Explainer: Double-Angle and Half-Angle Identities

In this explainer, we will learn how to use the Pythagorean identity and double-angle formulas to evaluate trigonometric values.

Recall the Pythagorean identity for trigonometric functions: sincos๏Šจ๏Šจ๐œƒ+๐œƒ=1.

Dividing through by cos๏Šจ๐œƒ gives the identity 1+๐œƒ=1๐œƒ=๐œƒtancossec๏Šจ๏Šจ๏Šจ. Using sin๏Šจ๐œƒ instead gives 1+๐œƒ=๐œƒcotcsc๏Šจ๏Šจ.

It is often a good strategy to rewrite more complicated trigonometric expressions in terms of sin๐œƒ and cos๐œƒ when seeking simplifications.

Example 1: Using Pythagorean Identities to Simplify Trigonometric Expressions

Simplify sincoscsccot๏Šจ๏Šจ๏Šจ๏Šจ๐œƒ+๐œƒ๐œƒโˆ’๐œƒ.

Answer

Using the Pythagorean identity on the numerator, we get sincoscsccotcsccot๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ๐œƒ+๐œƒ๐œƒโˆ’๐œƒ=1๐œƒโˆ’๐œƒ.

Since cscsin๐œƒ=1๐œƒ and cotcossin๐œƒ=๐œƒ๐œƒ, we get csccotsincossincossinsinsin๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ๐œƒโˆ’๐œƒ=1๐œƒโˆ’๐œƒ๐œƒ=1โˆ’๐œƒ๐œƒ=๐œƒ๐œƒ=1, resulting in 1๐œƒโˆ’๐œƒ=11=1.csccot๏Šจ๏Šจ

The double-angle formulas give sin(2๐œƒ) and cos(2๐œƒ) in terms of sin๐œƒ and cos๐œƒ. We recall these.

Definition: Double Angle Formulas

For any real number ๐œƒ, the following formulas hold: sinsincoscoscossin(2๐œƒ)=2๐œƒ๐œƒ,(2๐œƒ)=๐œƒโˆ’๐œƒ.๏Šจ๏Šจ

We can use these two to derive a formula for tan(2๐œƒ) too. In this case, purely in terms of tan๐œƒ, tansincossincoscossinthen,dividingthroughoutbycostantantan(2๐œƒ)=(2๐œƒ)(2๐œƒ)=2๐œƒ๐œƒ๐œƒโˆ’๐œƒ๐œƒ,==2(๐œƒ)(1)โˆ’=2๐œƒ1โˆ’๐œƒ,๏Šจ๏Šจ๏Šจ๏Šจ๏ผ๏ผ๏ผ๏ผ๏Šฑ๏ผ๏ผ๏ผ๏ผ๏ผ๏ผ๏Šจsincoscoscossincoscoscossincos๏Žก๏Žก๏Žก๏Žก๏Žก๏Žก๏Žก๏Žก giving us tantantan(2๐œƒ)=2๐œƒ1โˆ’๐œƒ๏Šจ and cotcotcot(2๐œƒ)=๐œƒโˆ’12๐œƒ.๏Šจ

Example 2: Simplifying Trigonometric Expressions Using Double-Angle Identities

Simplify sinsin๏Šจ๏Šช๐‘ฅโˆ’๐‘ฅ.

Answer

We notice that we can factor out a sin๏Šจ๐‘ฅ: sinsinsinsinsincoswhichremindsusofadouble-angleformulasincossincossinsin๏Šจ๏Šช๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ๐‘ฅโˆ’๐‘ฅ=๐‘ฅ๏€บ1โˆ’๐‘ฅ๏†=๐‘ฅ๐‘ฅ=(๐‘ฅ๐‘ฅ)=๏€ฝ2๐‘ฅ๐‘ฅ2๏‰=๏€ฝ2๐‘ฅ2๏‰=14(2๐‘ฅ).

Now, we will see examples on using the double-angle formulas in the opposite direction.

