Trigonometric functions and their properties

Using the formulas for \(\tan\left(\frac{\pi}{4}+x\right)\) and \(\tan\left(\frac{\pi}{4}-x\right)\), the product is (1). This is a useful standard result.

The first-quadrant condition tells us that both sine and cosine are positive. The ratio \\(\\tan x=\\frac{7}{24}\\) means perpendicular divided by base is 7 divided by 24 in a right-triangle model. Take the perpendicular as 7 units and the base as 24 units. By the Pythagorean theorem, the hypotenuse is \\(\\sqrt{7^2+24^2}=\\sqrt{49+576}=\\sqrt{625}=25\\).

Sine is defined as perpendicular divided by hypotenuse. Therefore, \\(\\sin x=\\frac{7}{25}\\), so option A is correct. The first-quadrant information ensures that the positive value is selected; there is no negative sign. Option B is the cosine value for this triangle, while option D is the tangent value already given. Option C is greater than 1, which cannot be the value of sine for a real angle.

The period of \\(\\cos x\\) is \\(2\\pi\\), because its values repeat after an increase of \\(2\\pi\\). For \\(\\cos 2x\\), the period is \\(\\pi\\), since increasing \\(x\\) by \\(\\pi\\) increases its angle by \\(2\\pi\\). The sum can repeat only at a common period of both terms.

Although \\(\\pi\\) is a period of \\(\\cos 2x\\), it is not a period of \\(\\cos x\\). The common fundamental period is therefore \\(2\\pi\\). Indeed, replacing \\(x\\) by \\(x+2\\pi\\) leaves both cosine terms unchanged. Hence option B is correct, not option C.

For any real (x), (-1\leq \sin x\leq 1). In exams, apply the basic range of sine and cosine first.

Since (\tan x=\frac{\sin x}{\cos x}), it is undefined when (\cos x=0). Thus (x=\frac{\pi}{2}+n\pi).

We use \(\sin x+\cos x=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)\). The maximum \(\sqrt{2}\) occurs at \(x=\frac{\pi}{4}\).

By the double angle formula, (2\sin x\cos x=\sin 2x). This identity is used frequently in exams.

The direct answer is cos 2x, so option C is correct. The double-angle identity follows from the standard formulas: cos 2x=cos^2 x−sin^2 x. Thus the expression already has exactly the form of cos 2x. Option A, sin 2x, is wrong because sin 2x=2sin x cos x, not a difference of squares. Option B, tan 2x, is wrong because tangent is a ratio, tan 2x=\frac{2tan x}{1−tan^2 x}, and it is not equal to this expression in general. Option C, cos 2x, is correct. Option D, 1+sin 2x, is also not generally equal; it contains an additional 1 and a sine term. For example, at x=0 the given expression is 1, cos 0 is 1, but sin 0 and tan 0 are 0, showing why the alternatives fail. Memory cue: remember the three equivalent forms: cos 2x=cos^2 x−sin^2 x=2cos^2 x−1=1−2sin^2 x.

Since \(\sec x=1/\cos x\), it is undefined when \(\cos x=0\), i.e. at \(x=\frac{(2n+1)\pi}{2}\). Also, \(|\sec x|\ge1\), giving the stated range. For \(\csc x\), the excluded values are \(n\pi\). Exam tip: first locate angles that make the denominator zero in reciprocal functions.

We have (\tan x+\cot x=\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}). Simplifying gives (\frac{1}{\sin x\cos x}=\sec x\cosec x).

The direct answer is option B, \(\frac{\pi}{3}\). The tangent function repeats after an angle change of \(\pi\), because \(\tan(\theta+\pi)=\tan\theta\). For \(f(x)=\tan 3x\), increase x by T. We need \(3(x+T)=3x+\pi\), so \(3T=\pi\), giving \(T=\frac{\pi}{3}\). Option A, \(3\pi\), is too large and is not the fundamental period. Option B is correct because it is the smallest positive repeating interval. Option C, \(\frac{2\pi}{3}\), is also a repeating interval, but it is twice the smallest one, so it is not fundamental. Option D, \(\pi\), is the period of \(\tan x\), not of \(\tan 3x\). Memory cue: multiplying x by 3 divides the tangent period by 3.

The period of (\cos ax) is ( \frac{2\pi}{|a|} ). Hence the period of (\cos 4x) is ( \frac{\pi}{2} ).

In the second quadrant, (\cos x) is negative. From (\sin^2 x+\cos^2 x=1), (\cos x=-\frac{12}{13}).

In the third quadrant, both (\sin x) and (\cos x) are negative, while (\tan x) is positive. Hence (\sin x=-\frac{4}{5}).

The standard triple-angle identity is \(\sin 3x=3\sin x-4\sin^3 x\), so option B is correct. Option C is the negative of option B and hence equals \(-\sin 3x\). Option D is related to the triple-angle formula for cosine, not sine. Exam tip: remember that the \(\sin^3 x\) term in \(\sin 3x\) has coefficient 4 and a minus sign.

The direct answer is option C: \(\cos 3x=4\cos^3x-3\cos x\). Start with the addition formula: \(\cos(2x+x)=\cos2x\cos x-\sin2x\sin x\). Using \(\cos2x=2\cos^2x-1\) and \(\sin2x=2\sin x\cos x\), we get \(\cos3x=(2\cos^2x-1)\cos x-2\sin^2x\cos x\). Since \(\sin^2x=1-\\cos^2x\), simplification gives \(4\cos^3x-3\cos x\). Option C is therefore correct. Option A has the terms reversed and equals \(-\cos3x\), not \(\cos3x\). Option B is the sine triple-angle expression, because \(\sin3x=3\sin x-4\sin^3x\), so its sign/order is also not the required cosine formula. Option D has an incorrect plus sign; the cubic and linear terms must be subtracted. Memory cue: cosine triple angle is “four cube minus three,” while sine is “three sine minus four sine-cube.”

(\cos x) is negative in the second and third quadrants. Hence (x=\frac{2\pi}{3},\frac{4\pi}{3}).

The period of (\sin 2x) is (\pi) and the period of (\sin 4x) is (\frac{\pi}{2}). Their common fundamental period is (\pi).

Since (\sin x(1+\sin x)=0), (\sin x=0) or (\sin x=-1). In the given interval, (x=0,\pi,\frac{3\pi}{2}).

In the fourth quadrant, (\sin \theta) is negative and (\cos \theta) is positive. Hence (\sin \theta=-\frac{15}{17}) and (\tan \theta=-\frac{15}{8}).