex]
(the denominator is not [tex]0[/tex], because [tex]ab+bc+ca\ge\frac{9abc}{a+b+c}=9[/tex]), so [tex]\alpha+\beta+\gamma=\pi[/tex]. Also [tex]\frac{1}{1+a^2}=\cos^2\alpha[/tex] and the root in the last term equals [tex]\frac{1}{\cos\alpha\cos\beta\cos\gamma}[/tex]. The expression is [tex]\cos^2\alpha+\cos^2\beta+\cos^2\gamma+2\cos\alpha\cos\beta\cos\gamma[/tex]. Now
[tex]\cos^2\alpha+\cos^2\beta=1+\cos(\alpha+\beta)\cos(\alpha-\beta)=1-\cos\gamma\cos(\alpha-\beta)[/tex]
[tex]\cos^2\gamma+2\cos\alpha\cos\beta\cos\gamma=\cos\gamma\left(-\cos(\alpha+\beta)+2\cos\alpha\cos\beta\right)=\cos\gamma\cos(\alpha-\beta)[/tex]
So the expression equals [tex]1[/tex].