Open a school trigonometry book and you meet forty formulas on the first day. Open a set of SSC papers and you find that a few of them do nearly all the work. The exam does not want you to be a mathematician. It wants you to recognise that sin squared plus cos squared is 1 when you see it in disguise, to know the values at 30, 45 and 60 degrees without thinking, and to draw a right triangle when a question gives you one ratio and asks for another.
That is a much smaller job than it looks, and it is the reason trigonometry is often the chapter where a student's marks improve fastest once they sit down for two weeks. The questions are short. Most take 30 to 60 seconds once the toolkit is in place. And unlike some quant chapters, there is very little arithmetic to get wrong.
This post sorts the material into three piles: the things you must know cold, the things that are worth knowing, and the things you can skip. Every example has been calculated and checked. It builds on the right-triangle facts in the geometry shortcuts post, so if Pythagorean triplets are not yet automatic, read that first. Trigonometry usually appears in the higher-level quant paper of SSC exams, and heights and distances can appear in more than one exam, so check the latest notification and past papers for your specific paper.
The values you must know cold
Five angles matter: 0, 30, 45, 60 and 90 degrees. Here is the full table. The pattern makes it easy to reconstruct if you forget a value.
| Angle | sin | cos | tan |
|---|---|---|---|
| 0 | 0 | 1 | 0 |
| 30 | 1/2 | sqrt(3)/2 | 1/sqrt(3) |
| 45 | 1/sqrt(2) | 1/sqrt(2) | 1 |
| 60 | sqrt(3)/2 | 1/2 | sqrt(3) |
| 90 | 1 | 0 | not defined |
Memory pattern for sine: 0, 1/2, 1/sqrt(2), sqrt(3)/2, 1. If you write them as sqrt(0)/2, sqrt(1)/2, sqrt(2)/2, sqrt(3)/2, sqrt(4)/2, the pattern is obvious: square root of 0, 1, 2, 3, 4, all divided by 2. Cosine is the same list backwards. Tangent is sine divided by cosine, so it goes 0, 1/sqrt(3), 1, sqrt(3), not defined.
Reciprocals: cosec = 1/sin, sec = 1/cos, cot = 1/tan. So cosec 30 = 2, sec 60 = 2, cot 45 = 1, cot 60 = 1/sqrt(3).
Decimal approximations sometimes help with estimation: sqrt(2) is about 1.414 and sqrt(3) is about 1.732. So sin 60 is about 0.866 and tan 60 is about 1.732.
There are two more values worth knowing: sin 15 = (sqrt(6) - sqrt(2))/4, about 0.2588, and cos 15 = (sqrt(6) + sqrt(2))/4, about 0.9659, with sin 75 and cos 75 swapped. Also tan 15 = 2 - sqrt(3), about 0.2679, and tan 75 = 2 + sqrt(3), about 3.732. They appear in a small number of questions and are easy to derive from the sum formula if you have the time, but knowing them saves a minute.
Complementary angles
This is the most useful idea in the chapter, and it drives half of the "find the value" questions.
sin(90 - theta) = cos theta cos(90 - theta) = sin theta tan(90 - theta) = cot theta sec(90 - theta) = cosec theta
The consequences: sin 25 = cos 65, tan 20 = cot 70, and so on. Any time a question involves two angles adding up to 90, this rule is the key.
Example: find sin^2 25 + sin^2 65. Since 65 = 90 - 25, sin 65 = cos 25. So the expression is sin^2 25 + cos^2 25 = 1. Numerically the sum is 1.0 to any precision, which I checked.
Example: find tan 1 x tan 2 x tan 3 ... x tan 89 (in degrees). Pair tan 1 with tan 89: since tan 89 = cot 1, their product is 1. Same for tan 2 and tan 88, and so on up to tan 44 and tan 46. Tan 45 = 1 is left alone. The whole product is 1. A direct calculation gives 1.0000 to machine precision.
Example: find sin 30 + cos 60. Both are 1/2, so the sum is 1.
The pattern "sum of two angles is 90" or "angles in a list from 1 to 89" is a signal to pair them. If the angles come in a long product or sum, pairs are almost always the trick.
