Trigonometric functions and their properties

Correct answer: C. ( 2\pi )

The periods of (\sin 3x) and (\cos 5x) are ( \frac{2\pi}{3} ) and ( \frac{2\pi}{5} ), whose common period is (2\pi). In exams, find each period separately first.

Correct answer: B. ( [1,5] )

Since (-1\leq \sin x \leq 1), (3+2\sin x) ranges from (1) to (5). In exams, write the base range first.

Correct answer: C. (2)

From (\sin x=\cos x), we get (\tan x=1), so (x=\frac{\pi}{4},\frac{5\pi}{4}). In exams, check the given interval carefully.

Correct answer: D. (1)

The minimum value of (\cos 2x) is (-1), so (4+3(-1)=1). In exams, remember the range of (\cos x) for maximum and minimum.

Correct answer: B. (1)

Since (\sin 70^\circ=\cos 20^\circ), the sum is (\sin^2 20^\circ+\cos^2 20^\circ=1). In exams, identify complementary angles.

Correct answer: C. ( 2\pi )

Both (\tan\frac{x}{2}) and (\cot\frac{x}{2}) have period (2\pi). In exams, use the period of (\tan ax) as ( \frac{\pi}{|a|} ).

Correct answer: D. \( [-\sqrt{2},\sqrt{2}] \)

We have \(\sin x+\cos x=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)\). Hence the range is \( [-\sqrt{2},\sqrt{2}] \).

Correct answer: B. Odd function

Given \(f(x)=\sin x\cos x\). Therefore, \(f(-x)=\sin(-x)\cos(-x)=(-\sin x)(\cos x)=-\sin x\cos x=-f(x)\). Hence, the function is odd. For an even function, we would need \(f(-x)=f(x)\), which is not true here. Exam tip: \(\sin x\) is odd and \(\cos x\) is even; the product of an odd and an even function is odd.

Correct answer: A. ( \frac{3}{4} )

We get (\cos x=\frac{4}{5}), so (\tan x=\frac{\sin x}{\cos x}=\frac{3}{4}). In exams, decide the sign from the quadrant.

Correct answer: C. (\cos x)

For \(\frac{\pi}{2}+x\), sine changes to cosine and remains positive in the second quadrant. Hence the value is (\cos x).

Correct answer: D. (-\cos x)

The form (\pi+x) is linked to the third quadrant where (\cos x) is negative. Hence (\cos(\pi+x)=-\cos x).

Correct answer: A. \(-\frac{3}{5}\)

Given \(\tan x=2\), we get \(\tan^2x=4\). Hence, \(\frac{1-\tan^2 x}{1+\tan^2 x}=\frac{1-4}{1+4}=\frac{-3}{5}=-\frac{3}{5}\). The value \(\frac{3}{5}\) results from a sign error in the numerator. Exam tip: square the given trigonometric value first, then substitute carefully in both numerator and denominator.

Correct answer: A. (2\sin 2x\cos x)

Direct answer: option A, \(2\sin 2x\cos x\). Use the sum-to-product identity \(\sin A+\sin B=2\sin\left((A+B)/2\right)\cos\left((A-B)/2\right)\). Put \(A=x\) and \(B=3x\). Then \((A+B)/2=(x+3x)/2=2x\), while \((A-B)/2=(x-3x)/2=-x\). Since \(\cos(-x)=\cos x\), the result is \(2\sin 2x\cos x\). Option A is correct. Option B, \(2\cos 2x\sin x\), corresponds to a different arrangement and is not generally equal to the given sum. Option C, \(2\sin x\cos 3x\), does not follow the sum-to-product formula and is generally different. Option D, \(2\cos x\sin 3x\), is also not the correct transformation. A quick check is to take a small convenient value such as \(x=30^\circ\); the correct expression matches the original sum. Memory cue: for sine plus sine, use sine of the average angle times cosine of half the difference.

Correct answer: B. (-2\sin 3x\sin x)

The formula is (\cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}). Hence the answer is (-2\sin 3x\sin x).

Correct answer: D. ( \pi )

In (|\sin x|), the negative part becomes positive, so the pattern repeats every (\pi). In exams, modulus can reduce the period.

Correct answer: A. ( \pi )

Since (\sin^2(x+\pi)=\sin^2 x), the fundamental period is (\pi). In exams, remember the period of squared sine and cosine.

Correct answer: B. ( \frac{1}{3} )

Since ((\sec x+\tan x)(\sec x-\tan x)=1), the other factor is (\frac{1}{3}). In exams, identify reciprocal pairs.

Correct answer: C. \(\tan x\)

Use the double-angle identities \(\sin 2x=2\sin x\cos x\) and \(1+\cos 2x=2\cos^2x\). Therefore, \(\frac{\sin 2x}{1+\cos 2x}=\frac{2\sin x\cos x}{2\cos^2x}=\frac{\sin x}{\cos x}=\tan x\). \(\cot x\) is the reciprocal of \(\tan x\), so it is not correct. Exam tip: first substitute the double-angle identities before cancelling common factors.

Correct answer: D. \(\tan x\)

Use the identities \(1-\cos 2x=2\sin^2x\) and \(\sin 2x=2\sin x\cos x\). Therefore, \(\frac{1-\cos 2x}{\sin 2x}=\frac{2\sin^2x}{2\sin x\cos x}=\frac{\sin x}{\cos x}=\tan x\). \(\cot x\) is the reciprocal of \(\tan x\), so it is not correct. Exam tip: first rewrite both numerator and denominator using double-angle identities.

Correct answer: A. (\sec^2 x)

The basic identity is (1+\tan^2 x=\sec^2 x). In exams, it can also be derived from (\sin^2 x+\cos^2 x=1).

Correct answer: B. 10

Use the identity \(\cosec^2 x=1+\cot^2 x\). Given \(\cot x=3\), we get \(\cot^2 x=9\). Therefore, \(\cosec^2 x=1+9=10\). Option 9 is only the value of \(\cot^2 x\), not of \(\cosec^2 x\). Exam tip: In the identity \(1+\cot^2 x=\cosec^2 x\), do not forget to add 1.

Correct answer: B. It is defined for all real \(x\) except \(x=\frac{(2n+1)\pi}{2}\), where \(n\in\mathbb{Z}\).

Since \(\tan x=\frac{\sin x}{\cos x}\), it is undefined wherever \(\cos x=0\). This occurs at \(x=\frac{(2n+1)\pi}{2}\). Option C excludes zeros of sine, which affect cotangent instead. Exam tip: check the denominator first.

Correct answer: A. \(x=2n\pi+y\) or \(x=(2n+1)\pi-y\), where \(n\in\mathbb Z\)

When \(\sin x=\sin y\), the two general families are \(x=2n\pi+y\) and \(x=(2n+1)\pi-y\), where \(n\in\mathbb Z\). The first follows from periodicity, while the second uses \(\sin(\pi-y)=\sin y\). Option B represents only some cases and misses the second family. Exam tip: for equal sine values, remember the angles \(y\) and \(\pi-y\).

Correct answer: B. (x=2n\pi\pm y)

For (\cos x=\cos y), angles are of the form (2n\pi\pm y). In exams, remember the even nature of cosine.

Correct answer: B. \(\cos 2x\)

Since \(\cos(-2x)=\cos 2x\), \(\cos 2x\) is even. Replacing \(x\) by \(x+\pi\) leaves it unchanged, and its range is \([-1,1]\). Although \(\sec 2x\) is even, its range excludes values between \(-1\) and \(1\). Exam tip: test \(f(-x)\) for parity.