MathematicsClass 9ICSE

Two Special Triangles Give Every Standard Angle Value

Derive the ratios of 0, 30, 45, 60 and 90 degrees from a square and an equilateral triangle, evaluate numerical expressions in them, verify given relations, and solve for a standard angle.

Where do the values of sin 30 and cos 45 actually come from?

They are not measured and they are not memorised facts about nature. They come out of two triangles you can draw in ten seconds, and if you can rebuild those two triangles you never need to trust your memory of the table.

The first is half a square. Draw a square of side

1

and cut it along a diagonal. Each half is a right-angled triangle with legs

1

and

1

, and by the Pythagoras theorem the diagonal is

2

. Since the two acute angles are equal and add to

9 0^{\circ}

, each is

4 5^{\circ}

. So

sin 4 5^{\circ} = \frac{1}{2}, cos 4 5^{\circ} = \frac{1}{2}, tan 4 5^{\circ} = \frac{1}{1} = 1

The second is half an equilateral triangle. Draw an equilateral triangle of side

2

and drop the perpendicular from the apex. From the isosceles chapter you know it bisects the base, so the two halves have base

1

, hypotenuse

2

, and height

2^{2} - 1^{2} = 3

The apex angle of

6 0^{\circ}

has been halved to

3 0^{\circ}

, so this triangle contains both

3 0^{\circ}

and

6 0^{\circ}

sin 3 0^{\circ} = \frac{1}{2}, cos 3 0^{\circ} = \frac{3}{2}, tan 3 0^{\circ} = \frac{1}{3}

sin 6 0^{\circ} = \frac{3}{2}, cos 6 0^{\circ} = \frac{1}{2}, tan 6 0^{\circ} = 3

Two drawings, six values, and no memory involved.

This page covers the second part of the ICSE Class 9 Mathematics chapter on trigonometrical ratios: the standard-angle values, evaluating expressions with them, verifying relations, and solving for an unknown standard angle.

Formula

What are the standard values of sine, cosine and tangent?

Five angles, and one pattern that holds them together.

The sines, in order of angle:

sin 0^{\circ} = 0, sin 3 0^{\circ} = \frac{1}{2}, sin 4 5^{\circ} = \frac{1}{2}, sin 6 0^{\circ} = \frac{3}{2}, sin 9 0^{\circ} = 1

The cosines are the same list read backwards:

cos 0^{\circ} = 1, cos 3 0^{\circ} = \frac{3}{2}, cos 4 5^{\circ} = \frac{1}{2}, cos 6 0^{\circ} = \frac{1}{2}, cos 9 0^{\circ} = 0

And the tangents follow from

tan = \frac{sin}{cos}

tan 0^{\circ} = 0, tan 3 0^{\circ} = \frac{1}{3}, tan 4 5^{\circ} = 1, tan 6 0^{\circ} = 3, tan 9 0^{\circ} is not defined

A cleaner way to remember the sines. Write them as

\frac{0}{4}, \frac{1}{4}, \frac{2}{4}, \frac{3}{4}, \frac{4}{4}

and the numerators simply count

0, 1, 2, 3, 4

. Check:

\frac{1}{4} = \frac{1}{2} = sin 3 0^{\circ}

and

\frac{2}{4} = \frac{1}{2} = sin 4 5^{\circ}

, as required.

The reciprocals need no new work.

cosec 3 0^{\circ} = 2, cosec 4 5^{\circ} = 2, cosec 6 0^{\circ} = \frac{2}{3}

sec 3 0^{\circ} = \frac{2}{3}, sec 4 5^{\circ} = 2, sec 6 0^{\circ} = 2

cot 3 0^{\circ} = 3, cot 4 5^{\circ} = 1, cot 6 0^{\circ} = \frac{1}{3}

And the two extreme angles are limiting cases, not triangles. At

0^{\circ}

the triangle has flattened so the opposite side has shrunk to nothing, giving

sin 0^{\circ} = 0

; at

9 0^{\circ}

the adjacent side has vanished, giving

cos 9 0^{\circ} = 0

. That is also why

tan 9 0^{\circ}

has no value: it would need division by zero.