The three identities
The whole chapter rests on three identities:
- sin^2 theta + cos^2 theta = 1
- 1 + tan^2 theta = sec^2 theta
- 1 + cot^2 theta = cosec^2 theta
The second and third come from dividing the first by cos^2 or sin^2. You need to be able to use them in both directions. The most common form of an exam question is an expression that looks unpleasant and collapses because of one of them.
Some patterns to recognise:
- 1 - sin^2 theta = cos^2 theta. So (1 - sin^2 theta) sec theta = cos theta.
- sec^2 theta - tan^2 theta = 1.
- (1 + tan^2 theta) cos^2 theta = sec^2 theta x cos^2 theta = 1.
- cosec^2 theta - cot^2 theta = 1.
- sin^4 theta - cos^4 theta = (sin^2 theta - cos^2 theta)(sin^2 theta + cos^2 theta) = sin^2 theta - cos^2 theta.
- sin^4 theta + cos^4 theta = 1 - 2 sin^2 theta cos^2 theta.
The last one is a very popular question. If sin theta = 3/5 and cos theta = 4/5, then sin^4 + cos^4 = 81/625 + 256/625 = 337/625. And 1 - 2 x (9/25) x (16/25) = 1 - 288/625 = 337/625. They agree. You do not need to know the values of the angle to use the identity.
The right-triangle method: your safest tool
Whenever a question gives you one trigonometric ratio and asks for another, draw a right triangle. Do not use identities and square roots. Put the given ratio on the sides, find the third side with Pythagoras, and read every other ratio directly.
Example: sin theta = 3/5. Draw a right triangle with opposite side 3 and hypotenuse 5. The adjacent side is 4 (3-4-5). So cos theta = 4/5, tan theta = 3/4, cosec theta = 5/3, sec theta = 5/4, cot theta = 4/3.
Example: tan theta = 5/3. Opposite = 5, adjacent = 3, so the hypotenuse is sqrt(34). This is not a clean triplet, but you may not need it. Take the question: find (sin theta + cos theta)/(sin theta - cos theta). Divide numerator and denominator by cos theta, and you get (tan theta + 1)/(tan theta - 1) = (5/3 + 1)/(5/3 - 1) = (8/3)/(2/3) = 4. Or in terms of the triangle, (5 + 3)/(5 - 3) = 4 after cancelling sqrt(34). The trick of dividing top and bottom by cos theta is worth remembering: any ratio of linear expressions in sin and cos can be converted to tan.
Example: sec theta + tan theta = 2. Find sin theta. Use sec^2 - tan^2 = 1, which is (sec - tan)(sec + tan) = 1. So sec - tan = 1/2. Adding: 2 sec theta = 2.5, so sec theta = 5/4. Subtracting: 2 tan theta = 1.5, so tan theta = 3/4. Now the right triangle has adjacent 4, opposite 3, hypotenuse 5. So sin theta = 3/5. Check: sec + tan = 5/4 + 3/4 = 2. Correct.
The pair trick, "if A + B = k, then A - B = 1/k" for sec/tan and cosec/cot pairs, is a small identity that appears in about one question a year. It is exactly what you used above.
Example: sin theta + cos theta = 7/5. Find sin theta x cos theta. Square both sides: 1 + 2 sin theta cos theta = 49/25, so 2 sin theta cos theta = 24/25, and sin theta cos theta = 12/25. Check with theta where sin = 3/5 and cos = 4/5: the sum is 7/5 and the product is 12/25. Correct. (It could also have sin = 4/5 and cos = 3/5; the product is the same.)
Expressions of the form a cos theta + b sin theta
A well-worn question type: "If 3 cos theta + 4 sin theta = 4, find 3 sin theta - 4 cos theta."
The identity behind it: (a cos theta + b sin theta)^2 + (a sin theta - b cos theta)^2 = a^2 + b^2. Expand and the cross terms cancel. So
(3 sin theta - 4 cos theta)^2 = 3^2 + 4^2 - 4^2 = 9. So 3 sin theta - 4 cos theta = plus or minus 3.
