Lesson Explainer: Double-Angle and Half-Angle Identities | Nagwa Lesson Explainer: Double-Angle and Half-Angle Identities | Nagwa

Lesson Explainer: Double-Angle and Half-Angle Identities Mathematics

In this explainer, we will learn how to use the double-angle and half-angle identities to evaluate trigonometric values.

The double-angle identities give cos2𝜃 and sin2𝜃 in terms of cos𝜃 and sin𝜃. They are a special case of the sum identities, coscoscossinsin(𝜃+𝜃)=𝜃𝜃𝜃𝜃 and sinsincoscossin(𝜃+𝜃)=𝜃𝜃+𝜃𝜃, namely when 𝜃=𝜃. The double-angle identities can be simply derived from them as shown in the following: coscoscoscossinsincossinsinsincossincossincossin2𝜃=(𝜃+𝜃)=𝜃𝜃𝜃𝜃=𝜃𝜃,2𝜃=(𝜃+𝜃)=𝜃𝜃+𝜃𝜃=2𝜃𝜃.

Identity: Double-Angle Identities

For any real number 𝜃, we have coscossinsincossin2𝜃=𝜃𝜃,2𝜃=2𝜃𝜃.

For 𝜃(45,135,225,315)+𝑛360𝜃𝜋4,3𝜋4,5𝜋4,7𝜋4+2𝑛𝜋, we have tantantan2𝜃=(2𝜃)1𝜃.

It is worth pausing here for a moment and reflecting on the respective signs of cos2𝜃, sin2𝜃, and tan2𝜃 as functions of 𝜃. As shown in the following figure, if we let 𝜃 be in a half-quadrant of the unit circle, then 2𝜃 is in a whole quadrant.

For instance, when 0𝜃45 (or 0𝜃𝜋4), we have 02𝜃90 (or 0𝜃𝜋2). In this case, cos𝜃, sin𝜃, tan𝜃, cos2𝜃, sin2𝜃, and tan2𝜃 are all nonnegative, and tan2𝜃 is not defined for 𝜃=45.

Let us check that this is consistent with the double-angle identities. The first identity is coscossin2𝜃=𝜃𝜃. Since |𝜃||𝜃|cossin, we have cossin𝜃𝜃, and thus, cossin𝜃𝜃0, which corresponds indeed to cos2𝜃 being nonnegative.

We derive from the second identity, sincossin2𝜃=2𝜃𝜃, that sin2𝜃 is nonnegative if cos𝜃 and sin𝜃 have the same sign or one of them is zero. This is the case for 0𝜃45.

Finally, since cossin𝜃>𝜃0 when 0𝜃<45, we have 0𝜃=𝜃𝜃<1tansincos, which leads to 1𝜃>0tan, and thus, tantantan2𝜃=2𝜃1𝜃0.

We can apply similar considerations for all 8 half-quadrants of the unit circle (for the 4 half-quadrants at the bottom of the circle, the values of 2𝜃 are then greater than 360 or 2𝜋 rad) and show that the double-angle identities hold for any values of 𝜃.

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

Let us look with our first example at how to use these double-angle identities to simplify trigonometric expressions.

Example 1: Simplifying Trigonometric Expressions Using Double-Angle Identities

Simplify 12𝑥1+2𝑥.coscos


Recall that coscossin2𝑥=𝑥𝑥, which we can use to simplify the numerator and denominator: 12𝑥=1𝑥𝑥=1𝑥+𝑥=2𝑥,coscossincossinsin where we used the Pythagorean identity 1𝑥=𝑥cossin. In the denominator, now with the identity 1𝑥=𝑥sincos, 1+2𝑥=1+𝑥𝑥=1+𝑥=2𝑥.coscossincossincos

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

Let us take a note of the two identities that we derived from coscossin2𝜃=𝜃𝜃 in the previous example.

Identity: Identities Derived from the Double-Angle Identity cos 2𝜃 = cos2 𝜃 - sin2 𝜃

For any real number 𝜃, we have 1+2𝜃=2𝜃,12𝜃=2𝜃.coscoscossin

Now, we will see an example of how we can simplify a trigonometric expression by using the double-angle identity in the opposite direction.

