Trigonometric functions and their properties

Using the subtraction formula, \(\sin(\pi-x)=\sin\pi\cos x-\cos\pi\sin x=0-(-1)\sin x=\sin x\). The angle \(\pi-x\) is in quadrant II, where sine is positive. Exam tip: check the quadrant sign for identities involving \(\pi\pm x\).

\\(\tan x=\frac{\sin x}{\cos x}\\). At \\(x=\frac{\pi}{2}+n\pi\\), \\(\cos x=0\\), so division by zero makes \\(\tan x\\) undefined. Exam tip: for a quotient function, first check where its denominator is zero.

The value of \(\sin x\) always lies from \(-1\) to \(1\), and both endpoints occur: \(\sin(\pi/2)=1\) and \(\sin(3\pi/2)=-1\). Hence the endpoints must be included. Exam tip: use inclusive inequality signs for the range of sine.

(\cos^2 x=1-\frac{1}{4}=\frac{3}{4}). In the first quadrant, (\cos x=\frac{\sqrt{3}}{2}) is taken.

\(\sin^2 x=1-\frac{9}{16}=\frac{7}{16}\). In the fourth quadrant, (\sin x) is negative.

The period of (\cos kx) is \(\frac{2\pi}{k}\). Here \(k=5\), so the period is \(\frac{2\pi}{5}\).

For a sine function of the form \(a\sin bx\), the amplitude is \(|a|\). Here, \(a=4\), so the amplitude is \(|4|=4\). The factor \(2\) affects the period of the function, not its amplitude. Exam tip: To find amplitude, take the absolute value of the coefficient outside \(\sin\) or \(\cos\).

Since \(\cos x\) lies between \(-1\) and \(1\), \(-3\cos x\) lies between \(-3\) and \(3\). When \(\cos x=1\), we get \(-3\cos x=-3\); hence the minimum value is \(-3\). The value \(3\) is the maximum, obtained when \(\cos x=-1\). Exam tip: multiplying a trigonometric function by a negative number reverses the order of its maximum and minimum values.

Since \(\sin x\) always lies between \(-1\) and \(1\), the minimum value of \(2+\sin x\) is \(2-1=1\) and its maximum value is \(2+1=3\). Hence, its range is \([1,3]\). The interval \([-1,1]\) is the range of \(\sin x\) alone, not of \(2+\sin x\). Exam tip: when a constant is added to a function, add that constant to both endpoints of its range.

Since \(\cos x\) lies in \([-1,1]\), the minimum value of \(1-\cos x\) is \(0\) when \(\cos x=1\), and its maximum value is \(2\) when \(\cos x=-1\). Hence, the range is \([0,2]\). The interval \([-1,1]\) is the range of \(\cos x\), not of \(1-\cos x\). Exam tip: for an expression such as \(a-\cos x\), substitute the endpoint values \(-1\) and \(1\) of \(\cos x\).

Dividing numerator and denominator by (\cos x) gives (\frac{\tan x+1}{\tan x-1}). Substituting (\tan x=3) gives (2).

The value is (\frac{2+\sqrt{3}}{2-\sqrt{3}}). Rationalising the denominator gives (7+4\sqrt{3}).

\(\tan x=\frac{\sin x}{\cos x}\). For \(x=\frac{(2n+1)\pi}{2}\), \(\cos x=0\), so \(\tan x\) is undefined. In contrast, \(\sin x\) and \(\cos x\) are defined for every real \(x\). Exam tip: for a quotient function, first check where its denominator becomes zero.

Using the identity \(\cosec^2 x-\cot^2 x=1\), we get \((\cosec x-\cot x)(\cosec x+\cot x)=1\). Given \(\cosec x-\cot x=\frac{1}{3}\), so \(\frac{1}{3}(\cosec x+\cot x)=1\). Therefore, \(\cosec x+\cot x=3\). The value \(\frac{1}{3}\) is the given expression; its reciprocal gives 3. Exam tip: recognise \(\cosec^2 x-\cot^2 x=1\) and multiply the conjugate expressions in such questions.

Using the difference of squares, \(\sin^4 x-\cos^4 x=(\sin^2 x-\cos^2 x)(\sin^2 x+\cos^2 x)\). Since \(\sin^2 x+\cos^2 x=1\), the simplified value is \(\sin^2 x-\cos^2 x\). Option B is its negative, so it is not correct. Exam tip: after applying \(a^2-b^2=(a-b)(a+b)\), use the fundamental trigonometric identity.

The numerator is (1+\tan^2 x=\sec^2 x) and the denominator is (1+\cot^2 x=\cosec^2 x). Their ratio is (\frac{\sec^2 x}{\cosec^2 x}=\tan^2 x).

Rationalising the denominator gives (\frac{(1+\sin x)^2}{\cos^2 x}). This equals ((\sec x+\tan x)^2).

By the half-angle identity, (\tan^2 \frac{x}{2}=\frac{1-\cos x}{1+\cos x}). Remember half-angle forms separately.

The standard half-angle identity is (\tan \frac{x}{2}=\frac{\sin x}{1+\cos x}). Identify the half-angle form quickly.

First, ((\sin x+\cos x)^2=1+2\sin x\cos x) gives (\sin x\cos x=\frac{5}{8}). Then ((\sin x-\cos x)^2=1-2\sin x\cos x=\frac{7}{4}).