Example 3: Simplifying Trigonometric Expressions Using Double-Angle Identities

Simplify 1โˆ’2๐‘ฅ1+2๐‘ฅ.coscos

Answer

We use the same formula in the numerator and denominator: 1โˆ’2๐‘ฅ=1โˆ’๏€บ๐‘ฅโˆ’๐‘ฅ๏†=1โˆ’๐‘ฅ+๐‘ฅ=2๐‘ฅ,coscossincossinsin๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ where we used the Pythagorean identity. In the denominator, now with the identity 1โˆ’๐‘ฅ=๐‘ฅsincos๏Šจ๏Šจ, 1+2๐‘ฅ=1+๏€บ๐‘ฅโˆ’๐‘ฅ๏†=1+โˆ’๐‘ฅ=2๐‘ฅ.coscossincossincos๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ

Therefore, 1โˆ’2๐‘ฅ1+2๐‘ฅ=2๐‘ฅ2๐‘ฅ=๐‘ฅ.coscossincostan๏Šจ๏Šจ๏Šจ

A more challenging example is the following.

Example 4: Using Trigonometric Identities to Find Exact Values of Trigonometric Expressions

Find the value of seccsc๐‘‹+๐‘‹ given sincos๐‘‹+๐‘‹=โˆ’67, where ๐œ‹2<๐‘‹<๐œ‹.

Answer

What we must evaluate is seccsccossinsincossincos๐‘‹+๐‘‹=1๐‘‹+1๐‘‹=๐‘‹+๐‘‹๐‘‹๐‘‹.

We are given the numerator: โˆ’67. It is therefore enough to know the value of sincos๐‘‹๐‘‹ for this specific ๐‘‹.

To do this, we can square the sum of the sine and cosine, because (๐‘‹+๐‘‹)=๐‘‹+2๐‘‹๐‘‹+๐‘‹=๏€บ๐‘‹+๐‘‹๏†+2๐‘‹๐‘‹=1+2๐‘‹๐‘‹,sincossinsincoscossincossincossincos๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ so that, using our known sum, ๏€ผโˆ’67๏ˆ=1+2๐‘‹๐‘‹๏Šจsincos which means sincos๐‘‹๐‘‹=12๏€ผ3649โˆ’1๏ˆ=โˆ’1398.

Then, seccscsincossincos๐‘‹+๐‘‹=๐‘‹+๐‘‹๐‘‹๐‘‹=โˆ’โˆ’=8413.๏Šฌ๏Šญ๏Šง๏Šฉ๏Šฏ๏Šฎ

A second example of such an evaluation follows.

Example 5: Using Double Angle Identities to Evaluate a Trigonometric Expression

Find, without using a calculator, the value of sincos2๐ต22๐ต given cos๐ต=45, where 3๐œ‹2<๐ต<2๐œ‹.

Answer

In order to apply the double-angle formulas, we need to know the value of sin๐ต. The Pythagorean identity sincos๏Šจ๏Šจ๐ต+๐ต=1 can be solved to give sincos๏Šจ๏Šจ๏Šจ๐ต=1โˆ’๐ต=1โˆ’๏€ผ45๏ˆ=925.

So sin๐ต is either 35 or โˆ’35. The information about the location of ๐ต tells us it is in the 4th quadrant, where the sine is negative. Therefore, sin๐ต=โˆ’35, and sincossincoscossin2๐ต22๐ต=2๐ต๐ต2๏€บ๐ตโˆ’๐ต๏†=2๏€ปโˆ’๏‡๏€ป๏‡2๏€ปโˆ’๏‡=โˆ’=โˆ’127.๏Šจ๏Šจ๏Šฉ๏Šซ๏Šช๏Šซ๏Šช๏Šซ๏Šฉ๏Šซ๏Šง๏Šจ๏Šจ๏Šซ๏Šญ๏Šจ๏Šซ๏Žก๏Žก๏Žก๏Žก

Of course, an alternative is to observe that we are being asked to evaluate tan2๐ต2 and use the double-angle formula for tangents.

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