Check the pattern against an identity. At

4 5^{\circ}

sin^{2} + cos^{2} = \frac{1}{2} + \frac{1}{2} = 1

; at

3 0^{\circ}

\frac{1}{4} + \frac{3}{4} = 1

; at

0^{\circ}

0 + 1 = 1

. The identity holds at every standard angle, which is a fast way to check that you have recalled a pair correctly.

How do you evaluate an expression in standard angles?

Substitute every value first, then simplify — and rationalise any surd denominators at the end.

Worked example 1. Evaluate

sin 3 0^{\circ} cos 6 0^{\circ} + cos 3 0^{\circ} sin 6 0^{\circ}

= (\frac{1}{2}) (\frac{1}{2}) + (\frac{3}{2}) (\frac{3}{2}) = \frac{1}{4} + \frac{3}{4} = 1

**The answer is exactly

sin 9 0^{\circ}

**, and

3 0^{\circ} + 6 0^{\circ} = 9 0^{\circ}

. That is not a coincidence — it is the addition formula

sin (A + B) = sin A cos B + cos A sin B

, which you meet in Class 11. This chapter is quietly rehearsing it.

Worked example 2. Evaluate

2 tan^{2} 4 5^{\circ} + cos^{2} 3 0^{\circ} - sin^{2} 6 0^{\circ}

= 2 (1)^{2} + (\frac{3}{2})^{2} - (\frac{3}{2})^{2} = 2 + \frac{3}{4} - \frac{3}{4} = 2

The last two terms cancelled because

cos 3 0^{\circ}

and

sin 6 0^{\circ}

are the same number. Spotting that saves the arithmetic entirely.

Worked example 3. Evaluate

4 (sin^{4} 6 0^{\circ} + cos^{4} 3 0^{\circ}) - 3 (cos^{2} 4 5^{\circ} - sin^{2} 9 0^{\circ})

.

Both

sin 6 0^{\circ}

and

cos 3 0^{\circ}

are

\frac{3}{2}

, whose fourth power is

\frac{9}{16}

4 (\frac{9}{16} + \frac{9}{16}) - 3 (\frac{1}{2} - 1) = 4 (\frac{9}{8}) - 3 (- \frac{1}{2}) = \frac{9}{2} + \frac{3}{2} = 6

Worked example 4 — a product that collapses. Evaluate

tan 3 0^{\circ} \times tan 4 5^{\circ} \times tan 6 0^{\circ}

= \frac{1}{3} \times 1 \times 3 = 1

Worked example 5 — rationalising. Evaluate

\frac{tan 6 0 ^{\circ} - tan 3 0 ^{\circ}}{1 + tan 6 0 ^{\circ} tan 3 0 ^{\circ}}

= \frac{3 - \frac{1}{3}}{1 + 3 \times \frac{1}{3}} = \frac{\frac{3 - 1}{3}}{2} = \frac{2}{2 3} = \frac{1}{3} = \frac{3}{3}

**And the answer is

tan 3 0^{\circ}

**, while

6 0^{\circ} - 3 0^{\circ} = 3 0^{\circ}

— another Class 11 formula appearing early, this time for

tan (A - B)

.

The habit that prevents most errors here is writing the substituted line in full before simplifying anything. Trying to evaluate

cos^{2} 3 0^{\circ}

mentally and then combining is how

\frac{3}{2}

becomes

\frac{3}{2}

. Square the whole ratio, including the denominator.

How do you verify a given relation at a standard angle?

Work out the two sides separately and compare. A verification is not a proof — it checks one angle, and the question is asking whether you can substitute accurately.

Worked example 1. Verify that

sin 2 θ = 2 sin θ cos θ

for

θ = 3 0^{\circ}

.

Left side:

sin 6 0^{\circ} = \frac{3}{2}

.

