Trigonometric functions and their properties

Since \(\cos x\) lies between \(-1\) and \(1\), \(5-2\cos x\) is smallest when \(\cos x=1\). Thus, its minimum value is \(5-2(1)=3\). The value \(7\) is the maximum, obtained when \(\cos x=-1\). Exam tip: For a trigonometric term with a negative coefficient, use the maximum value of that trigonometric function to find the minimum of the expression.

\(\cos x=\frac{12}{13}\) and \(\tan x=\frac{\sin x}{\cos x}\). Hence \(\tan x=\frac{5}{12}\).

(\sin x=\frac{15}{17}) and (\cot x=\frac{\cos x}{\sin x}). Therefore, (\cot x=\frac{8}{15}).

From (\tan x=\frac{3}{4}), the hypotenuse in a right triangle is (5). Hence (\sin x=\frac{3}{5}).

For (\cot x=\frac{5}{12}), take adjacent as (5) and opposite as (12). The hypotenuse is (13), so (\cos x=\frac{5}{13}).

Using allied-angle identities, \(\sin(\pi+x)=-\sin x\) and \(\sin(\pi-x)=\sin x\). Therefore, \(\sin(\pi+x)+\sin(\pi-x)=-\sin x+\sin x=0\). The result \(2\sin x\) would occur only if both terms had the same sign, which they do not here. Exam tip: for angles of the form \(\pi\pm x\), determine the sign by identifying the quadrant.

Using allied-angle identities, \(\cos(\pi+x)=-\cos x\) and \(\cos(\pi-x)=-\cos x\). Hence, \(\cos(\pi+x)+\cos(\pi-x)=-\cos x-\cos x=-2\cos x\). The value \(0\) may occur only for particular values such as when \(\cos x=0\); it is not the general simplified form. Exam tip: For angles of the form \(\pi\pm x\), first check the sign of the trigonometric function.

The period of \(\tan\) is \(\pi\), so \(\tan(\pi+x)=\tan x\). Also, \(\tan(\pi-x)=\tan(-x)=-\tan x\). Hence, \(\tan(\pi+x)-\tan(\pi-x)=\tan x-(-\tan x)=2\tan x\). The value \(-2\tan x\) would result if the order of the two terms were reversed. Exam tip: carefully retain the negative sign in \(\tan(\pi-x)\).

In the form (\frac{\pi}{2}+x), (\sin) changes to (\cos) and the sign remains positive. Hence the answer is (\cos x).

In the form (\frac{\pi}{2}+x), (\cos) changes to (\sin) with a negative sign. Hence it becomes (-\sin x).

At (\frac{\pi}{2}+x), (\tan) changes to (\cot) with a negative sign. Hence (\tan(\frac{\pi}{2}+x)=-\cot x).

(\cos(\frac{\pi}{2}-x)=\sin x), so (\sec(\frac{\pi}{2}-x)=\frac{1}{\sin x}=\cosec x). Complementary angles give cofunctions.

(\sin(\frac{\pi}{2}-x)=\cos x), so (\cosec(\frac{\pi}{2}-x)=\frac{1}{\cos x}=\sec x). Use reciprocal and cofunction identities together.

The identity gives \(|\cos x|=\frac{24}{25}\). In the second quadrant, \(\cos x\) is negative.

From (\sec^2 x=1+\tan^2 x), (|\sec x|=\frac{41}{40}). In the third quadrant, (\sec x) is negative.

The identity gives \(|\sin x|=\frac{8}{17}\). In the fourth quadrant, \(\sin x\) is negative.

From (\cosec^2 x=1+\cot^2 x), (|\cosec x|=\frac{5}{4}). In the second quadrant, (\cosec x) is positive.

Using the identity \(\sec^2 x=1+\tan^2 x\), we get \(\sec^2 x-1=\tan^2 x\). Therefore, \(\frac{\sec^2 x-1}{\tan^2 x}=\frac{\tan^2 x}{\tan^2 x}=1\), wherever the given expression is defined. Option \(0\) is incorrect because the numerator and denominator are equal; when they are zero, the fraction is undefined rather than zero. Exam tip: Replace \(\sec^2 x-1\) directly with \(\tan^2 x\).

Using the Pythagorean trigonometric identity \(\cosec^2 x=1+\cot^2 x\), we get \(\cosec^2 x-1=\cot^2 x\). Therefore, \(\frac{\cosec^2 x-1}{\cot^2 x}=\frac{\cot^2 x}{\cot^2 x}=1\), wherever the expression is defined. \(\cot^2 x\) is not the final answer because it is only the simplified numerator. Exam tip: remember \(\cosec^2 x=1+\cot^2 x\) alongside \(\sec^2 x=1+\tan^2 x\).

Since \(\cos(-x)=\cos x\), \(\cos x\) is an even function. In contrast, \(\sin(-x)=-\sin x\), so \(\sin x\), \(\tan x\), and \(\cot x\) are odd. Exam tip: test parity by replacing \(x\) with \(-x\).