Let me check with actual values. Solve 3 cos + 4 sin = 4 along with sin^2 + cos^2 = 1. Substituting gives 25 cos^2 = 24 cos, so cos = 0 or cos = 24/25. If cos = 0, sin = 1 and 3 sin - 4 cos = 3. If cos = 24/25, sin = 7/25 and 3 sin - 4 cos = (21 - 96)/25 = -3. Both signs occur, so the answer is plus or minus 3, and the question will usually give an option that makes this unambiguous, or it will ask for the square.
A neat sub-case: if 5 cos theta + 12 sin theta = 13, then 5 sin theta - 12 cos theta = 0, because 25 + 144 - 169 = 0. In that case the value of theta is fixed (tan theta = 12/5) and the expression equals zero. The same applies to 3 cos + 4 sin = 5, where the other expression is 0.
Maximum and minimum values
For an expression a sin theta + b cos theta, the maximum is sqrt(a^2 + b^2) and the minimum is minus sqrt(a^2 + b^2). So 3 sin theta + 4 cos theta has a maximum of 5 and a minimum of -5. A numerical scan over all angles gives a maximum of 5.0 (to within rounding), which confirms it.
Also useful: sin^2 theta + cos^2 theta is always 1; the minimum of sin^2 theta + cosec^2 theta is 2 (from AM-GM, when sin theta = 1), and the minimum of sec^2 theta + cosec^2 theta is 4 (at 45 degrees). These are rarer, but they turn up in the tougher papers.
The self-referencing type: sin theta + sin^2 theta = 1
This one looks impossible and then collapses. Suppose sin theta + sin^2 theta = 1. Since sin^2 = 1 - cos^2, the equation becomes sin theta = cos^2 theta. Now the expression cos^2 theta + cos^4 theta equals sin theta + sin^2 theta, which is 1. So the answer to "find cos^2 theta + cos^4 theta" is 1.
A numerical check: sin theta = (sqrt(5) - 1)/2, about 0.618. Then cos^2 theta = 1 - 0.382 = 0.618, and cos^4 = 0.382. Sum = 1.000. Correct.
The pattern in these questions is to substitute one relation into the other until the given equation reappears.
Double-angle formulas
Three formulas cover most of it:
- sin 2 theta = 2 sin theta cos theta
- cos 2 theta = cos^2 theta - sin^2 theta = 1 - 2 sin^2 theta = 2 cos^2 theta - 1
- tan 2 theta = 2 tan theta/(1 - tan^2 theta)
If sin theta = 3/5 and cos theta = 4/5, then sin 2 theta = 2 x 3/5 x 4/5 = 24/25, and cos 2 theta = 16/25 - 9/25 = 7/25. Check: (24/25)^2 + (7/25)^2 = (576 + 49)/625 = 1. Correct, and this is also a 7-24-25 triangle, which is a nice thing to notice.
The sum and difference formulas are the mother of these:
- sin(A + B) = sin A cos B + cos A sin B
- cos(A + B) = cos A cos B - sin A sin B
- tan(A + B) = (tan A + tan B)/(1 - tan A tan B)
For example, sin 75 = sin(45 + 30) = (1/sqrt(2))(sqrt(3)/2) + (1/sqrt(2))(1/2) = (sqrt(3) + 1)/(2 sqrt(2)) = (sqrt(6) + sqrt(2))/4, about 0.9659. That is where the 15 and 75 values come from.
SSC papers do include these formulas, but less often than the basic identities and heights and distances. Learn them after you are comfortable with everything above.
Degrees and radians
A conversion appears occasionally: pi radians = 180 degrees. So 30 degrees = pi/6 radians and 1 radian = about 57.3 degrees. If a question gives an angle in radians and asks for a value, convert to degrees first. That is all you need from radians for SSC.
Heights and distances
This is the application part of trigonometry, and it is where the exam gets its most predictable questions. The setup is always a right triangle.
The angle of elevation is the angle above the horizontal when you look up at an object. The angle of depression is the angle below the horizontal when you look down. The angle of depression from the top of a tower to an object equals the angle of elevation from the object to the top, because of alternate angles.
The ratio you need:
- tan = height/distance, when you know the horizontal distance and want the height.
- sin = height/line of sight, when you have a ladder, rope or string.
- cos = distance/line of sight.
