1The idea in one paragraph

Sine and cosine are the two coordinates of a point on a circle of radius 1, and Pythagoras says those coordinates satisfy x² + y² = 1. That single fact, cos²θ + sin²θ = 1, ties sine and cosine together so tightly that knowing one fixes the size of the other; the quadrant then fixes the sign, and tan θ follows by division. The double angle identities go one step further: they write sin 2θ and cos 2θ in terms of sin θ and cos θ, so knowing the ratios of an angle gives you the ratios of twice the angle too. None of this needs a calculator or the angle itself, which is exactly why Paper 1 loves it.

2The Pythagorean identity

On the unit circle P = (cos θ, sin θ), and every point on the circle satisfies x² + y² = 1. So

cos²θ + sin²θ = 1 for every angle θ. From the formula booklet.

Figure 1 shows why it holds in every quadrant, not only the first. The right-angled triangle under P has legs of length |cos θ| and |sin θ| and a hypotenuse of 1. Squaring removes the signs, so Pythagoras gives the identity wherever P is.

Notation. sin²θ means (sin θ)², the sine of θ, squared. It does not mean sin(θ²), the sine of θ squared. Your calculator needs the brackets: type (sin(θ))².

An identity is not an equation. An identity is true for every value of θ; an equation such as sin θ = 0.3 is true only for particular values, which you solve for. Some books write ≡ for an identity to make the difference visible. You never "solve" cos²θ + sin²θ = 1: every θ satisfies it.

It is used in three rearranged forms, and you should see all three at a glance:

Checking with an exact value. θ = π/6: cos²(π/6) + sin²(π/6) = (√3/2)² + (1/2)² = 3/4 + 1/4 = 1. ✓

3One ratio from another, without finding the angle

The guide's phrase is the relationship between the trigonometric ratios. The method has two halves:

The identity gives the size. The quadrant gives the sign.

Worked example 1 (a quadrant given). sin θ = 2/3 and π/2 < θ < π. Find the exact values of cos θ and tan θ.

Figure 2 shows the same thing as a triangle. sin θ = 2/3 means y = 2 on a circle of radius 3; Pythagoras gives x² = 9 − 4 = 5, and in the second quadrant x = −√5. Then cos θ = x ÷ r and tan θ = y ÷ x. Some students find the triangle faster; the algebra is what you write down.

Worked example 2 (no quadrant given). sin θ = 1/4. Find the possible values of tan θ.

The word possible is the signal. Without a quadrant, you keep both signs, and the answer is two values. Giving only +√15/15 loses a mark.

Worked example 3 (starting from tan). tan θ = −2 and 3π/2 < θ < 2π. Find sin θ and cos θ.

The triangle route is quicker here: tan θ = 2/1 in size means opposite 2, adjacent 1, hypotenuse √5, and then the fourth quadrant makes the sine negative.

4The double angle identities

The formula booklet gives

sin 2θ = 2 sin θ cos θ and cos 2θ = cos²θ − sin²θ = 2cos²θ − 1 = 1 − 2sin²θ.

Where they come from. The guide suggests a simple geometric diagram, and Figure 3 is the one to carry. Take an isosceles triangle OAB with OA = OB = 1 and angle AOB = 2θ. Its line of symmetry OM splits it into two right-angled triangles, each with hypotenuse 1 and angle θ at O, so OM = cos θ and AM = MB = sin θ.

• Area two ways. By ½ab sin C, the area is ½ × 1 × 1 × sin 2θ. As two right-angled triangles, the area is 2 × ½ × cos θ × sin θ. Setting these equal gives sin 2θ = 2 sin θ cos θ.
• The cosine rule on AB. AB = 2 sin θ, so (2 sin θ)² = 1² + 1² − 2 × 1 × 1 × cos 2θ. That is 4sin²θ = 2 − 2cos 2θ, so cos 2θ = 1 − 2sin²θ.

The picture only covers 0 < 2θ < π, but the identities hold for every angle, which a graphing package will confirm for you by drawing both sides and seeing one curve.

The three forms of cos 2θ. They are the same identity, rewritten with the Pythagorean identity:

Choose the form that matches what you are given or what you want. If you know sin θ, use 1 − 2sin²θ. If you know cos θ, use 2cos²θ − 1. If an equation contains cos 2x and sin x, the 1 − 2sin²x form makes everything sine, and 3.8 turns that into a quadratic.

"Double" is relative. The identities work for any angle and its double: sin 6x = 2 sin 3x cos 3x, and cos x = 1 − 2sin²(x/2). Spotting the pattern is often the whole question.

The trap Figure 4 shows. sin 2x is not 2 sin x. Doubling the angle makes the wave go twice as fast; doubling the sine makes it twice as tall. They agree only where both are zero.

Running the identities backwards. Rearranging the cos 2θ forms gives

Figure 5 draws the first one: the graph of cos²x is the graph of cos 2x halved and lifted by ½. These forms matter in calculus, where cos²x and sin²x cannot be integrated as they stand but (1 + cos 2x) ÷ 2 can. They also give exact values of half-angles: cos²(π/12) = (1 + cos(π/6)) ÷ 2 = (2 + √3)/4, and since π/12 is acute, cos(π/12) = √(2 + √3)/2.

5Double angles from a single ratio

This is the most common Paper 1 question on the page. Given one ratio of x, find the ratios of 2x.

Worked example 4. cos x = 1/3 and x is acute. Find the exact values of sin 2x, cos 2x and tan 2x.

Two sanity checks that take seconds. First, sin²2x + cos²2x = 32/81 + 49/81 = 1, so the pair is consistent. Second, the signs make sense: cos x = 1/3 means x is a little over 70°, so 2x is a little over 140°, in the second quadrant, where sine is positive and cosine and tangent are negative. That matches.

Worked example 5 (the quadrant matters twice). sin θ = −3/5 and π < θ < 3π/2. Find sin 2θ and cos 2θ.

Notice that sin 2θ is positive although both sin θ and cos θ are negative: θ is between π and 3π/2, so 2θ is between 2π and 3π, which puts it back in the first or second quadrant. Never assume 2θ is in the same quadrant as θ.

6Showing that an identity is true

A "show that" identity question gives you both sides. Your job is a chain of equal expressions from one side to the other.

1. Start from the more complicated side, usually the one with a double angle or a fraction, and work towards the simpler one.
2. Do not treat it as an equation. Do not add, multiply or square both sides. Transform one side until it becomes the other.
3. Replace the double angles with single-angle forms, choosing the form of cos 2θ that will cancel.
4. Look for cos²θ + sin²θ to replace by 1, and for a common factor to cancel.
5. Finish by writing the target expression, so the examiner sees you have arrived.

Worked example 6. Show that (sin θ + cos θ)² = 1 + sin 2θ.

Worked example 7. Show that sin 2θ ÷ (1 + cos 2θ) = tan θ.

The choice in the first line is the skill. With 1 − 2sin²θ instead, the denominator would be 2 − 2sin²θ, which still works but takes two more lines. Choose the form that kills the constant.