\(\sin^2\alpha(1-\cot^2\alpha)=\sin^2\alpha-\cos^2\alpha=-\cos 2\alpha\). Answer: E.
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2
Task
One correct answer
Which interval contains the value of \(\sin 410^\circ\)?
Solution
\(\sin410^\circ=\sin50^\circ\). Since sine is increasing on \((0^\circ,90^\circ)\) and \(45^\circ<50^\circ<60^\circ\), \(\dfrac{\sqrt2}{2}<\sin50^\circ<\dfrac{\sqrt3}{2}\). Answer: D.
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3
Task
One correct answer
Choose the interval containing \(\cot25^\circ\).
Solution
On \((0^\circ,90^\circ)\), \(\cot x\) is decreasing. Since \(25^\circ<30^\circ\), \(\cot25^\circ>\cot30^\circ=\sqrt3\). Thus \(\cot25^\circ\in(\sqrt3,+\infty)\). Answer: E.
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4
Task
One correct answer
The unit circle shows the point \(P(-0.8,0.6)\) and angle \(\alpha\). Find \(\cos\alpha\).
Solution
On the unit circle, cosine equals the x-coordinate of the corresponding point. Hence \(\cos\alpha=-0.8\). Answer: A.
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5
Task
One correct answer
Simplify \(2\cos(450^\circ+\alpha)-\sin\alpha\).
Solution
\(\cos(450^\circ+\alpha)=\cos(90^\circ+\alpha)=-\sin\alpha\). Thus \(2(-\sin\alpha)-\sin\alpha=-3\sin\alpha\). Answer: B.
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6
Task
One correct answer
Calculate \(\sin210^\circ\).
Solution
\(210^\circ=180^\circ+30^\circ\), so \(\sin210^\circ=-\sin30^\circ=-\dfrac12\). Answer: A.
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7
Task
One correct answer
The expression \(\sin^2 2x\) is identically equal to:
Solution
Using \(\sin2x=2\sin x\cos x\), we get \(\sin^22x=4\sin^2x\cos^2x\). Answer: C.
Since \(\sin^2\beta+\cos^2\beta=1\), the expression equals \(\dfrac{7-1}{3\cdot1}=2\). Answer: E.
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9
Task
One correct answer
Calculate \(4\sin^2\alpha\) if \(4\cos^2\alpha=1\).
Solution
From \(4\cos^2\alpha=1\), \(\cos^2\alpha=\dfrac14\). Hence \(\sin^2\alpha=\dfrac34\), so \(4\sin^2\alpha=3\). Answer: D.
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10
Task
One correct answer
Find the least positive period of \(y=\dfrac12\tan(4x)\).
Solution
For \(\tan(kx)\), the least positive period is \(T=\dfrac\pi{|k|}\). With \(k=4\), \(T=\dfrac\pi4\). The factor \(\dfrac12\) does not affect the period. Answer: A.
The principal value of \(\arccos x\) lies in \([0,\pi]\). Since \(\cos\dfrac{\pi}{6}=\dfrac{\sqrt3}{2}\), \(\arccos\left(-\dfrac{\sqrt3}{2}\right)=\pi-\dfrac{\pi}{6}=\dfrac{5\pi}{6}\). Answer: E.
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12
Task
Numeric answer
Find the value of \(\tan^2\alpha+\cot^2\alpha\) if \(\tan\alpha-\cot\alpha=4\).
Solution
Square the equation: \((\tan\alpha-\cot\alpha)^2=16\). Thus \(\tan^2\alpha+\cot^2\alpha-2\tan\alpha\cot\alpha=16\). Since \(\tan\alpha\cot\alpha=1\), the required value is \(18\).
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13
Task
One correct answer
Arrange in increasing order: \(a=\tan36^\circ\), \(b=\tan93^\circ\), \(c=\tan180^\circ\).
Solution
\(a=\tan36^\circ>0\). Also \(b=\tan93^\circ=\tan(90^\circ+3^\circ)=-\cot3^\circ<0\), while \(c=\tan180^\circ=0\). Hence \(b
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14
Task
Numeric answer
Find \(\tan\alpha+\cot\alpha\) if \(\alpha=15^\circ\).
Solution
\(\tan\alpha+\cot\alpha=\dfrac{1}{\sin\alpha\cos\alpha}=\dfrac{2}{\sin2\alpha}\). For \(\alpha=15^\circ\): \(2/\sin30^\circ=4\).
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15
Task
One correct answer
Given \(\cot\alpha\lt0\) and \(\cos\alpha\gt0\), which value can \(\sin\alpha\) take?