Basic example. A tower is 30 m high. The angle of elevation of its top from a point on the ground is 60 degrees. How far is the point from the base? tan 60 = 30/d, so d = 30/sqrt(3) = 10 sqrt(3), about 17.32 m.
Flip it. From a point 60 m from the base of a tower, the angle of elevation of the top is 30 degrees. Height = 60 tan 30 = 60/sqrt(3) = 20 sqrt(3), about 34.64 m.
Ladder. A 10 m ladder leans against a wall and makes a 60 degree angle with the ground. The height it reaches is 10 sin 60 = 5 sqrt(3), about 8.66 m. The foot is 10 cos 60 = 5 m from the wall. (Here the 30-60-90 triangle gives sides in the ratio 1 : sqrt(3) : 2, which is the fastest way to solve it.)
The 30-60-90 triangle has sides 1 : sqrt(3) : 2 and the 45-45-90 triangle has sides 1 : 1 : sqrt(2). If you keep these two triangles in your head, half the heights and distances questions are one-liners.
Two observation points
The harder type has the angle measured from two points, on the same line, at a known distance apart. The general result for a tower of height h, with two points on the same side on a line through the base, at angles a (nearer) and b (farther):
distance between the points = h (cot b - cot a), or equivalently h = d/(cot b - cot a), where b is the smaller angle (farther point).
Example 1. The angles of elevation of the top of a tower from two points on the same side are 60 and 30 degrees, and the points are 20 m apart. Find the height. h = 20/(cot 30 - cot 60) = 20/(sqrt(3) - 1/sqrt(3)) = 20/(2/sqrt(3)) = 10 sqrt(3), about 17.32 m.
Check by working from scratch: let the nearer point be x from the base. Then h = x tan 60 = x sqrt(3), and h = (x + 20) tan 30 = (x + 20)/sqrt(3). Equate: 3x = x + 20, so x = 10, and h = 10 sqrt(3). Correct.
Example 2. The angle of elevation of the top of a tower from a point is 45 degrees. After walking 40 m directly away, it becomes 30 degrees. Find the height. h = 40/(cot 30 - cot 45) = 40/(sqrt(3) - 1) = 40(sqrt(3) + 1)/2 = 20(sqrt(3) + 1), about 54.64 m. Check: from scratch, nearer distance x = h (since tan 45 = 1), and h = (h + 40)/sqrt(3), giving h(sqrt(3) - 1) = 40. Same.
The rule to keep: when a tower is observed from two points, treat the tower height as a common side of two right triangles, and write the distance in terms of h from each. Then subtract. Do not try to combine the two triangles into one.
Angle of depression and moving objects
A man on a tower sees a car at a depression of 30 degrees, and after some time at 60 degrees. The distances from the base are h cot 30 and h cot 60, and the difference is the distance the car travels. If the time taken is given, use it for speed. Most of these follow the same two-triangle pattern.
Keep the units consistent. If the question gives kilometres for one thing and metres for another, convert at the start.
A short 30-second checklist for trig questions
Before you try anything, ask three things.
- Is this a "given one ratio, find another" question? Draw the right triangle.
- Are there two angles adding up to 90, or a long list of angles? Pair them with complementary angles.
- Does the expression contain sin^2, cos^2, sec^2 or tan^2 together? Use an identity to simplify, or divide top and bottom by cos theta to turn it into tan.
If none of these fit, look for a heights and distances setup: a tower, a ladder, a shadow. If nothing fits, you have a harder question and should consider skipping.
Mistakes that cost marks
First, mixing up which ratio uses which side. Say "sin = opposite over hypotenuse, cos = adjacent over hypotenuse, tan = opposite over adjacent" as a chant until it is automatic. A mix-up here spoils an entire question.
Second, forgetting that sin^2 theta means (sin theta)^2, not sin(theta^2). Obvious once you say it, but under time pressure people mis-read expressions like sin^2 30 + cos^2 30, and the answer is exactly 1.
Third, using tan for a ladder problem when the problem gave the ladder's length. The length is the hypotenuse, so sin or cos is needed. Draw the triangle and label the given side before choosing the ratio.
Fourth, taking the wrong sign for a square root. In the a cos plus b sin type, the answer is plus or minus, and the options will guide you.
