MathematicsClass 11CBSE

Find sin 75° Exactly, Without a Calculator, by Splitting It Into 45° and 30°

Expand sin and cos of sums and differences, use the tan and cot compound-angle formulae, convert between sums and products, and apply the double and triple angle identities for sine, cosine and tangent.

Why do we need formulas for sums of angles?

You know

sin 4 5^{\circ}

and

sin 3 0^{\circ}

exactly, but not

sin 7 5^{\circ}

. **It is tempting to add them — but

sin (4 5^{\circ} + 3 0^{\circ}) \neq = sin 4 5^{\circ} + sin 3 0^{\circ}

.** The correct formulas combine sines and cosines of both angles.

From them come formulas for tangents, sum-product conversions and multiple angles.

This part covers each family of formulas with worked values you can check.

How do you expand sin and cos of x + y and x - y, and use them for non-standard angles?

sin (x \pm y) = sin x cos y \pm cos x sin y

and

cos (x \pm y) = cos x cos y \mp sin x sin y

, so splitting an angle into standard angles gives exact values.

Worked example 1.**

sin 7 5^{\circ} = sin (4 5^{\circ} + 3 0^{\circ}) = \frac{2}{2} \cdot \frac{3}{2} + \frac{2}{2} \cdot \frac{1}{2} = \frac{6 + 2}{4} \approx 0.9659

Worked example 2.

cos 10 5^{\circ} = cos (6 0^{\circ} + 4 5^{\circ}) = \frac{1}{2} \cdot \frac{2}{2} - \frac{3}{2} \cdot \frac{2}{2} = \frac{2 - 6}{4} \approx - 0.2588

Worked example 3.

sin x = \frac{3}{5}

and

cos y = \frac{12}{13}

, both acute, so

cos x = \frac{4}{5}

and

sin y = \frac{5}{13}

sin (x + y) = \frac{36}{65} + \frac{20}{65} = \frac{56}{65} cos (x + y) = \frac{48}{65} - \frac{15}{65} = \frac{33}{65}

Check:

5 6^{2} + 3 3^{2} = 4225 = 6 5^{2}

.

An everyday example. **An architect marking a

7 5^{\circ}

angle for a temple spire** can get its exact slope from the known values at

4 5^{\circ}

and

3 0^{\circ}

.

The trap. **

sin 6 0^{\circ} \neq = sin 3 0^{\circ} + sin 3 0^{\circ}

**: the left side is about

0.866

, the right side is

1

How do you deduce and apply the formulas for tan and cot of x + y and x - y?

**Dividing the sine formula by the cosine formula gives

tan (x \pm y) = \frac{t a n x \pm t a n y}{1 \mp t a n x t a n y}

, and similarly

cot (x \pm y) = \frac{c o t x c o t y \mp 1}{c o t y \pm c o t x}

.

Worked example 1.**

tan 1 5^{\circ} = tan (4 5^{\circ} - 3 0^{\circ}) = \frac{1 - \frac{1}{3}}{1 + \frac{1}{3}} = \frac{3 - 1}{3 + 1} = 2 - 3 \approx 0.268

In the same way,

tan 7 5^{\circ} = 2 + 3

.

Worked example 2.

tan (\frac{π}{4} + x) = \frac{1 + tan x}{1 - tan x}

Worked example 3.

tan A = \frac{1}{2}

and

tan B = \frac{1}{3}

, with

A

and

B

acute.

tan (A + B) = \frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{6}} = \frac{\frac{5}{6}}{\frac{5}{6}} = 1 A + B = 4 5^{\circ}

An everyday example. **Two stretches of a hill road with gradients

\frac{1}{2}

and

\frac{1}{3}

** make angles with the horizontal that add up to exactly

4 5^{\circ}

.

The boundary case. **When

tan x tan y = 1

tan (x + y)

is undefined**, because then

x + y

is an odd multiple of

9 0^{\circ}

How do you convert between sums and products of sines and cosines?

**Adding and subtracting the compound-angle formulas gives product-to-sum rules, and substituting

x + y = C

and

x - y = D

turns them into sum-to-product rules.

Product to sum:**

2 sin x cos y = sin (x + y) + sin (x - y) 2 cos x cos y = cos (x + y) + cos (x - y)

2 sin x sin y = cos (x - y) - cos (x + y)

Sum to product:

sin C + sin D = 2 sin \frac{C + D}{2} cos \frac{C - D}{2} cos C + cos D = 2 cos \frac{C + D}{2} cos \frac{C - D}{2}

cos C - cos D = - 2 sin \frac{C + D}{2} sin \frac{C - D}{2}

Worked example 1.

sin 7 5^{\circ} + sin 1 5^{\circ} = 2 sin 4 5^{\circ} cos 3 0^{\circ} = \frac{6}{2} \approx 1.2247

Worked example 2.

cos 7 5^{\circ} - cos 1 5^{\circ} = - 2 sin 4 5^{\circ} sin 3 0^{\circ} = - \frac{2}{2} \approx - 0.7071

Worked example 3. Prove

\frac{s i n 5 x + s i n 3 x}{c o s 5 x + c o s 3 x} = tan 4 x

\frac{2 sin 4 x cos x}{2 cos 4 x cos x} = tan 4 x

An everyday example. Two veena strings slightly out of tune produce a slow throbbing sound, because the sum of two close sine waves becomes a product with a slowly changing loudness.

