Exercise practice: Trigonometric Functions

5.1 How the exercises are laid out

Exercise 3.1 covers degree and radian conversion, arc length and real objects such as wheels and pendulums. Exercise 3.2 covers the six functions, quadrant signs, domains, ranges and reduction of large angles. Exercise 3.3 covers identity proofs built from the sum, difference, double angle and transformation formulas. The miscellaneous exercise mixes proofs, half angle values and conditional identities.

This page takes a representative set from each section and writes the full solution. The remaining items follow the same method with different numbers.

5.2 Exercise 3.1: conversion and arc length

Question 1 asks for radian measures. For 25 degrees, multiply by pi over 180 to get 5 pi by 36. For minus 47 degrees 30 minutes, first write 30 minutes as half a degree, so the angle is minus 47.5 degrees, then multiply to get minus 19 pi by 72. For 240 degrees the result is 4 pi by 3, and for 520 degrees it is 26 pi by 9.

Question 4 gives an arc of 22 cm on a circle of radius 100 cm. The angle is l over r, which is 0.22 radians. In degrees that is 0.22 times 180 divided by pi, about 12.6 degrees, or 12 degrees 36 minutes.

Question 6 has equal arcs subtending 60 degrees and 75 degrees. Converting to radians gives pi by 3 and 5 pi by 12. Since the arc lengths are equal, r times theta is the same for both circles, so the radii are inversely proportional to the angles and the ratio is 5 to 4.

Question 3 is the wheel: 360 revolutions per minute is 360 times 2 pi radians in 60 seconds, which is 12 pi radians per second.

5.3 Exercise 3.2: values, signs and reduction

Questions 1 to 5 give one value and a quadrant. For cos x equal to minus 1 by 2 in the third quadrant, the reference angle is pi by 3 and the angle is pi plus pi by 3, that is 4 pi by 3. Then sin x is minus square root of 3 by 2, tan x is square root of 3, cot x is 1 by square root of 3, sec x is minus 2 and cosec x is minus 2 by square root of 3.

For sec x equal to 13 by 5 in the fourth quadrant, cos x is 5 by 13. The fourth quadrant makes sine negative, so sin x is minus 12 by 13. Then tan x is minus 12 by 5, cot x is minus 5 by 12 and cosec x is minus 13 by 12.

Questions 6 to 10 reduce large angles. For sin 765 degrees, subtract two full turns to get 45 degrees, so the value is 1 by square root of 2. For cosec of minus 1410 degrees, add four full turns to get minus 210 degrees, which is a third quadrant angle; the reference angle is 30 degrees and cosec is minus 2. For cot of minus 15 pi by 4, add 4 pi to get pi by 4, so the value is 1.

5.4 Exercise 3.3: identity proofs

The exercise begins with numerical proofs. Question 1 asks for sin squared of pi by 6 plus cos squared of pi by 3 minus tan squared of pi by 4. Each term is a standard value: 1 by 4, 1 by 4 and 1. The result is minus 1 by 2, which matches the right side.

Question 5 asks for sin 75 degrees and tan 15 degrees. Write 75 as 45 plus 30 and apply the sine addition formula to get (square root of 3 plus 1) over (2 times square root of 2). Write 15 as 45 minus 30 and apply the tangent subtraction formula to get 2 minus square root of 3.

Question 6 proves that cos(pi by 4 minus x) cos(pi by 4 minus y) minus sin(pi by 4 minus x) sin(pi by 4 minus y) equals sin(x plus y). The left side is the cosine of a sum, with the two angles being pi by 4 minus x and pi by 4 minus y. That sum is pi by 2 minus (x plus y), whose cosine is the sine of x plus y. This is the cleanest route; expanding everything also works but takes longer.

Question 12 proves sin squared 6x minus sin squared 4x equals sin 2x sin 10x. Use the half angle forms on both squares, combine into a single cosine difference, and apply the transformation formula for cosine. The angle average is 5x and the half difference is x, which gives the product on the right.

5.5 Miscellaneous exercise patterns

Question 5 asks for sin x plus sin 3x plus sin 5x plus sin 7x as a product. Group the first and last terms and the two middle terms. The first pair gives 2 sin 4x cos 3x and the second gives 2 sin 4x cos x. Adding gives 2 sin 4x times the sum cos 3x plus cos x, and the transformation formula turns that sum into 2 cos 2x cos x, so the final answer is 4 sin 4x cos 2x cos x.

Question 8 gives tan x equal to minus 4 by 3 with x in the second quadrant and asks for the half angle values. In the second quadrant, x lies between pi by 2 and pi, so x by 2 lies between pi by 4 and pi by 2, which fixes the signs: sin half is positive and cos half is negative. The half angle forms then give the two values and their ratio gives tan half.

Question 2 proves that (sin 3x plus sin x) sin x plus (cos 3x minus cos x) cos x equals 0. Expand the brackets and use the double angle forms: the first bracket is 2 sin 2x cos x and the second is minus 2 sin 2x sin x. Multiplying by sin x and cos x respectively gives two terms that differ only in sign and cancel.

5.6 Common mistakes in these exercises

Four slips repeat. Using degrees inside l = r theta. Forgetting the quadrant when a square root appears. Converting a radian answer to degrees only at the end and losing the exact form. Expanding both sides of an identity instead of simplifying one side into the other.

There is one more subtle slip in reduction questions: subtracting a multiple of 2 pi is allowed for sine and cosine, but tangent and cotangent have period pi, so subtracting the smaller multiple is often quicker. Check the period of the function before choosing the multiple.

Finally, keep the answer in the form the question requests. A radian answer can be exact, such as 5 pi by 36, or approximate, such as 0.436; the exact form is almost always preferred unless the question gives a decimal value for pi.