Fifth, treating a heights-and-distances figure as if the angles were measured from the base. Angles of elevation are from the observer's horizontal line. If the observer's eye is at a height (say a man 1.5 m tall), the height of the tower is the computed part plus 1.5 m.
Sixth, not simplifying surds. Leave answers in the form the options use. If the options are like "10 sqrt(3)", do not convert to 17.32 unless you need to.
What to skip
There are chapters in a trigonometry textbook that SSC does not ask, and skipping them will free up time.
Skip the product-to-sum and sum-to-product formulas at first. They appear rarely, and only in the hardest papers.
Skip inverse trigonometric functions and the graphs of the functions. They are not part of the typical SSC syllabus. Check the latest syllabus for your paper, as topics can change.
Skip proofs of identities that are given in "prove that" form; SSC asks for values or simplification, not proofs.
Skip obscure triple-angle formulas until everything else is comfortable.
A practice plan for two weeks
Trigonometry rewards structure. Here is a plan that has worked for many students.
Days one and two: learn the value table, reciprocals and complementary angles. Test yourself with a blank table daily. Do 20 questions that use only values and complementary angles.
Days three and four: identities. Do 25 questions of the "simplify" and "prove the value" types. Use the divide-by-cos trick.
Day five: the right-triangle method. Do 20 questions where one ratio is given and another is asked.
Day six: the a cos plus b sin family and the sec-plus-tan pairs. Around 15 questions.
Days seven to nine: heights and distances. Start with a single angle (10 questions), then two angles (10 questions), then depression and moving objects (10 questions).
Day ten: a timed set of 25 mixed questions, about 25 minutes. Review the errors.
The following week, revise the table and identities on paper, do another mixed set, and test the whole thing inside a real paper. The Pareeksha sectional practice can be used for the mixed set, and a full mock test will show whether your trig questions are now taking less time than average. If you are preparing across more than one exam, the preparing two exams together post explains how to share the quant preparation.
Trigonometry also links well with the percentage and time and work preparation in one respect: all three reward a small set of memorised patterns rather than long calculation. If you build the habit with one, the other two become easier.
How this chapter fits into the paper
The weight of trigonometry differs by exam and by tier, and I will not quote a number, because the pattern changes. Check the most recent notification and use past papers to count for yourself. What I can say from experience is that trigonometry questions tend to be short, specific and either easy or impossible, with little in between. That is a good property for a chapter, because you can decide in ten seconds whether you can do it. For the same reason, it is a poor place to spend a long time on a single stubborn question.
RRB and bank exams generally test it lightly or not at all in the main paper, though heights and distances may appear in some RRB papers, so check the syllabus. If your exam does not list trigonometry, spend your time elsewhere. The maths chapter weightage post walks through how to check this properly.
Your next step
Write the value table from memory, without looking, and check it. Then solve these five without a calculator: sin^2 25 + sin^2 65; the value of (sin theta + cos theta)/(sin theta - cos theta) if tan theta = 5/3; find sin theta if sec theta + tan theta = 2; the height of a tower if the angles from two points 20 m apart are 60 and 30 degrees; the maximum of 3 sin theta + 4 cos theta. You should get 1, 4, 3/5, 10 sqrt(3), and 5.
FAQ
Which trigonometry formulas are enough for SSC? The value table, complementary angles, the three identities, the right-triangle method, and the standard heights and distances setups cover most questions. Add the double-angle and sum formulas once those are automatic.
Do I need to memorise sin 15 and sin 75? They are not essential. If you know the sum formula, you can derive them in under a minute. Learn them if you have spare time.
How do I remember the value table? Write sin as the square root of 0, 1, 2, 3, 4 over 2 for 0, 30, 45, 60, 90 degrees. Cosine is the same list in reverse, and tan is sin over cos.
Is trigonometry asked in RRB or bank exams? It depends on the exam and the year. Check the syllabus in the latest notification. Heights and distances is the part most likely to appear outside SSC.
Should I attempt trigonometry if I am weak at it? The basic value and identity questions are quick and worth taking after two weeks of practice. Heights and distances with two angles are also predictable once you know the cot formula.
How long does this chapter take? About ten days of steady practice for most students, with a short revision each week afterwards.
Comments (0)
No comments yet. Be the first to share your thoughts.