The substance. Sum-to-product turns an expression into factors, which is what makes proofs and equations easy to simplify.

How do you derive and apply the formulas for sin 2x, cos 2x, tan 2x, sin 3x, cos 3x and tan 3x?

**Put

y = x

in the compound-angle formulas to get the double-angle identities, and expand

x + 2 x

to get the triple-angle identities.

Double angles:**

sin 2 x = 2 sin x cos x cos 2 x = cos^{2} x - sin^{2} x = 2 cos^{2} x - 1 = 1 - 2 sin^{2} x

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

Triple angles:

sin 3 x = 3 sin x - 4 sin^{3} x cos 3 x = 4 cos^{3} x - 3 cos x tan 3 x = \frac{3 tan x - tan ^{3} x}{1 - 3 tan ^{2} x}

Worked example 1.

sin x = \frac{3}{5}

x

acute, so

cos x = \frac{4}{5}

sin 2 x = \frac{24}{25} cos 2 x = 1 - \frac{18}{25} = \frac{7}{25} tan 2 x = \frac{24}{7}

**Worked example 2 — check at

3 0^{\circ}

.**

sin 9 0^{\circ} = 3 \cdot \frac{1}{2} - 4 \cdot \frac{1}{8} = 1

.

Worked example 3.

tan x = \frac{1}{2}

tan 2 x = \frac{1}{1 - \frac{1}{4}} = \frac{4}{3} tan 3 x = \frac{\frac{3}{2} - \frac{1}{8}}{1 - \frac{3}{4}} = \frac{11}{2}

An everyday example. A cricketer throwing from the boundary gets the longest throw near

4 5^{\circ}

, because the range of a projectile depends on

sin 2 θ

, which is largest at

2 θ = 9 0^{\circ}

.

The substance. **Choose the form of

cos 2 x

that matches what you know** — use

1 - 2 sin^{2} x

when only

sin x

is given.

Exam tip

What earns full marks on compound and multiple angle formulae?

Write the formula with the signs in place before substituting, simplify surds fully, and check a value numerically when possible.

- **

sin (x \pm y)

: signs match;

cos (x \pm y)

: signs swap
-

tan (x + y)

: minus sign in the denominator
- Product to sum**:

2 sin x cos y = sin (x + y) + sin (x - y)

- Sum to product: half-sum and half-difference angles
- **

cos 2 x

: three forms; pick the one that fits

The trap.** Writing

cos (x + y) = cos x cos y + sin x sin y

. For a sum, the cosine formula has a minus sign.

Did you know

Why can sin 2x never be more than 1, even though it equals 2 sin x cos x?

Doubling something usually makes it bigger, so

2 sin x cos x

looks as if it could exceed

1

.

But

sin x

and

cos x

cannot both be large at once. When one is near

1

, the other is near

0

. Their product is largest when they are equal, at

x = 4 5^{\circ}

2 \times \frac{2}{2} \times \frac{2}{2} = 1

So the maximum of

sin 2 x

is exactly

1

— as it must be, since

sin 2 x

is itself a sine.

Exam relevance

How are compound angle formulae tested in JEE Main and JEE Advanced?

Compound, multiple and sum-product formulae are central to Trigonometric Functions in JEE Main and JEE Advanced, and they are used in Trigonometric Equations, Inverse Trigonometric Functions and Integrals. In Physics, the same identities appear in projectile motion, superposition of waves and alternating current, part of both JEE Main and NEET.

What gets asked. Exact values of non-standard angles, simplifying products of cosines, and the **maximum and minimum of

a sin x + b cos x

**, which lie between

- a^{2} + b^{2}

and

a^{2} + b^{2}

.

Question types. Multiple-choice and numerical-value questions, often needing two formulas in sequence.

The trap that costs marks. **A sign error in

cos (x \pm y)

or in the denominator of

tan (x \pm y)

.**

Key takeaways

What must you be able to do from this part?

- **

sin (x \pm y) = sin x cos y \pm cos x sin y

;

cos (x \pm y) = cos x cos y \mp sin x sin y

sin 7 5^{\circ} = \frac{6 + 2}{4}

**;

cos 10 5^{\circ} = \frac{2 - 6}{4}

- **

tan (x \pm y) = \frac{t a n x \pm t a n y}{1 \mp t a n x t a n y}

**;

tan 1 5^{\circ} = 2 - 3

- Product to sum and sum to product;

sin 7 5^{\circ} + sin 1 5^{\circ} = \frac{6}{2}

- Double angles:

sin 2 x = 2 sin x cos x

cos 2 x

in three forms,

tan 2 x = \frac{2 t a n x}{1 - t a n ^{2} x}

- Triple angles:

sin 3 x = 3 sin x - 4 sin^{3} x

cos 3 x = 4 cos^{3} x - 3 cos x

Find

cos 1 5^{\circ}

exactly in two different ways — from

4 5^{\circ} - 3 0^{\circ}

and from the formula for

cos 2 x

— and check that the answers agree.