Trig Identities and Proofs

Use sin²θ + cos²θ = 1 and tan θ = sin θ/cos θ (with the sec² and cosec² identities for 0606/4037) to simplify expressions and prove identities line by line.

In the exam for IGCSE Additional Maths 0606, Edexcel Further Pure Maths 4PM1 · 15 real past-paper questions below

اقرأ هذا الدرس مشروحاً بالعربي ←

True for every angle

The equation $sin θ = \frac{1}{2}$ is true only for some angles. An identity is different: it is true for every angle. It lets you swap one form for another, like changing one 100 note into five 20 notes. The value stays the same.

On the circle of radius 1, the point P is $(cos θ, sin θ)$ . Its distance from the origin is 1. Pythagoras on the across and up distances gives the most important identity in trig:

Formula

s i n^{2} θ + c o s^{2} θ = 1

$sin^{2} θ$ — means $(sin θ)^{2}$ : find the sine, then square it.
$cos^{2} θ$ — means $(cos θ)^{2}$ .
$θ$ — any angle at all, in degrees or radians.

Worked example

1
Check the identity with $θ = 3 0^{\circ}$ .
2
$sin 3 0^{\circ} = \frac{1}{2}, cos 3 0^{\circ} = \frac{3}{2}$ — Exact values.
3
$(\frac{1}{2})^{2} + (\frac{3}{2})^{2} = \frac{1}{4} + \frac{3}{4}$ — Square each one.
4
$= 1$ — It works, as it must for every angle.

You will often use it rearranged, to swap a square of one ratio for the other:

Formula

sin^{2} θ = 1 - cos^{2} θ

$sin^{2} θ$ — take the cos-squared part away from both sides.

Formula

cos^{2} θ = 1 - sin^{2} θ

$cos^{2} θ$ — take the sin-squared part away from both sides.

The second identity you met already: tan is sin over cos.

Formula

tan θ = \frac{sin θ}{cos θ}

$θ$ — any angle where $cos θ$ is not 0.

Rule

0606 and 4037 print $sin^{2} θ + cos^{2} θ = 1$ in the List of formulas, but not the tan one. 4PM1 prints the tan one on the formula sheet, but you must learn $sin^{2} θ + cos^{2} θ = 1$ .

Common mistake

Reading $sin^{2} θ$ as $sin (θ^{2})$ . The 2 squares the answer, not the angle.

Key idea

The co in cosine stands for complement: the angle that makes $9 0^{\circ}$ with yours. The cosine of an angle is the sine of its complement: $cos 3 0^{\circ} = sin 6 0^{\circ}$ .

Two more (0606 and 4037)

0606 and 4037 also use identities with sec, cosec and cot. 4PM1 does not, so 4PM1 students can skip this step. Both come from the first identity by dividing.

Divide every term by $cos^{2} θ$ :

$sin^{2} θ + cos^{2} θ = 1$ — start from the main identity.

$\frac{s i n ^{2} θ}{c o s ^{2} θ} + \frac{c o s ^{2} θ}{c o s ^{2} θ} = \frac{1}{c o s ^{2} θ}$ — divide every term by cos squared.

$tan^{2} θ + 1 = sec^{2} θ$ — sin over cos is tan, and 1 over cos is sec.

Formula

1 + t a n^{2} θ = s e c^{2} θ

$tan^{2} θ$ — the square of $tan θ$ .
$sec^{2} θ$ — the square of $\frac{1}{c o s θ}$ .

Now divide every term by $sin^{2} θ$ instead:

$\frac{s i n ^{2} θ}{s i n ^{2} θ} + \frac{c o s ^{2} θ}{s i n ^{2} θ} = \frac{1}{s i n ^{2} θ}$ — divide every term by sin squared.

$1 + cot^{2} θ = cosec^{2} θ$ — cos over sin is cot, and 1 over sin is cosec.

Formula

1 + c o t^{2} θ = cosec^{2} θ

$cot^{2} θ$ — the square of $cot θ$ .
$cosec^{2} θ$ — the square of $\frac{1}{s i n θ}$ .

Rule

Both are printed in the List of formulas. If you ever forget one, divide $sin^{2} θ + cos^{2} θ = 1$ again.

