Trigonometric functions and their properties

Using the supplementary-angle identity, \(\cos(\pi-x)=-\cos x\). The cosine of \(\pi-x\) has the same magnitude as \(\cos x\), but the opposite sign. Note that \(\sin(\pi-x)=\sin x\), so the sine options do not apply here. Exam tip: for \(\pi-x\), remember that sine remains positive while cosine changes sign.

The period of (\tan x) is (\pi), so (\tan(\pi+x)=\tan x). Using period gives the answer quickly.

Sine is an odd function. Therefore, changing the sign of the angle changes the sign of its value: \(\sin(-x)=-\sin x\). Choosing \(\sin x\) would apply the even-function rule, so it is incorrect here. Exam tip: \(\sin\) and \(\tan\) are odd functions, whereas \(\cos\) is an even function.

(\cos x) is an even function, so (\cos(-x)=\cos x). Use even-odd properties for negative angle questions.

(\tan x) is an odd function, so (\tan(-x)=-\tan x). This can also be understood from (\frac{\sin x}{\cos x}).

In the first quadrant, all trigonometric functions are positive. Remember the quadrant sign table.

In the second quadrant, only (\sin x) and (\cosec x) are positive. Therefore, (\cos x) is negative.

In the third quadrant, (\tan x) and (\cot x) are positive. The sign rule gives the answer directly.

In the fourth quadrant, the y-coordinate of a point is negative. Since \(\sin x=\frac{y}{r}\) and \(r\) is always positive, \(\sin x\) is negative. It is zero only on the x-axis, not within a quadrant. Exam tip: In the fourth quadrant, remember that cosine is positive and sine is negative.

The sine function has period \(2\pi\). Hence, adding \(2\pi\) to any angle \(x\) does not change its sine: \(\sin(2\pi+x)=\sin x\). The value \(-\sin x\) occurs on adding \(\pi\), since \(\sin(\pi+x)=-\sin x\). Exam tip: remember \(\sin(x+2n\pi)=\sin x\).

The cosine function has period \(2\pi\). Therefore, adding \(2\pi\) to any angle \(x\) does not change its cosine: \(\cos(x+2\pi)=\cos x\). Hence, \(\cos x\) is correct. The expression \(-\cos x\) occurs on adding \(\pi\), since \(\cos(x+\pi)=-\cos x\). Exam tip: Remember that both \(\sin x\) and \(\cos x\) have period \(2\pi\).

By the complementary angle identity, (\sin(\frac{\pi}{2}-x)=\cos x). With (\frac{\pi}{2}), the function changes.

By the complementary angle identity, (\cos(\frac{\pi}{2}-x)=\sin x). In such questions, (\sin x) and (\cos x) interchange.

For a complementary angle with (\frac{\pi}{2}), (\tan x) changes to (\cot x). Remember cofunction identities.

The period of (\sec x) is (2\pi), like (\cos x). The reciprocal function keeps the related basic period.

(\cot x) repeats its value after every (\pi). Both (\tan x) and (\cot x) have period (\pi).

(\cosec x) is the reciprocal of (\sin x) and its period is (2\pi). Use reciprocal relation to identify the period.

(\sec x=\frac{1}{\cos x}) and (\cos x) lies in ([-1,1]). Hence (\sec x) is greater than or equal to (1) or less than or equal to (-1).

\(\cosec x=\frac{1}{\sin x}\). Since \(-1\leq\sin x\leq 1\), and \(\sin x\neq0\) wherever cosecant is defined, its reciprocal satisfies \(\cosec x\leq-1\) or \(\cosec x\geq1\). Hence, the range is \((-\infty,-1]\cup[1,\infty)\). The interval \([-1,1]\) is the range of \(\sin x\), not of \(\cosec x\). Exam tip: reciprocal trigonometric functions cannot take values between \(-1\) and \(1\).

(\tan x=\frac{\sin x}{\cos x}), so it is undefined when (\cos x=0). A fraction is not defined when the denominator is zero.