Right side:

2 (\frac{1}{2}) (\frac{3}{2}) = \frac{3}{2}

.

The two sides agree, so the relation is verified at

3 0^{\circ}

.

Worked example 2. Verify that

cos 2 θ = 2 cos^{2} θ - 1

for

θ = 3 0^{\circ}

.

Left:

cos 6 0^{\circ} = \frac{1}{2}

. Right:

2 (\frac{3}{4}) - 1 = \frac{3}{2} - 1 = \frac{1}{2}

. Verified.

Worked example 3. Verify that

1 + tan^{2} θ = sec^{2} θ

for

θ = 4 5^{\circ}

.

Left:

1 + 1^{2} = 2

. Right:

(2)^{2} = 2

. Verified.

Worked example 4. Verify that

\frac{2 tan θ}{1 - tan ^{2} θ} = tan 2 θ

for

θ = 3 0^{\circ}

.

Left:

\frac{2/ 3}{1 - \frac{1}{3}} = \frac{2/ 3}{2/3} = \frac{2}{3} \times \frac{3}{2} = 3

. Right:

tan 6 0^{\circ} = 3

. Verified.

Now the misconception this section exists to destroy. It is tempting to read

sin 2 θ

2 sin θ

. Test it at

θ = 3 0^{\circ}

sin 6 0^{\circ} = \frac{3}{2} \approx 0.866 but 2 sin 3 0^{\circ} = 2 (\frac{1}{2}) = 1

**They are different, so

sin 2 θ \neq = 2 sin θ

.** The same test kills

cos 2 θ = 2 cos θ

and

sin (A + B) = sin A + sin B

: at

A = B = 3 0^{\circ}

the left side is

sin 6 0^{\circ} \approx 0.866

while the right side is

1

.

And one case where the trap almost works. At

θ = 0^{\circ}

sin 2 θ = 0

and

2 sin θ = 0

, so the false rule gives the right answer. One angle agreeing proves nothing — which is exactly why a verification question says verify for this angle and never prove.

How do you find the angle when the ratio is given?

Recognise the value in the standard table, write down the angle it belongs to, and then solve the little equation inside the bracket.

Worked example 1. Solve

sin 2 θ = \frac{3}{2}

, where

θ

is acute.

The value

\frac{3}{2}

sin 6 0^{\circ}

, so

2 θ = 6 0^{\circ} ⟹ θ = 3 0^{\circ}

Check:

sin (2 \times 3 0^{\circ}) = sin 6 0^{\circ} = \frac{3}{2}

, as required.

Worked example 2. Solve

2 sin 3 θ = 3

.

Divide first:

sin 3 θ = \frac{3}{2}

, so

3 θ = 6 0^{\circ}

and

θ = 2 0^{\circ}

.

The answer need not be a standard angle itself — only the combination inside the sine has to be.

Worked example 3. Solve

tan (\frac{θ}{2}) = 1

tan 4 5^{\circ} = 1

, so

\frac{θ}{2} = 4 5^{\circ}

and

θ = 9 0^{\circ}

.

Worked example 4. Solve

sin (θ + 3 0^{\circ}) = 1

sin 9 0^{\circ} = 1

, so

θ + 3 0^{\circ} = 9 0^{\circ}

and

θ = 6 0^{\circ}

.

Worked example 5 — two ratios in one equation. Solve

3 tan 2 θ = 3

tan 2 θ = \frac{3}{3} = 3 ⟹ 2 θ = 6 0^{\circ} ⟹ θ = 3 0^{\circ}

Notice the surd simplification that had to happen first:

\frac{3}{3} = 3

, from the surds chapter. Many of these equations hide a rationalisation in the first line.

Worked example 6 — an equation with no solution. Solve

sin θ = 2

.

There is none. A sine can never exceed

1

, so no angle satisfies this equation — and saying so is the complete answer. Similarly

sec θ = \frac{1}{2}

is impossible, since a secant is never less than

1

.

