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

Solution:
[tex]x+y+z=\pi[/tex], so [tex]{\frac{x}{2}}+{\frac{y}{2}}={\frac{\pi}{2}}-{\frac{z}{2}}; {\frac{z}{2}}={\frac{\pi}{2}}-({\frac{x}{2}}+{\frac{y}{2}})[/tex]. Because of [tex]cot\alpha={\frac{1}{tan\alpha}}[/tex], we get [tex]cot\alpha={\frac{cos\alpha}{sin\alpha}}[/tex]. Then [tex]cot{\frac{x}{2}}+cot{\frac{y}{2}}+cot{\frac{z}{2}}={\frac{cos{\frac{x}{2}}}{sin{\frac{x}{2}}}}+{\frac{cos{\frac{y}{2}}}{sin{\frac{y}{2}}}}+{\frac{cos{\frac{z}{2}}}{sin{\frac{z}{2}}}}={\frac{cos{\frac{x}{2}}sin{\frac{y}{2}}+cos{\frac{y}{2}}sin{\frac{x}{2}}}{sin{\frac{x}{2}}sin{\frac{y}{2}}}}+{\frac{cos{\frac{z}{2}}}{sin{\frac{z}{2}}}}={\frac{sin({\frac{x}{2}}+{\frac{y}{2}})}{sin{\frac{x}{2}}sin{\frac{y}{2}}}}+{\frac{cos{\frac{z}{2}}}{sin{\frac{z}{2}}}}={\frac{cos{\frac{z}{2}}}{sin{\frac{x}{2}}sin{\frac{y}{2}}}}+{\frac{cos{\frac{z}{2}}}{sin{\frac{z}{2}}}}=cos{\frac{z}{2}}\cdot{\frac{sin{\frac{z}{2}}+sin{\frac{x}{2}}sin{\frac{y}{2}}}{sin{\frac{x}{2}}sin{\frac{y}{2}}sin{\frac{z}{2}}}}={\frac{cos{\frac{z}{2}}}{sin{\frac{z}{2}}}}\cdot{\frac{cos({\frac{x}{2}}+{\frac{y}{2}})+sin{\frac{x}{2}}sin{\frac{y}{2}}}{sin{\frac{x}{2}}sin{\frac{y}{2}}}}={\frac{cos{\frac{z}{2}}}{sin{\frac{z}{2}}}}\cdot{\frac{cos{\frac{x}{2}}cos{\frac{y}{2}}}{sin{\frac{x}{2}}sin{\frac{y}{2}}}}=cot{\frac{x}{2}}cot{\frac{y}{2}}cot{\frac{z}{2}}[/tex].

Solution:
[tex]5\cos A + 12\sin A + 12 = 13\left(\frac{5}{13}\cos A + \frac{12}{13}\sin A\right) + 12[/tex]
Since [tex]\left(\frac{5}{13}\right)^2 + \left(\frac{12}{13}\right)^2 = 1[/tex], there is an angle [tex]B[/tex] with [tex]\sin B = \frac{5}{13}[/tex] and [tex]\cos B = \frac{12}{13}[/tex]. Then
[tex]5\cos A + 12\sin A + 12 = 13(\sin B\cos A + \cos B\,\sin A) + 12 = 13\sin(A+B) + 12[/tex]
Since [tex]-1 \le \sin(A+B) \le 1[/tex], the maximum value is [tex]13 \times 1 + 12 = 25[/tex]

Solution:
[tex]5\cos A + 12\sin A + 12 = 13\left(\frac{5}{13}\cos A + \frac{12}{13}\sin A\right) + 12[/tex]
Since [tex]\left(\frac{5}{13}\right)^2 + \left(\frac{12}{13}\right)^2 = 1[/tex], there is an angle [tex]B[/tex] with [tex]\sin B = \frac{5}{13}[/tex] and [tex]\cos B = \frac{12}{13}[/tex]. Then
[tex]5\cos A + 12\sin A + 12 = 13(\sin B\cos A + \cos B\,\sin A) + 12 = 13\sin(A+B) + 12[/tex]
Since [tex]-1 \le \sin(A+B) \le 1[/tex], the minimum value is [tex]13 \times (-1) + 12 = -1[/tex]

Solution:
From the condition
[tex]\sin A=1-\sin ^2 A=\cos^2 A\;\Rightarrow\;\cos^4 A=\sin ^2 A\;\Rightarrow\;\cos^2 A+\cos^4 A=\sin A+\sin ^2 A=1[/tex]
The given expression is a perfect cube:
[tex]\cos^{12}A+3\cos^{10}A+3\cos^8 A+\cos^6 A=\left(\cos^4 A+\cos^2 A\right)^3=1^3=1[/tex]

Solution:
Since [tex]\frac{7\pi}{16}=\frac{\pi}{2}-\frac{\pi}{16}[/tex] and [tex]\frac{5\pi}{16}=\frac{\pi}{2}-\frac{3\pi}{16}[/tex], we get [tex]\sin \frac{7\pi}{16}=\cos\frac{\pi}{16}[/tex] and [tex]\sin \frac{5\pi}{16}=\cos\frac{3\pi}{16}[/tex]. For every [tex]t[/tex]:
[tex]\sin ^4 t+\cos^4 t=1-2\sin ^2 t\cos^2 t=1-\frac{1}{2}\sin ^2 2t[/tex]
So the sum equals
[tex]2-\frac{1}{2}\left(\sin ^2\frac{\pi}{8}+\sin ^2\frac{3\pi}{8}\right)=2-\frac{1}{2}=\frac{3}{2}[/tex]
because [tex]\sin \frac{3\pi}{8}=\cos\frac{\pi}{8}[/tex] and the bracket equals [tex]1[/tex].

