Trigonometric Expressions and Identities: Difficult Problems with Solutions

Solution:
Let [tex]p=\sin ^2\alpha\cos^2\alpha[/tex]. Then
[tex]\sin ^4\alpha+\cos^4\alpha=1-2p, \quad\sin ^8\alpha+\cos^8\alpha=(1-2p)^2-2p^2=1-4p+2p^2[/tex]
The fraction equals [tex]\frac{-4p+2p^2}{p}=-4+2p[/tex], and the whole expression is [tex]-4+2p+1-2p=-3[/tex].

Solution:
Simplify step by step:
[tex]\frac{\sin 4\alpha}{1+\cos 4\alpha}=\frac{2\sin 2\alpha\cos 2\alpha}{2\cos^2 2\alpha}=\tan 2\alpha[/tex]
[tex]\tan 2\alpha\times \frac{\cos 2\alpha}{1+\cos 2\alpha}=\frac{\sin 2\alpha}{2\cos^2\alpha}=\tan \alpha[/tex]
[tex]\tan \alpha\times \frac{\cos\alpha}{1+\cos\alpha}=\frac{\sin \alpha}{1+\cos\alpha}=\frac{2\sin \frac{\alpha}{2}\cos\frac{\alpha}{2}}{2\cos^2\frac{\alpha}{2}}=\tan \frac{\alpha}{2}[/tex]

Solution:
[tex]\cos 2\alpha=\frac{1-\tan ^2\alpha}{1+\tan ^2\alpha}=\frac{1-2}{1+2}=-\frac{1}{3}, \quad\cos 4\alpha=2\cos^2 2\alpha-1=-\frac{7}{9}[/tex]
By the product-to-sum formula:
[tex]\cos\alpha\cos 3\alpha=\frac{1}{2}(\cos 4\alpha+\cos 2\alpha)=\frac{1}{2}\left(-\frac{7}{9}-\frac{3}{9}\right)=-\frac{5}{9}[/tex]

Solution:
Since [tex]\sin 30^{\circ}=\frac{1}{2}[/tex] and [tex]\sin x=\cos(90^{\circ}-x)[/tex], the expression equals [tex]8\cos 20^{\circ}\cos 40^{\circ}\cos 80^{\circ}[/tex]. Multiply and divide by [tex]\sin 20^{\circ}[/tex] and use [tex]2\sin x\cos x=\sin 2x[/tex] three times:
[tex]8\sin 20^{\circ}\cos 20^{\circ}\cos 40^{\circ}\cos 80^{\circ}=4\sin 40^{\circ}\cos 40^{\circ}\cos 80^{\circ}=2\sin 80^{\circ}\cos 80^{\circ}=\sin 160^{\circ}=\sin 20^{\circ}[/tex]
So the value is [tex]\frac{\sin 20^{\circ}}{\sin 20^{\circ}}=1[/tex].

Solution:
Transform the numerator:
[tex]\sin 2\alpha+\sin 6\alpha+\sin 4\alpha=2\sin 4\alpha\cos 2\alpha+2\sin 2\alpha\cos 2\alpha=2\cos 2\alpha(\sin 4\alpha+\sin 2\alpha)=4\cos\alpha\cos 2\alpha\sin 3\alpha[/tex]
So the fraction equals [tex]4[/tex].

Solution:
Bring the first bracket to a common denominator:
[tex]\frac{(1-\sin \alpha)-(1+\sin \alpha)}{\sqrt{1-\sin ^2\alpha}}=\frac{-2\sin \alpha}{|\cos\alpha|}[/tex]
In the same way the second bracket equals [tex]\frac{-2\cos\alpha}{|\sin \alpha|}[/tex]. The product is
[tex]\frac{4\sin \alpha\cos\alpha}{|\sin \alpha|\times |\cos\alpha|}[/tex]
For [tex]\frac{\pi}{2}\lt\alpha\lt\pi[/tex] we have [tex]\sin \alpha\gt 0[/tex] and [tex]\cos\alpha\lt 0[/tex], so the value is [tex]-4[/tex].

Solution:
[tex]\sin ^2 17^{\circ}=\frac{1-\cos 34^{\circ}}{2}, \quad\cos^2 47^{\circ}=\frac{1+\cos 94^{\circ}}{2}, \quad\sin 17^{\circ}\cos 47^{\circ}=\frac{1}{2}\left(\sin 64^{\circ}-\sin 30^{\circ}\right)[/tex]
Also [tex]\cos 94^{\circ}-\cos 34^{\circ}=-2\sin 64^{\circ}\sin 30^{\circ}=-\sin 64^{\circ}[/tex]. The sum is
[tex]1+\frac{\cos 94^{\circ}-\cos 34^{\circ}}{2}+\frac{1}{2}\sin 64^{\circ}-\frac{1}{4}=1-\frac{1}{2}\sin 64^{\circ}+\frac{1}{2}\sin 64^{\circ}-\frac{1}{4}=\frac{3}{4}[/tex]