Solution
Since \(\cot\alpha=\dfrac{\cos\alpha}{\sin\alpha}\lt0\) and \(\cos\alpha\gt0\), we must have \(\sin\alpha\lt0\). The value \(\sin\alpha=-1\) is impossible when \(\cos\alpha\gt0\). Thus only \(-\dfrac12\) works. Answer: B.
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16
Task
Matching
Match each expression (1–4) with an identically equal expression (A–E).
А
\(\cos^2\alpha\) Б
\(\cos2\alpha\) В
\(\sin2\alpha\) Г
\(-\cos2\alpha\) Д
\(\sin^2\alpha\)
Solution
1) \(1-\cos^2\alpha=\sin^2\alpha\) — E. 2) \(2\sin\alpha\cos\alpha=\sin2\alpha\) — C. 3) \(\cos^2\alpha-\sin^2\alpha=\cos2\alpha\) — B. 4) \(\cos^4\alpha+\sin^2\alpha\cos^2\alpha=\cos^2\alpha\) — A. Matching: 1–E, 2–C, 3–B, 4–A.
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17
Task
One correct answer
Simplify \((1-\cos^2\alpha)\cot^2\alpha\).
Solution
\(1-\cos^2\alpha=\sin^2\alpha\) and \(\cot^2\alpha=\dfrac{\cos^2\alpha}{\sin^2\alpha}\). Hence the expression simplifies to \(\cos^2\alpha\). Answer: A.
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18
Task
Numeric answer
Calculate \(2\sin\alpha\cos\alpha\) if \(\sin\alpha+\cos\alpha=1.2\).
Solution
Square the condition: \((\sin\alpha+\cos\alpha)^2=1.44\). Since \(\sin^2\alpha+\cos^2\alpha=1\), we get \(1+2\sin\alpha\cos\alpha=1.44\), so \(2\sin\alpha\cos\alpha=0.44\).
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19
Task
Numeric answer
Calculate \(\cos\alpha\) if \(\sin\alpha=0.8\) and \(\dfrac{\pi}{2}\lt\alpha\lt\pi\).
Solution
From \(\sin^2\alpha+\cos^2\alpha=1\), \(\cos^2\alpha=1-0.8^2=0.36\), so \(\cos\alpha=\pm0.6\). Since \(\alpha\) is in quadrant II, cosine is negative; hence \(\cos\alpha=-0.6\).
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20
Task
One correct answer
Simplify \(\dfrac{1}{1+\tan^2\alpha}\).
Solution
Using \(1+\tan^2\alpha=\dfrac{1}{\cos^2\alpha}\), we get \(\dfrac{1}{1+\tan^2\alpha}=\cos^2\alpha\). Answer: A.
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21
Task
One correct answer
Which interval contains the value of \(\sin 415^\circ\)?
Solution
\(415^\circ=360^\circ+55^\circ\), so \(\sin415^\circ=\sin55^\circ\). Sine is increasing in the first quadrant, and \(45^\circ<55^\circ<60^\circ\). Hence \(\dfrac{\sqrt2}{2}<\sin55^\circ<\dfrac{\sqrt3}{2}\). Answer: D.
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22
Task
One correct answer
Calculate \(\tan\alpha\) if \(4\sin\alpha-\cos\alpha=2\cos\alpha-\sin\alpha\).
Solution
Rearranging gives \(5\sin\alpha=3\cos\alpha\). Since \(\cos\alpha=0\) does not satisfy this equation, divide by \(\cos\alpha\): \(5\tan\alpha=3\). Thus \(\tan\alpha=\dfrac35\). Answer: A.
Using \(\sin(-x)=-\sin x\), \(\cos(-x)=\cos x\), \(\tan(-x)=-\tan x\), and \(\cot(-x)=-\cot x\), the numerator is \((-\tfrac12)(\tfrac12)-1=-\tfrac54\), while the denominator is \((-\tfrac1{\sqrt3})(\tfrac{\sqrt3}{2})-(-\tfrac{\sqrt3}{2})(-\tfrac1{\sqrt3})=-1\). Thus the value is \(\dfrac54=1.25\).
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24
Task
One correct answer
Calculate \(\dfrac{2\cot\alpha}{\tan\alpha}\) if \(\tan\alpha=\dfrac15\).
Solution
Since \(\cot\alpha=\dfrac1{\tan\alpha}=5\), \(\dfrac{2\cot\alpha}{\tan\alpha}=\dfrac{2\cdot5}{1/5}=50\). Answer: A.
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25
Task
One correct answer
Simplify \(1-\sin^2\alpha-\cos^2\alpha\).