Solution:
Let [tex]s=\sin x\in\left[-1, 1\right][/tex]. Using [tex]\sin 3x=3s-4s^3[/tex] and [tex]\cos 2x=1-2s^2[/tex]:
[tex]f=-4s^3-8s^2+11s-3=-(2s-1)^2(s+3)[/tex]
For [tex]s\ge-1[/tex] we have [tex](2s-1)^2\ge 0[/tex] and [tex]s+3\gt 0[/tex], so [tex]f\le 0[/tex]. Equality holds for [tex]s=\frac{1}{2}[/tex], for example at [tex]x=\frac{\pi}{6}[/tex]. So the largest value is [tex]0[/tex].

Solution:
Let [tex]c=\cos x\in\left[-1, 1\right][/tex]. Then
[tex]\sin ^2 x-2\cos x-1=-c^2-2c=1-(c+1)^2[/tex]
which takes all values in [tex]\left[-3, 1\right][/tex]. So we need [tex]ku\le 3[/tex] for every [tex]u\in\left[-3, 1\right][/tex]. Since [tex]ku[/tex] is linear in [tex]u[/tex], it is enough to check the endpoints: [tex]u=1[/tex] gives [tex]k\le 3[/tex], and [tex]u=-3[/tex] gives [tex]-3k\le 3[/tex], i.e. [tex]k\ge-1[/tex]. So [tex]k\in\left[-1, 3\right][/tex].

Solution:
A number [tex]m[/tex] is a value of [tex]y[/tex] exactly when the equation [tex]m(3+\sin x)=\sqrt{3}\cos x+1[/tex] has a solution, i.e.
[tex]\sqrt{3}\cos x-m\sin x=3m-1[/tex]
An equation [tex]a\cos x+b\sin x=c[/tex] has a solution exactly when [tex]c^2\le a^2+b^2[/tex]. So
[tex](3m-1)^2\le 3+m^2\iff 8m^2-6m-2\le 0\iff(4m+1)(m-1)\le 0[/tex]
Hence [tex]m\in\left[-\frac{1}{4}, 1\right][/tex].

Solution:
By Vieta's formulas, [tex]\tan \alpha+\tan \beta=-p[/tex] and [tex]\tan \alpha\tan \beta=q[/tex]. If [tex]q=1[/tex], then [tex]\cos(\alpha+\beta)=\cos\alpha\cos\beta(1-\tan \alpha\tan \beta)=0[/tex] and the expression equals [tex]\sin ^2(\alpha+\beta)=1=q[/tex]. If [tex]q\ne 1[/tex], let
[tex]t=\tan (\alpha+\beta)=\frac{-p}{1-q}=\frac{p}{q-1}[/tex]
Since [tex]\cos^2(\alpha+\beta)=\frac{1}{1+t^2}[/tex], the expression equals [tex]\frac{t^2+pt+q}{1+t^2}[/tex]. Substituting [tex]t[/tex] and multiplying the numerator and the denominator by [tex](q-1)^2[/tex]:
[tex]\frac{p^2+p^2(q-1)+q(q-1)^2}{(q-1)^2+p^2}=\frac{q\left(p^2+(q-1)^2\right)}{p^2+(q-1)^2}=q[/tex]

Solution:
Since [tex]|\sin x|\le 1[/tex], every root satisfies [tex]|x|\le 13[/tex]. Both sides are odd functions, so the roots are symmetric about [tex]0[/tex], and [tex]x=0[/tex] is a root. Let [tex]x\gt 0[/tex] and [tex]h(x)=\sin x-\frac{x}{13}[/tex].
On [tex]\left[0, \pi\right][/tex] the function [tex]h[/tex] is concave, [tex]h(0)=0[/tex], [tex]h'(0)=\frac{12}{13}\gt 0[/tex] and [tex]h(\pi)\lt 0[/tex], so there is exactly one root in [tex]\left(0, \pi\right)[/tex]. On [tex]\left[\pi, 2\pi\right][/tex] and [tex]\left[3\pi, 4\pi\right][/tex] we have [tex]\sin x\le 0\lt\frac{x}{13}[/tex], so there are no roots. On [tex]\left[2\pi, 3\pi\right][/tex] the function [tex]h[/tex] is concave, negative at both ends, and [tex]h\left(\frac{5\pi}{2}\right)=1-\frac{5\pi}{26}\gt 0[/tex], so there are exactly two roots. On [tex]\left(4\pi, 13\right][/tex]:
[tex]\sin x\le\sin (13-4\pi)\lt\sin \frac{\pi}{6}=\frac{1}{2}\lt\frac{4\pi}{13}\lt\frac{x}{13}[/tex]
so there are no roots there. So there are [tex]3[/tex] positive roots, [tex]3[/tex] negative roots and [tex]x=0[/tex]: [tex]7[/tex] solutions in total.