Solution:
[tex]\sin 78^{\circ}-\sin 42^{\circ}=2\cos 60^{\circ}\sin 18^{\circ}=\sin 18^{\circ}, \quad\sin 6^{\circ}-\sin 66^{\circ}=-2\cos 36^{\circ}\sin 30^{\circ}=-\cos 36^{\circ}[/tex]
So [tex]A=\sin 18^{\circ}-\sin 54^{\circ}=-2\cos 36^{\circ}\sin 18^{\circ}[/tex]. Multiplying by [tex]\cos 18^{\circ}[/tex]:
[tex]A\cos 18^{\circ}=-\cos 36^{\circ}\sin 36^{\circ}=-\frac{1}{2}\sin 72^{\circ}=-\frac{1}{2}\cos 18^{\circ}[/tex]
Hence [tex]A=-\frac{1}{2}[/tex].

Solution:
Using [tex]\sin u\cos v=\frac{1}{2}\left[\sin (u+v)+\sin (u-v)\right][/tex] with [tex]u+v=2x+\frac{\pi}{12}[/tex] and [tex]u-v=\frac{\pi}{8}+\frac{\pi}{24}=\frac{\pi}{6}[/tex]:
[tex]\sin \left(x+\frac{\pi}{8}\right)\cos\left(x-\frac{\pi}{24}\right)=\frac{1}{2}\sin \left(2x+\frac{\pi}{12}\right)+\frac{1}{4}[/tex]
The smallest value of the sine is [tex]-1[/tex], so the minimum is [tex]-\frac{1}{2}+\frac{1}{4}=-\frac{1}{4}[/tex].

Solution:
Let [tex]u=\alpha-\frac{\beta}{2}[/tex] and [tex]v=\frac{\alpha}{2}-\beta[/tex]; then [tex]u-v=\frac{\alpha+\beta}{2}[/tex]. Both angles are acute, so
[tex]\sin u=\sqrt{1-\frac{1}{81}}=\frac{4\sqrt{5}}{9}, \quad\cos v=\sqrt{1-\frac{4}{9}}=\frac{\sqrt{5}}{3}[/tex]
[tex]\cos\frac{\alpha+\beta}{2}=\cos u\cos v+\sin u\sin v=\frac{1}{9}\times \frac{\sqrt{5}}{3}+\frac{4\sqrt{5}}{9}\times \frac{2}{3}=\frac{9\sqrt{5}}{27}=\frac{\sqrt{5}}{3}[/tex]

Solution:
Let [tex]s=\frac{\alpha+\beta}{2}[/tex] and [tex]d=\frac{\alpha-\beta}{2}[/tex]. By the sum-to-product formulas, [tex]a=2\cos s\cos d[/tex] and [tex]b=2\sin s\cos d[/tex]. So
[tex]a^2+b^2=4\cos^2 d, \quad a^2-b^2=4\cos^2 d\left(\cos^2 s-\sin ^2 s\right)=4\cos^2 d\cos(\alpha+\beta)[/tex]
Since [tex]a^2+b^2\ne 0[/tex], we get [tex]\cos(\alpha+\beta)=\frac{a^2-b^2}{a^2+b^2}[/tex].

Solution:
[tex]\cot x-\tan x=\frac{\cos^2 x-\sin ^2 x}{\sin x\cos x}=\frac{2\cos 2x}{\sin 2x}=2\cot 2x[/tex]
Apply this three times:
[tex]\cot \alpha-\tan \alpha=2\cot 2\alpha, \quad 2\cot 2\alpha-2\tan 2\alpha=4\cot 4\alpha, \quad 4\cot 4\alpha-4\tan 4\alpha=8\cot 8\alpha[/tex]

Solution:
Since [tex]\gamma=180^{\circ}-(\alpha+\beta)[/tex], we have [tex]\sin \gamma=\sin (\alpha+\beta)=\sin \alpha\cos\beta+\cos\alpha\sin \beta[/tex]. The condition becomes
[tex]\sin \alpha\cos\beta+\cos\alpha\sin \beta=2\sin \alpha\cos\beta\iff\sin \alpha\cos\beta-\cos\alpha\sin \beta=0\iff\sin (\alpha-\beta)=0[/tex]
Since [tex]-180^{\circ}\lt\alpha-\beta\lt 180^{\circ}[/tex], we get [tex]\alpha-\beta=0[/tex]. So [tex]\alpha=\beta[/tex] and the triangle is isosceles.