The habit that makes these quick. Learn to read the table backwards: seeing

\frac{1}{2}

should make you think

4 5^{\circ}

immediately, and seeing

3

should say tangent of sixty. Nine out of ten of these equations are solved by recognition, not algebra, and the algebra is only the one line that follows.

Exam tip

What layout keeps standard-angle work accurate?

Write the substituted line in full before you simplify anything. These questions are pure substitution, so the marks live in the accuracy of that one line.

- Sketch the two triangles in the margin at the start — the

1

1

2

and the

1

3

2

. Then every value is available and nothing is recalled from memory
- Square the whole ratio.

cos^{2} 3 0^{\circ} = (\frac{3}{2})^{2} = \frac{3}{4}

, not

\frac{3}{4}

and not

\frac{3}{2}

- Rationalise at the end.

\frac{1}{3}

is acceptable working;

\frac{3}{3}

is the finished answer
- Watch for terms that cancel:

cos 3 0^{\circ}

and

sin 6 0^{\circ}

are the same number, as are

sin 3 0^{\circ}

and

cos 6 0^{\circ}

- **Check with

sin^{2} + cos^{2} = 1

if you are unsure whether you have the right pair for an angle
- When solving for an angle, solve the bracket as a separate line**: *

2 θ = 6 0^{\circ}

, so

θ = 3 0^{\circ}

*
- Say so when an equation is impossible —

sin θ = 2

has no solution, and that is a full answer
- **Never write

tan 9 0^{\circ}

as a number**; write not defined

The misconception to name.

sin 2 θ

is not

2 sin θ

, and the quickest disproof is the one from the previous section: at

θ = 3 0^{\circ}

the two sides are

0.866

and

1

. The angle inside the bracket cannot be taken out, and every doubling formula you meet later exists precisely because it cannot.

Did you know

Why is thirty degrees the angle a roof and a ramp both prefer?

sin 3 0^{\circ} = \frac{1}{2}

exactly. Among all the standard angles only

0^{\circ}

3 0^{\circ}

and

9 0^{\circ}

give a sine you can write without a surd, and

3 0^{\circ}

is the only one of those that makes a usable slope.

That exactness has a practical consequence. A rafter set at

3 0^{\circ}

to the horizontal rises exactly half its own length:

rise = rafter \times sin 3 0^{\circ} = \frac{rafter}{2}

So a

6

m rafter lifts the ridge exactly

3

m, and a carpenter can set out the whole roof with a measuring tape and no protractor. The horizontal run is

6 cos 3 0^{\circ} = 33 \approx 5.20

m, and the check is that

3^{2} + 5.2 0^{2} \approx 36

.

**The

4 5^{\circ}

case is even easier to spot but harder to build.** There

tan 4 5^{\circ} = 1

, so the rise equals the run — a slope of one in one, which is far too steep for a road or a ramp. That is why gradient signs read 1 in 12 or 1 in 20 rather than 45 degrees: those are tangents of about

4. 8^{\circ}

and

2. 9^{\circ}

.

**And

6 0^{\circ}

carries the equilateral triangle with it. Because the two special triangles come from the square and the equilateral triangle — the two most symmetric figures in the plane — their angles are the ones that keep appearing in tiling, in truss design and in the hexagon you constructed two chapters ago. Symmetry is the reason these particular five angles have exact values and every other angle needs a table.

A neat consequence to test yourself on.** Since

tan 6 0^{\circ} = 3

and

tan 3 0^{\circ} = \frac{1}{3}

, their product is exactly

1

. So the two halves of an equilateral triangle have reciprocal slopes — steepen a line from

3 0^{\circ}

6 0^{\circ}

and its gradient does not double, it triples.

Exam relevance

How are standard-angle values used in JEE and NEET questions?

This is foundation work whose values are needed on sight in three different papers.

Where it leads. Class 10 uses them throughout heights and distances, and Class 11 extends them to all angles in Trigonometric Functions, a JEE Main topic. Every standard value you learn here reappears as a limit of an integral, an entry in a table of inverse trigonometric values, and the argument of a complex number in polar form. In Calculus the standard angles are the points at which almost every definite integral is evaluated.

