Trigonometric functions and their properties

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.

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

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.

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.

We use ((\tan \theta+\cot \theta)^2=\tan^2\theta+\cot^2\theta+2). Hence the value is (16-2=14).

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

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|} ).

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

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.

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.

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

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

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.

Use (\sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2}). Here the answer is (2\sin 2x\cos 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).

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

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

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

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.

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.