Example 2: Simplifying Trigonometric Expressions Using Double-Angle Identities

Simplify sinsin𝑥𝑥.


We notice that we can factor out a sin𝑥: sinsinsinsin𝑥𝑥=𝑥1𝑥. The Pythagorean identity cossin𝑥+𝑥=1 gives 1𝑥=𝑥sincos. Substituting this in leads to sinsinsincossincos𝑥𝑥=𝑥𝑥=(𝑥𝑥). Using the double-angle identity sinsincos2𝑥=2𝑥𝑥, we find that sincossin𝑥𝑥=2𝑥2. Substituting in gives sinsinsinsin𝑥𝑥=2𝑥2=14(2𝑥).

Let us now evaluate a trigonometric expression using the double-angle identity.

Example 3: 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.


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.

Using the double-angle identities sinsincos2𝐵=2𝐵𝐵 and coscossin2𝐵=𝐵𝐵, we find that 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.

So far, we have used the double-angle identities. The half-angle identities can be derived from them simply by realizing that the difference between considering one angle and its double and considering an angle and its half is just a matter of perspective.

Using the identity 2𝜃=1+2𝜃coscos that we derived from the double-angle identity coscossin2𝜃=𝜃𝜃 and substituting in 2𝜃=𝜑 and 𝜃=𝜑2, we get 2𝜑2=1+𝜑𝜑2=±1+𝜑2.coscoscoscos

Starting from the other identity derived from the double-angle identity coscossin2𝜃=𝜃𝜃 (i.e., 2𝜃=12𝜃sincos), we find that 2𝜑2=1𝜑𝜑2=±1𝜑2.sincossincos

Finally, to find tan𝜑2, we use the expressions we just established for cos𝜑2 and sin𝜑2 by writing tansincoscoscoscoscos𝜑2==±1𝜑2±21+𝜑=±1𝜑1+𝜑.

With the three half-angle identities we have derived above, we see that the absolute values of the three trigonometric functions of half an angle depend only on the value of the cosine of this angle. However, the “±” shows that their signs are not determined. This means that we will be able to write the correct sign in the identity only if we know in which quadrant the half angle is located.

This comes from the fact that an angle is not fully determined by its cosine value. Take for instance 𝜑=140, 𝜑=220, 𝜑=500, and 𝜑=580. While we have coscoscoscos𝜑=𝜑=𝜑=𝜑, we see that coscoscoscos𝜑2=𝜑2=𝜑2=𝜑2 and sinsinsinsin𝜑2=𝜑2=𝜑2=𝜑2, as shown in the following diagram.

Let us now see how the tangent half-angle identity can be rewritten so that the sign is determined.

Starting with tansincos𝜑2=, we multiply the numerator and denominator of the fraction on the right-hand side by 2𝜑2cos, which gives tansincoscos𝜑2=22.

The numerator can be rewritten as sin𝜑 using the double-angle identities 2𝜃𝜃=2𝜃sincossin (with 2𝜃=𝜑 and 𝜃=𝜑2) and the denominator as 1+𝜑cos using the identity 1+2𝜃=2𝜃coscos (with 2𝜃=𝜑 and 𝜃=𝜑2) derived from the double-angle identity coscossin2𝜃=𝜃𝜃.

Hence, we have tansincos𝜑2=𝜑1+𝜑.

Starting again from tansincos𝜑2=, we now multiply the numerator and denominator of the fraction on the right-hand side by 2𝜑2sin. We get tansinsincos𝜑2=22.

Using here the identity 12𝜃=2𝜃cossin, we can rewrite 2𝜑2sin as 1𝜑cos and, as before, 2𝜑2𝜑2sincos as sin𝜑. Hence, we find that tancossin𝜑2=1𝜑𝜑.