Worked example

1
Simplify $(sec^{2} θ - 1) cos^{2} θ$ .
2
$sec^{2} θ - 1 = tan^{2} θ$ — Rearrange 1 + tan² = sec².
3
$tan^{2} θ cos^{2} θ = \frac{s i n ^{2} θ}{c o s ^{2} θ} \times cos^{2} θ$ — Write tan as sin over cos.
4
$= sin^{2} θ$ — The cos squared cancels.

Common mistake

Getting a sign wrong: $sec^{2} θ = 1 - tan^{2} θ$ is false. Both squares are added: $1 + tan^{2} θ = sec^{2} θ$ . A quick test with $θ = 0$ catches this: $sec^{2} 0 = 1$ and $tan^{2} 0 = 0$ .

Key idea

One identity, divided two ways, gives the other two. Plus signs on the side with the 1.

Simplifying expressions

Before proofs, practise tidying. The same few moves come up again and again.

Steps

1
1. Change tan (and sec, cosec, cot) into sin and cos.
2. Look for $1 - sin^{2} θ$ , $1 - cos^{2} θ$ or $sin^{2} θ + cos^{2} θ$ , and swap them.
3. Factorise: take out a common factor, or use the difference of two squares.
4. Cancel only whole factors, never single terms of a sum.

Worked example

1
Simplify $\frac{1 - c o s ^{2} θ}{s i n θ c o s θ}$ .
2
$1 - cos^{2} θ = sin^{2} θ$ — The rearranged identity.
3
$\frac{s i n ^{2} θ}{s i n θ c o s θ} = \frac{s i n θ}{c o s θ}$ — One factor of sin θ cancels from top and bottom.
4
$= tan θ$ — Sin over cos is tan.

Worked example

1
Show that $sin^{4} θ - cos^{4} θ = sin^{2} θ - cos^{2} θ$ .
2
$sin^{4} θ - cos^{4} θ = (sin^{2} θ)^{2} - (cos^{2} θ)^{2}$ — Two squares with a minus between them.
3
$= (sin^{2} θ - cos^{2} θ) (sin^{2} θ + cos^{2} θ)$ — Difference of two squares.
4
$= (sin^{2} θ - cos^{2} θ) \times 1$ — The second bracket is the identity.
5
$= sin^{2} θ - cos^{2} θ$ — As required.

Worked example

1
Write $3 cos^{2} θ + 2 sin θ$ in terms of $sin θ$ only.
2
$3 (1 - sin^{2} θ) + 2 sin θ$ — Swap cos squared for 1 − sin squared.
3
$= 3 + 2 sin θ - 3 sin^{2} θ$ — Expand the bracket.

Common mistake

Cancelling part of a sum: $\frac{s i n θ + c o s θ}{s i n θ}$ is NOT $1 + cos θ$ . Split it first: $1 + \frac{c o s θ}{s i n θ}$ .

Key idea

Sin and cos first, swap the squares, then factorise. That order solves most simplifying questions.

Proving an identity

“Prove” or “Show that” means the answer is already printed. So every mark is for the working: each line must follow clearly from the line before.

Steps

1
1. Start with the more complicated side, usually the one with fractions.
2. Write everything in sin and cos.
3. Combine fractions over one common denominator.
4. Use an identity to swap squares.
5. Simplify until you reach the other side exactly. Then stop.

Worked example

1
Prove that $\frac{c o s θ}{1 - s i n θ} + \frac{c o s θ}{1 + s i n θ} = \frac{2}{c o s θ}$ .
2
$LHS = \frac{c o s θ ( 1 + s i n θ ) + c o s θ ( 1 - s i n θ )}{( 1 - s i n θ ) ( 1 + s i n θ )}$ — The left side is harder, so start there. Common denominator: multiply the two bottoms.
3
$= \frac{c o s θ + c o s θ s i n θ + c o s θ - c o s θ s i n θ}{1 - s i n ^{2} θ}$ — Expand the top. The bottom is a difference of two squares.
4
$= \frac{2 c o s θ}{c o s ^{2} θ}$ — The cos θ sin θ terms cancel. And 1 − sin² θ is cos² θ.
5
$= \frac{2}{c o s θ} = RHS$ — Cancel one cos θ. This is the right side.

In 0606 and 4037, the same answer may be written $2 sec θ$ , since $\frac{1}{c o s θ} = sec θ$ .

