Lesson 3 of 5Trigonometric Identities & Equations

Unit 5 · Lesson 3 of 5

Double Angle Formulas

Setting $B = A$ in the compound angle formulas produces the double angle formulas, including three interchangeable versions of $cos 2 θ$ — and choosing the right version is most of the skill.

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Key things to remember

1 $sin 2 θ = 2 sin θ cos θ$ .
2 $cos 2 θ = cos^{2} θ - sin^{2} θ = 2 cos^{2} θ - 1 = 1 - 2 sin^{2} θ$ — all three are equal.
3 $tan 2 θ = \frac{2 tan θ}{1 - tan ^{2} θ}$ .
4The formulas generalize: $sin 6 x = 2 sin 3 x cos 3 x$ , because the half-angle is whatever is half of the angle in front of you.
5Rearranged for proofs: $1 + cos 2 θ = 2 cos^{2} θ$ and $1 - cos 2 θ = 2 sin^{2} θ$ .

Double angle formulas answer a very practical question: if you know something about $θ$ , what can you say about $2 θ$ ? They are not new facts — every one of them falls out of the compound angle formulas by setting the two angles equal. That is worth knowing, because if you blank on a double angle formula in the middle of a test you can rebuild it in about ten seconds.

Deriving them from compound angles

Since $2 θ = θ + θ$ , just take the addition formulas and replace both $A$ and $B$ with $θ$ . Everything collapses because the two angles are now the same.

sin 2 θ = sin (θ + θ) = sin θ cos θ + cos θ sin θ = 2 sin θ cos θ

The sine double angle formula, derived in one line.

cos 2 θ = cos (θ + θ) = cos θ cos θ - sin θ sin θ = cos^{2} θ - sin^{2} θ

The cosine double angle formula. Note the minus sign it inherits from the cosine addition formula.

Fig 1Setting the two angles equal in each compound angle formula collapses it into a double angle formula.

VideoAn Ontario Grade 12 lesson on the double angle formulas, which come from the compound angle formulas by setting the two angles equal.Watch on YouTube

Three faces of the cosine formula

The cosine version is special because you can feed the Pythagorean identity into it in two different ways. Replace $sin^{2} θ$ with $1 - cos^{2} θ$ and everything becomes cosine; replace $cos^{2} θ$ with $1 - sin^{2} θ$ and everything becomes sine. All three expressions are equal — they are the same number wearing different clothes.

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

All three forms are equal. Memorize the pattern: the cosine version has the 1 at the end, the sine version has the 1 at the front.

Situation	Best form of the cosine formula	Why
You know only $sin θ$	$cos 2 θ = 1 - 2 sin^{2} θ$	No cosine appears, so you never need its sign.
You know only $cos θ$	$cos 2 θ = 2 cos^{2} θ - 1$	No sine appears, so you never need its sign.
You know both	$cos 2 θ = cos^{2} θ - sin^{2} θ$	Fastest to substitute; also factors as a difference of squares.
The expression contains $1 + cos 2 θ$	$cos 2 θ = 2 cos^{2} θ - 1$	The $- 1$ cancels the $+ 1$ , leaving $2 cos^{2} θ$ .
The expression contains $1 - cos 2 θ$	$cos 2 θ = 1 - 2 sin^{2} θ$	The ones cancel, leaving $2 sin^{2} θ$ .

Choosing the form that kills the most work. The last two rows show up constantly in proofs.

The tangent double angle formula

The same substitution in $tan (A + B)$ gives the tangent version. It is the one students forget, and the one place where a stray square is easy to drop.

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

Tangent double angle. The denominator is one minus tangent SQUARED.

Working from a single given ratio

The standard question gives you one ratio and a quadrant, then asks for $sin 2 θ$ , $cos 2 θ$ and $tan 2 θ$ . The routine never changes: use the Pythagorean identity to find the missing ratio, use the quadrant to pick its sign, then substitute. Watch out — the quadrant tells you the sign of $sin θ$ and $cos θ$ , NOT the sign of $sin 2 θ$ , because doubling the angle may well move it into a different quadrant.

The formulas run backwards too

Any expression of the form two sine theta cosine theta is a disguised sine of two theta, and any expression of the form cosine squared minus sine squared is a disguised cosine of two theta. Spotting these in reverse is what turns a four-line proof into a one-line proof.

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

Given

sin θ = \frac{3}{5}

with

θ

in quadrant II, find the exact values of

sin 2 θ

cos 2 θ

and

tan 2 θ

Example 2

Find the exact value of

2 cos^{2} \frac{π}{8} - 1

without a calculator.

Example 3

Simplify

\frac{sin 2 θ}{1 + cos 2 θ}

Where marks get lost

Writing $sin 2 θ = 2 sin θ$ . Doubling the angle is not doubling the ratio — the correct formula is $2 sin θ cos θ$ , and the extra cosine changes everything.
Dropping the $2$ in front: writing $sin 2 θ = sin θ cos θ$ . The two identical terms from the compound formula add, they do not merge.
Mixing up the two one-sided forms of $cos 2 θ$ and writing $2 sin^{2} θ - 1$ or $1 - 2 cos^{2} θ$ . Both are wrong: the COSINE form ends in $- 1$ , the SINE form starts with $1 -$ .
Using $tan 2 θ = \frac{2 tan θ}{1 - tan θ}$ without the square. The denominator must be $1 - tan^{2} θ$ .
Assuming $2 θ$ lies in the same quadrant as $θ$ and forcing the signs of $sin 2 θ$ and $cos 2 θ$ to match. Use the quadrant only to fix $sin θ$ and $cos θ$ , then let the formulas produce the doubled signs.

Flashcards 9

sin 2 θ

cos 2 θ

(three forms)

tan 2 θ

Where do the double angle formulas come from?

+ 5 more

Practice

Which is the correct double angle formula for sine?

You are told only the value of

sin θ

and asked for

cos 2 θ

. Which form saves the most work?

Given

sin θ = \frac{3}{5}

with

θ

in quadrant II, find

cos 2 θ

as a decimal.

cos θ = \frac{1}{4}

, find

cos 2 θ

as a decimal.

tan θ = \frac{3}{4}

, find

tan 2 θ

as a decimal to four places.

Given

sin θ = \frac{3}{5}

with

θ

in quadrant I, find

sin 2 θ

as a decimal.

Simplify

2 sin 3 x cos 3 x

to a single trigonometric ratio.

Simplify

1 - 2 sin^{2} 4 x

to a single trigonometric ratio.

Put these steps in order to derive

cos 2 θ = 1 - 2 sin^{2} θ

1Replace $cos^{2} θ$ with $1 - sin^{2} θ$ using the Pythagorean identity
2Collect like terms to get $cos 2 θ = 1 - 2 sin^{2} θ$
3Write $cos 2 θ = (1 - sin^{2} θ) - sin^{2} θ$
4Start from $cos 2 θ = cos^{2} θ - sin^{2} θ$

Match each expression to its simplified form.

A. $cos 2 θ$
B. $sin 2 θ$
C. $tan 2 θ$
D. $2 cos^{2} θ$
E. $2 sin^{2} θ$

2 sin θ cos θ

1 + cos 2 θ

1 - cos 2 θ

cos^{2} θ - sin^{2} θ

\frac{2 tan θ}{1 - tan ^{2} θ}

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