The two formulas this chapter foreshadows. Worked examples 1 and 5 above were instances of

sin (A + B) = sin A cos B + cos A sin B and tan (A - B) = \frac{tan A - tan B}{1 + tan A tan B}

Both are examined directly in JEE Main, and the standard angles are how you check that you have recalled them with the signs the right way round. **Substituting

3 0^{\circ}

and

6 0^{\circ}

into a formula you are unsure of is a legitimate exam technique — if the two sides disagree, the recalled formula is wrong.

Where it appears in Physics.** Resolution of forces and velocities uses these values constantly: a projectile launched at

4 5^{\circ}

has equal components because

sin 4 5^{\circ} = cos 4 5^{\circ}

, and that is why

4 5^{\circ}

gives the maximum range. An inclined plane at

3 0^{\circ}

gives a component of exactly

\frac{m g}{2}

along the slope. Both JEE Main and NEET set numericals built on precisely these three angles, because they keep the arithmetic exact.

Question types to expect. At this level: evaluate, verify, solve for the angle. In competitive papers: evaluate a compound expression, or identify the angle satisfying a trigonometric equation — and the trigonometric equation chapter of Class 11 begins with exactly the recognise the value step you practised above.

The single trap that costs marks. Taking a coefficient through a function: writing

sin 2 θ = 2 sin θ

, or

cos (A + B) = cos A + cos B

. It is the most heavily punished error in trigonometry because it wrecks an entire question rather than one line. **Test any doubtful identity at

3 0^{\circ}

before using it.

Board versus competitive emphasis. ICSE wants the substituted line, exact surds and a rationalised answer; a competitive paper wants the value in seconds. The transferable asset is the two triangles** — draw them once at the top of your rough sheet and the whole table is yours for the rest of the paper.

Key takeaways

What should you know about standard angles before Part 3?

Two triangles, five angles, and one pattern that ties the values together.

- Half a square gives

4 5^{\circ}

: sides

1

1

2

, so

sin 4 5^{\circ} = cos 4 5^{\circ} = \frac{1}{2}

and

tan 4 5^{\circ} = 1

- Half an equilateral triangle gives

3 0^{\circ}

and

6 0^{\circ}

: sides

1

3

2

- **The sines are

\frac{0}{4}, \frac{1}{4}, \frac{2}{4}, \frac{3}{4}, \frac{4}{4}

, and the cosines are the same list reversed
-

tan 9 0^{\circ}

is not defined, because it would need division by zero
-

sin 3 0^{\circ} = cos 6 0^{\circ}

and

cos 3 0^{\circ} = sin 6 0^{\circ}

— pairs that often cancel in an expression
- Square the whole ratio**:

cos^{2} 3 0^{\circ} = \frac{3}{4}

- To verify a relation, evaluate both sides separately and compare
- To find an angle, recognise the value, then solve the bracket in one line
- **

sin θ = 2

has no solution**, and neither does

sec θ = \frac{1}{2}

- **

sin 2 θ \neq = 2 sin θ

** — test at

3 0^{\circ}

and the two sides differ

The quickest self-test is worked example 3: evaluate

4 (sin^{4} 6 0^{\circ} + cos^{4} 3 0^{\circ}) - 3 (cos^{2} 4 5^{\circ} - sin^{2} 9 0^{\circ})

from the two triangles alone, and see whether you get

6

without looking anything up.

Where do the values of sin 30 and cos 45 actually come from?

What are the standard values of sine, cosine and tangent?

How do you evaluate an expression in standard angles?

How do you verify a given relation at a standard angle?

How do you find the angle when the ratio is given?

What layout keeps standard-angle work accurate?

Why is thirty degrees the angle a roof and a ramp both prefer?

How are standard-angle values used in JEE and NEET questions?

What should you know about standard angles before Part 3?

Related study notes

Ready to put this into practice?