Solution
By the fundamental identity \(\sin^2\alpha+\cos^2\alpha=1\). Therefore \(1-\sin^2\alpha-\cos^2\alpha=0\). Answer: B.
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26
Task
One correct answer
If \(2\sin\alpha=\cos\alpha\), then \(\tan\alpha=\ldots\)
Solution
From \(2\sin\alpha=\cos\alpha\), \(\cos\alpha\ne0\). Divide by \(\cos\alpha\): \(2\tan\alpha=1\), so \(\tan\alpha=\dfrac12=0.5\). Answer: D.
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27
Task
One correct answer
Choose the correct inequality if \(a=\sin120^\circ\) and \(b=\cos120^\circ\).
Solution
\(120^\circ\) lies in quadrant II. Thus \(a=\sin120^\circ=\dfrac{\sqrt3}{2}>0\) and \(b=\cos120^\circ=-\dfrac12<0\). Hence \(b\lt0\lt a\). Answer: E.
Using the difference of squares, \(\cos^4x-\sin^4x=(\cos^2x-\sin^2x)(\cos^2x+\sin^2x)=\cos2x\). For \(x=\dfrac{\pi}{12}\), this is \(\cos\dfrac{\pi}{6}=\dfrac{\sqrt3}{2}\). Answer: B.
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29
Task
One correct answer
Find the least positive period of the function \(y=2\cot(3x)\).
Solution
\(\cot x\) has period \(\pi\). For \(\cot(kx)\), the least positive period is \(T=\dfrac{\pi}{|k|}\). With \(k=3\), \(T=\dfrac{\pi}{3}\). The factor 2 does not affect the period. Answer: C.
Since \(\sqrt{u^2}=|u|\) and \(\sin45^\circ=\cos45^\circ=\dfrac{\sqrt2}{2}\), the expression becomes \(|\sqrt2+1|-|1-\sqrt2|\). As \(1-\sqrt2<0\), this equals \(\sqrt2+1-(\sqrt2-1)=2\). Answer: E.
By the reduction formula, \(\cos(90^\circ+\alpha)=-\sin\alpha\). Hence \(\dfrac{-\sin\alpha}{\sin\alpha}=-1\). Answer: A.
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34
Task
One correct answer
Which interval contains the value of \(\sin\dfrac{7\pi}{6}-1\)?
Solution
\(\sin\dfrac{7\pi}{6}=\sin(\pi+\dfrac{\pi}{6})=-\sin\dfrac{\pi}{6}=-\dfrac12\). Thus the value is \(-\dfrac32\), and \(-\dfrac32\in[-2;-1)\). Answer: B.
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35
Task
One correct answer
Simplify \((1-\sin^2\alpha)\tan^2\alpha\).
Solution
\(1-\sin^2\alpha=\cos^2\alpha\) and \(\tan^2\alpha=\dfrac{\sin^2\alpha}{\cos^2\alpha}\). Therefore the expression equals \(\sin^2\alpha\). Answer: D.
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36
Task
One correct answer
Simplify \(2\sin^2\alpha\cdot\cot\alpha\).
Solution
\(2\sin^2\alpha\cdot\cot\alpha=2\sin^2\alpha\cdot\dfrac{\cos\alpha}{\sin\alpha}=2\sin\alpha\cos\alpha=\sin2\alpha\). Answer: E.
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37
Task
One correct answer
Calculate \(\cos210^\circ\).
Solution
\(210^\circ=180^\circ+30^\circ\), so \(\cos210^\circ=-\cos30^\circ=-\dfrac{\sqrt3}{2}\). Answer: D.
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38
Task
One correct answer
If \(2\cos\alpha-5\sin\alpha=0\), then \(\tan\alpha=\ldots\)
Solution
From \(2\cos\alpha=5\sin\alpha\). Since \(\cos\alpha
e0\), divide by \(5\cos\alpha\): \(\tan\alpha=\dfrac25\). Answer: A.
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39
Task
One correct answer
Simplify \((1+\tan^2\alpha)\sin^2\alpha\).
Solution
Using \(1+\tan^2\alpha=\dfrac1{\cos^2\alpha}\), \((1+\tan^2\alpha)\sin^2\alpha=\dfrac{\sin^2\alpha}{\cos^2\alpha}=\tan^2\alpha\). Answer: E.
Using \(\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}\), \(\dfrac{\cos\alpha\cdot\tan\alpha}{\sin^2\alpha}=\dfrac{\sin\alpha}{\sin^2\alpha}=\dfrac1{\sin\alpha}\). Answer: C.