Worked example

1
Prove that $(sin θ + cos θ)^{2} = 1 + 2 sin θ cos θ$ .
2
$LHS = sin^{2} θ + 2 sin θ cos θ + cos^{2} θ$ — Expand the square: first squared, twice the product, last squared.
3
$= (sin^{2} θ + cos^{2} θ) + 2 sin θ cos θ$ — Group the two squares.
4
$= 1 + 2 sin θ cos θ = RHS$ — Use the identity.

Rule

Mark schemes give a mark for using an identity correctly, marks for the combining steps, and the last mark only when every line is correct. Write $θ$ after every sin and cos: a bare “sin” can cost that last mark.

Common mistake

Treating it like an equation and moving terms across the = sign. You are not solving; you are showing that one side turns into the other.

Common mistake

Starting from the printed answer and working back to something true, like 1 = 1. That proves nothing unless every step can be reversed. Start from one side instead.

Key idea

Pick the harder side, go to sin and cos, use one common denominator, and finish at the other side.

Proofs with sec, cosec, cot

For 0606 and 4037, proofs often include sec, cosec and cot. 4PM1 students can skip this step. Use the same plan, with two extra tricks: change them to sin and cos, or use the sec² and cosec² identities.

Worked example

1
Prove that $tan θ + cot θ = sec θ cosec θ$ .
2
$LHS = \frac{s i n θ}{c o s θ} + \frac{c o s θ}{s i n θ}$ — Write tan and cot in sin and cos.
3
$= \frac{s i n ^{2} θ + c o s ^{2} θ}{s i n θ c o s θ}$ — Common denominator sin θ cos θ.
4
$= \frac{1}{s i n θ c o s θ}$ — The top is the identity.
5
$= \frac{1}{c o s θ} \times \frac{1}{s i n θ} = sec θ cosec θ = RHS$ — Split the fraction into the two reciprocals.

Worked example

1
Prove that $\frac{s e c ^{2} θ - 1}{cosec ^{2} θ - 1} = tan^{4} θ$ .
2
$sec^{2} θ - 1 = tan^{2} θ$ — From 1 + tan² θ = sec² θ.
3
$cosec^{2} θ - 1 = cot^{2} θ$ — From 1 + cot² θ = cosec² θ.
4
$LHS = \frac{t a n ^{2} θ}{c o t ^{2} θ} = tan^{2} θ \times tan^{2} θ$ — Dividing by cot² θ is the same as multiplying by tan² θ.
5
$= tan^{4} θ = RHS$ — Multiply the powers together: 2 + 2 = 4.

Worked example

1
Prove that $cosec θ - sin θ = cos θ cot θ$ .
2
$LHS = \frac{1}{s i n θ} - sin θ = \frac{1 - s i n ^{2} θ}{s i n θ}$ — Write cosec as 1 over sin, then use one denominator.
3
$= \frac{c o s ^{2} θ}{s i n θ}$ — 1 − sin² θ is cos² θ.
4
$= cos θ \times \frac{c o s θ}{s i n θ} = cos θ cot θ = RHS$ — Split off one cos θ. Cos over sin is cot.

Common mistake

$\frac{1}{s i n θ + c o s θ}$ is NOT $cosec θ + sec θ$ . One over a sum cannot be split into two fractions.

Key idea

When you see $sec^{2} θ - 1$ or $cosec^{2} θ - 1$ , swap it straight away for $tan^{2} θ$ or $cot^{2} θ$ .

Which method? Read the question

Look at what the expression contains. That tells you the first move.

You see	You do
$tan θ$ mixed with sin and cos	Write $tan θ = \frac{s i n θ}{c o s θ}$
$1 - sin^{2} θ$ or $1 - cos^{2} θ$	Swap it for $cos^{2} θ$ or $sin^{2} θ$
Two fractions added or subtracted	One common denominator, then simplify the top
$(1 - sin θ) (1 + sin θ)$	Difference of two squares: $1 - sin^{2} θ = cos^{2} θ$
A fourth power like $sin^{4} θ - cos^{4} θ$	Difference of two squares, then $sin^{2} θ + cos^{2} θ = 1$
“in terms of $sin θ$ ”	Replace every $cos^{2} θ$ by $1 - sin^{2} θ$
$sec^{2} θ - 1$ or $cosec^{2} θ - 1$ (0606 and 4037)	Swap for $tan^{2} θ$ or $cot^{2} θ$

Common mistake

Writing several steps in one line. If the examiner cannot see where the identity was used, the mark for it is lost.

Key idea

Every identity here comes from one fact: the point on the circle is 1 away from the centre.

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