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: sincos𝜃+𝜃=1.

Dividing through by cos𝜃 gives the identity 1+𝜃=1𝜃=𝜃tancossec. Using sin𝜃 instead gives 1+𝜃=𝜃cotcsc.

It is often a good strategy to rewrite more complicated trigonometric expressions in terms of sin𝜃 and cos𝜃 when seeking simplifications.

Example 1: Using Pythagorean Identities to Simplify Trigonometric Expressions

Simplify sincoscsccot𝜃+𝜃𝜃𝜃.

Answer

Using the Pythagorean identity on the numerator, we get sincoscsccotcsccot𝜃+𝜃𝜃𝜃=1𝜃𝜃.

Since cscsin𝜃=1𝜃 and cotcossin𝜃=𝜃𝜃, we get csccotsincossincossinsinsin𝜃𝜃=1𝜃𝜃𝜃=1𝜃𝜃=𝜃𝜃=1, resulting in 1𝜃𝜃=11=1.csccot

The double-angle formulas give sin(2𝜃) and cos(2𝜃) in terms of sin𝜃 and cos𝜃. We recall these.

Definition: Double Angle Formulas

For any real number 𝜃, the following formulas hold: sinsincoscoscossin(2𝜃)=2𝜃𝜃,(2𝜃)=𝜃𝜃.

We can use these two to derive a formula for tan(2𝜃) too. In this case, purely in terms of tan𝜃, tansincossincoscossinthen,dividingthroughoutbycostantantan(2𝜃)=(2𝜃)(2𝜃)=2𝜃𝜃𝜃𝜃𝜃,==2(𝜃)(1)=2𝜃1𝜃,sincoscoscossincoscoscossincos giving us tantantan(2𝜃)=2𝜃1𝜃 and cotcotcot(2𝜃)=𝜃12𝜃.

Example 2: Simplifying Trigonometric Expressions Using Double-Angle Identities

Simplify sinsin𝑥𝑥.

Answer

We notice that we can factor out a sin𝑥: sinsinsinsinsincoswhichremindsusofadouble-angleformulasincossincossinsin𝑥𝑥=𝑥1𝑥=𝑥𝑥=(𝑥𝑥)=2𝑥𝑥2=2𝑥2=14(2𝑥).

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 12𝑥1+2𝑥.coscos

Answer

We use the same formula in the numerator and denominator: 12𝑥=1𝑥𝑥=1𝑥+𝑥=2𝑥,coscossincossinsin where we used the Pythagorean identity. In the denominator, now with the identity 1𝑥=𝑥sincos, 1+2𝑥=1+𝑥𝑥=1+𝑥=2𝑥.coscossincossincos

Therefore, 12𝑥1+2𝑥=2𝑥2𝑥=𝑥.coscossincostan

A more challenging example is the following.

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

Find the value of seccsc𝑋+𝑋 given sincos𝑋+𝑋=67, where 𝜋2<𝑋<𝜋.

Answer

What we must evaluate is seccsccossinsincossincos𝑋+𝑋=1𝑋+1𝑋=𝑋+𝑋𝑋𝑋.

We are given the numerator: 67. It is therefore enough to know the value of sincos𝑋𝑋 for this specific 𝑋.

To do this, we can square the sum of the sine and cosine, because (𝑋+𝑋)=𝑋+2𝑋𝑋+𝑋=𝑋+𝑋+2𝑋𝑋=1+2𝑋𝑋,sincossinsincoscossincossincossincos so that, using our known sum, 67=1+2𝑋𝑋sincos which means sincos𝑋𝑋=1236491=1398.

Then, seccscsincossincos𝑋+𝑋=𝑋+𝑋𝑋𝑋==8413.

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 sincos2𝐵22𝐵 given cos𝐵=45, where 3𝜋2<𝐵<2𝜋.

Answer

In order to apply the double-angle formulas, we need to know the value of sin𝐵. The Pythagorean identity sincos𝐵+𝐵=1 can be solved to give sincos𝐵=1𝐵=145=925.

So sin𝐵 is either 35 or 35. The information about the location of 𝐵 tells us it is in the 4th quadrant, where the sine is negative. Therefore, sin𝐵=35, and sincossincoscossin2𝐵22𝐵=2𝐵𝐵2𝐵𝐵=22==127.

Of course, an alternative is to observe that we are being asked to evaluate tan2𝐵2 and use the double-angle formula for tangents.

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