Solution:
Since [tex]\sin (\alpha+\beta)=\sin (180^{\circ}-\gamma)=\sin \gamma[/tex] and [tex]\cos\gamma=-\cos(\alpha+\beta)[/tex]:
[tex]\sin 2\alpha+\sin 2\beta=2\sin \gamma\cos(\alpha-\beta), \quad\sin 2\gamma=2\sin \gamma\cos\gamma=-2\sin \gamma\cos(\alpha+\beta)[/tex]
So the left side equals
[tex]2\sin \gamma\left[\cos(\alpha-\beta)-\cos(\alpha+\beta)\right]=2\sin \gamma\times 2\sin \alpha\sin \beta=4\sin \alpha\sin \beta\sin \gamma[/tex]

Solution:
[tex]\sin \alpha+\sin \beta=2\sin \frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}, \quad\sin \gamma=\sin (\alpha+\beta)=2\sin \frac{\alpha+\beta}{2}\cos\frac{\alpha+\beta}{2}[/tex]
Since [tex]\sin \frac{\alpha+\beta}{2}\ne 0[/tex], the condition gives [tex]\cos\frac{\alpha-\beta}{2}=2\cos\frac{\alpha+\beta}{2}[/tex]. Expanding:
[tex]\cos\frac{\alpha}{2}\cos\frac{\beta}{2}+\sin \frac{\alpha}{2}\sin \frac{\beta}{2}=2\cos\frac{\alpha}{2}\cos\frac{\beta}{2}-2\sin \frac{\alpha}{2}\sin \frac{\beta}{2}[/tex]
So [tex]3\sin \frac{\alpha}{2}\sin \frac{\beta}{2}=\cos\frac{\alpha}{2}\cos\frac{\beta}{2}[/tex], i.e. [tex]\tan \frac{\alpha}{2}\tan \frac{\beta}{2}=\frac{1}{3}[/tex]. The required value is [tex]\frac{2}{3}[/tex].

Solution:
Write the product as
[tex]\frac{\sin \alpha\sin (60^{\circ}-\alpha)}{\cos\alpha\cos(60^{\circ}-\alpha)}=\frac{\cos(2\alpha-60^{\circ})-\cos 60^{\circ}}{\cos(2\alpha-60^{\circ})+\cos 60^{\circ}}[/tex]
Let [tex]u=\cos(2\alpha-60^{\circ})[/tex]; then [tex]\frac{1}{2}\lt u\le 1[/tex]. The expression equals
[tex]\frac{u-\frac{1}{2}}{u+\frac{1}{2}}=1-\frac{1}{u+\frac{1}{2}}[/tex]
and it increases with [tex]u[/tex]. The maximum is at [tex]u=1[/tex], i.e. [tex]\alpha=30^{\circ}[/tex]: [tex]1-\frac{2}{3}=\frac{1}{3}[/tex].

Solution:
[tex]\sin 3\alpha+\sin 3\beta=2\sin \frac{3(\alpha+\beta)}{2}\cos\frac{3(\alpha-\beta)}{2}, \quad\sin 3\gamma=\sin 3(\alpha+\beta)=2\sin \frac{3(\alpha+\beta)}{2}\cos\frac{3(\alpha+\beta)}{2}[/tex]
So the sum equals
[tex]2\sin \frac{3(\alpha+\beta)}{2}\left[\cos\frac{3(\alpha-\beta)}{2}+\cos\frac{3(\alpha+\beta)}{2}\right]=4\sin \frac{3(\alpha+\beta)}{2}\cos\frac{3\alpha}{2}\cos\frac{3\beta}{2}[/tex]
and [tex]\sin \frac{3(\alpha+\beta)}{2}=\sin \left(270^{\circ}-\frac{3\gamma}{2}\right)=-\cos\frac{3\gamma}{2}[/tex]. Hence [tex]\cos\frac{3\alpha}{2}\cos\frac{3\beta}{2}\cos\frac{3\gamma}{2}=0[/tex]. Every [tex]\frac{3x}{2}[/tex] lies in [tex]\left(0^{\circ}, 270^{\circ}\right)[/tex], so one of the angles satisfies [tex]\frac{3x}{2}=90^{\circ}[/tex], i.e. it equals [tex]60^{\circ}[/tex]. The other two angles have sum [tex]120^{\circ}[/tex], so one of them is at most [tex]60^{\circ}[/tex] and the other is at least [tex]60^{\circ}[/tex]. So the middle angle is [tex]60^{\circ}[/tex].

Solution:
Let [tex]t=\sin \alpha[/tex]:
[tex]15t^2-29t+12=0\iff t=\frac{29\pm 11}{30}[/tex]
i.e. [tex]t=\frac{4}{3}[/tex] (impossible) or [tex]t=\frac{3}{5}[/tex]. In the second quadrant [tex]\cos\alpha=-\frac{4}{5}[/tex]. Then
[tex]\tan \frac{\alpha}{2}=\frac{\sin \alpha}{1+\cos\alpha}=\frac{\frac{3}{5}}{\frac{1}{5}}=3[/tex]

Solution:
The period of the tangent is [tex]180^{\circ}[/tex] and [tex]2015^{\circ}=11\times 180^{\circ}+35^{\circ}[/tex], so [tex]\tan 2015^{\circ}=\tan 35^{\circ}[/tex]. Since [tex]35^{\circ}=45^{\circ}-10^{\circ}[/tex]:
[tex]\tan 35^{\circ}=\frac{\tan 45^{\circ}-\tan 10^{\circ}}{1+\tan 45^{\circ}\tan 10^{\circ}}=\frac{1-a}{1+a}[/tex]