Trigonometric functions and their properties

Squaring both sides gives (1+2\sin x\cos x=2). Hence (\sin x\cos x=\frac{1}{2}); squaring is useful in such questions.

By definition, \(\tan x=\frac{\sin x}{\cos x}\). Since \(\tan x=2\) is given, \(\frac{\sin x}{\cos x}=2\). The value \(\frac{1}{2}\) is the reciprocal and represents \(\cot x\), not \(\tan x\). Exam tip: Remember \(\tan x\) as \(\sin x/\cos x\).

\(\tan x=\frac{\sin x}{\cos x}\). It is undefined when \(\cos x=0\), i.e. at \(x=\frac{(2n+1)\pi}{2}\). At \(x=n\pi\), tan equals 0, so option A is not correct. Exam tip: link tan’s domain to non-zero cosine.

(\cosec x=\frac{1}{\sin x}). Hence (\sin x=\frac{1}{4}); learn each reciprocal function as a pair.

Using the Pythagorean identity \(\sin^2 x+\cos^2 x=1\), we get \(1-\cos^2 x=\sin^2 x\). Therefore, \(\frac{1-\cos^2 x}{\sin^2 x}=\frac{\sin^2 x}{\sin^2 x}=1\), provided \(\sin x\ne 0\). Neither \(\sin x\), \(\cos x\), nor \(\tan x\) is the simplified value of this ratio. Exam tip: first replace \(1-\cos^2 x\) with \(\sin^2 x\) in such questions.

The direct answer is 1, so option B is correct, subject to the usual domain condition that the denominator is defined. Begin with the fundamental identity (\sin^2 x+\cos^2 x=1). Rearranging gives 1−\sin^2 x=\cos^2 x. Substitute this into the fraction: \frac{1−\sin^2 x}{\cos^2 x}=\frac{\cos^2 x}{\cos^2 x}=1, whenever \cos x\ne0. Option A, \sec x, is wrong because the quotient cancels to 1, not to the reciprocal of cosine. Option B, 1, is correct. Option C, \tan x, is wrong because tangent is \frac{\sin x}{\cos x}; no such remaining factor exists after cancellation. Option D, \cot x, is also wrong because cotangent is \frac{\cos x}{\sin x}. If \cos x=0, the original fraction is undefined, so one should not claim a value there. Memory cue: when you see 1−\sin^2 x, replace it with \cos^2 x.

The governing transformation rule is that the period of \(\sin(kx)\), for nonzero k, is \(2\pi/|k|\). To verify it here, require \(\sin[3(x+T)]=\sin(3x)\). The sine argument must increase by one full cycle, so \(3T=2\pi\), giving \(T=2\pi/3\). Thus option C is correct. The coefficient 3 compresses the graph horizontally by a factor of 3, so its period is one-third of the basic sine period \(2\pi\). Option A ignores the coefficient, option B is only half of the required period, and option D corresponds to \(\pi/3\), which does not generally reproduce the same sine value.

The governing rule is that the fundamental period of \(\tan(kx)\) is \(\pi/|k|\), because tangent repeats after an argument increase of \(\pi\). For \(f(x)=\tan 4x\), require \(4(x+T)=4x+\pi\). Subtracting \(4x\) gives \(4T=\pi\), hence \(T=\pi/4\). Therefore, option D is correct. The factor 4 compresses the ordinary tangent graph horizontally by four, changing its period from \(\pi\) to one-fourth of that value. Option A is the period of \(\tan x\) without the coefficient; options B and C result from incorrect division and do not reproduce the function for every x.

Since \(\cos x\) lies between \(-1\) and \(1\), multiplying by 2 gives values of \(2\cos x\) from \(-2\) to \(2\). The endpoints are included because \(2\cos 0=2\) and \(2\cos \pi=-2\). Thus, \([-1,1]\) is the range of \(\cos x\), not of \(2\cos x\). Exam tip: the range of \(a\cos x\) is \([-|a|,|a|]\).

Since \(\sin x\) lies between \(-1\) and \(1\), its maximum value is \(1\). Therefore, the maximum of \(3\sin x+1\) is \(3\times1+1=4\). Option 3 is only the maximum of \(3\sin x\); adding the constant \(1\) shifts the maximum to 4. Exam tip: for \(a\sin x+b\), when \(a>0\), the maximum value is \(a+b\).

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.

The direct answer is option C, 5/12. In a right-triangle interpretation, sin x=opposite/hypotenuse=5/13. The Pythagorean relation gives the adjacent side: adjacent²=13²−5²=169−25=144, so adjacent=12. Because x is in the first quadrant, all basic ratios are positive, hence cos x=12/13. Now tan x=opposite/adjacent=5/12, or tan x=sin x/cos x=(5/13)/(12/13)=5/12. Option A, 12/13, is cos x, not tan x. Option B, 13/12, is sec x, the reciprocal of cosine. Option C is correct. Option D, 12/5, is cot x, the reciprocal of tan x. The first-quadrant condition matters because it fixes the positive sign. Memory cue: from the 5-12-13 triangle, sin=5/13, cos=12/13, tan=5/12.

The direct answer is option A, \(\frac{8}{15}\). In a right-triangle interpretation, cosine means adjacent side divided by hypotenuse: \(\cos x=8/17\). Therefore the opposite side is found by Pythagoras: \(8^2+b^2=17^2\), so \(b^2=289-64=225\), hence b=15. The first-quadrant condition makes sine and cosine positive, so \(\sin x=15/17\). Since \(\cot x=\cos x/\sin x\), we get \((8/17)/(15/17)=8/15\). Option A is correct. Option B, \(15/8\), is tan x, not cot x. Option C, \(17/8\), is the reciprocal of cosine, sec x. Option D, \(8/17\), is the given cosine, not cotangent. First quadrant is important because it fixes the positive sign.

The tangent ratio is \\(\\tan x=\\frac{\\text{opposite}}{\\text{adjacent}}=\\frac{3}{4}\\). In a right triangle, take the opposite side as 3 units and the adjacent side as 4 units. By the Pythagorean theorem, the hypotenuse is \\(\\sqrt{3^2+4^2}=\\sqrt{25}=5\\). Therefore, \\(\\sin x=\\frac{\\text{opposite}}{\\text{hypotenuse}}=\\frac{3}{5}\\). Since x is in the first quadrant, sine is positive, so option B is correct.

The first-quadrant condition removes any sign ambiguity: both sine and cosine are positive there. One may also use \\(1+\\tan^2x=\\sec^2x\\), but the 3-4-5 triangle gives the result directly. The values \\(4/5\\), \\(5/3\\), and \\(5/4\\) correspond to other ratios or are greater than 1, so they cannot be the sine value here. The supplied answer B and its explanation are correct.

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 direct answer is option C, −24/25. Begin with sin²x+cos²x=1. Since sin x=7/25, cos²x=1−(7/25)²=1−49/625=576/625, so |cos x|=24/25. The second quadrant decides the sign: sine is positive there, but cosine is negative. Therefore cos x=−24/25. Option A, +24/25, has the correct magnitude but the wrong sign. Option B, −7/25, simply repeats the sine magnitude with a negative sign and does not follow from the identity. Option C has both the correct magnitude and the correct second-quadrant sign. Option D, +7/25, is the given sine value’s magnitude and is not cosine. A common mistake is to calculate the square root and forget the quadrant. Memory cue: ASTC says in quadrant II only sine is positive; cosine is negative.

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