It is worth noting that rearranging tancoscos𝜑2=±1𝜑1+𝜑 by multiplying the right-hand side of the equation by either 1+𝜑1+𝜑coscos or 1𝜑1𝜑coscos leads to tansincos𝜑2=±|𝜑|1+𝜑 and tancossin𝜑2=±1𝜑|𝜑|. Studying the respective signs of tan𝜑2 and sin𝜑 as shown in the following table, we conclude that tan𝜑2 and sin𝜑 always have the same sign. This allows us to establish the identities tansincos𝜑2=𝜑1+𝜑 and tancossin𝜑2=1𝜑𝜑.

𝜑 ()(0–90)(90–180)(180–270)(270–360)(360–450)(450–540)(540–630)(630–720)
𝜑2 ()(0–45)(45–90)(90–135)(135–180)(180–225)(225–270)(270–315)(315–360)

Let us recap the half-angles identities that we have just derived.

Identity: Half-Angle Identities

For any real number 𝜃, we have coscossincos𝜃2=±1+𝜃2,𝜃2=±1𝜃2.

For 𝜃180+𝑛360(𝜃𝜋+2𝑛𝜋), where 𝑛 is an integer, we have tancoscostansincos𝜃2=±1𝜃1+𝜃,𝜃2=𝜃1+𝜃.

For 𝜃(0,180)+𝑛360(𝜃(0,𝜋)+2𝑛𝜋), we have tancossin𝜃2=1𝜃𝜃.

Let us look at how to use the half-angle identities in the following example.

Example 4: Evaluating the Cosine Function for a Half Angle given the Cosine Function and Quadrant of the Angle

Find the value of cos𝜃2 given cos𝜃=1517, where 0<𝜃<90, without using a calculator.


Recall the half-angle identity coscos𝜃2=±1+𝜃2.

We are told that 0<𝜃<90. Hence, 0<𝜃2<45 and cos𝜃2>0. This means that we have coscos𝜃2=1+𝜃2 since 1+𝜃20.cos

Let us work out the value of cos𝜃2 by substituting the value of cos𝜃 into the above equation: cos𝜃2=1+2=322×17=1617=417.

We can rationalize this fraction by multiplying it by 1717, which gives cos𝜃2=41717.

Finally, let us see with our last example how the double-angle and half-angle identities can be used to find the exact value of some trigonometric functions.

Example 5: Finding the Exact Value of a Trigonometric Expression

Using the half-angle formulas, or otherwise, find the exact value of tan𝜋8.


The angle 𝜋8 radians is not one of the special angles whose trigonometric ratios are well known. However, we notice that it is half the value of these special angles, namely half of 𝜋4 radians. Using the half-angle identity tansincos𝜃2=𝜃1+𝜃, we find that tansincos𝜋8=1+.

Since sincos𝜋4=𝜋4=12, we have tan𝜋8=1+=12+1.

Multiplying the right-hand side by 2121 gives tan𝜋8=2121=21.21 or 1+2 is the exact value of tan𝜋8.

We would of course find the same result using tancossin𝜃2=1𝜃𝜃.

We could also use tancoscos𝜃2=±1𝜃1+𝜃, which gives tancoscos𝜋8=11+ since we know that tan𝜋8 is positive.

Let us now summarize what we have learned in this explainer.

Key Points

  • The double-angle identities state that, for any real number 𝜃, we have coscossinsincossin2𝜃=𝜃𝜃,2𝜃=2𝜃𝜃, and for 𝜃(45,135,225,315)+𝑛360𝜃𝜋4,3𝜋4,5𝜋4,7𝜋4+2𝑛𝜋, we have tantantan2𝜃=2𝜃1𝜃.
  • From the above identity coscossin2𝜃=𝜃𝜃, we can derive the following two identities for any real number 𝜃: 1+2𝜃=2𝜃,12𝜃=2𝜃.coscoscossin
  • The half-angle identities state that, for any real number 𝜃, we have coscossincos𝜃2=±1+𝜃2,𝜃2=±1𝜃2. For 𝜃180+𝑛360(𝜃𝜋+2𝑛𝜋), we have tancoscostansincos𝜃2=±1𝜃1+𝜃,𝜃2=𝜃1+𝜃, and for 𝜃(0,180)+𝑛360(𝜃(0,𝜋)+2𝑛𝜋), we have tancossin𝜃2=1𝜃𝜃.

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