Lesson Explainer: Double-Angle and Half-Angle Identities Mathematics • 10th Grade

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 cos2πœƒ and sin2πœƒ in terms of cosπœƒ and sinπœƒ. They are a special case of the sum identities, coscoscossinsin(πœƒ+πœƒβ€²)=πœƒπœƒβ€²βˆ’πœƒπœƒβ€² and sinsincoscossin(πœƒ+πœƒβ€²)=πœƒπœƒβ€²+πœƒπœƒβ€², namely when πœƒ=πœƒβ€². The double-angle identities can be simply derived from them as shown in the following: coscoscoscossinsincossinsinsincossincossincossin2πœƒ=(πœƒ+πœƒ)=πœƒπœƒβˆ’πœƒπœƒ=πœƒβˆ’πœƒ,2πœƒ=(πœƒ+πœƒ)=πœƒπœƒ+πœƒπœƒ=2πœƒπœƒ.

Identity: Double-Angle Identities

For any real number πœƒ, we have coscossinsincossin2πœƒ=πœƒβˆ’πœƒ,2πœƒ=2πœƒπœƒ.

For πœƒβ‰ (45,135,225,315)+𝑛360ο€Όπœƒβ‰ ο€Όπœ‹4,3πœ‹4,5πœ‹4,7πœ‹4+2π‘›πœ‹οˆβˆ˜βˆ˜βˆ˜βˆ˜βˆ˜, we have tantantan2πœƒ=(2πœƒ)ο€Ή1βˆ’πœƒο….

It is worth pausing here for a moment and reflecting on the respective signs of cos2πœƒ, sin2πœƒ, and tan2πœƒ as functions of πœƒ. As shown in the following figure, if we let πœƒ be in a half-quadrant of the unit circle, then 2πœƒ is in a whole quadrant.

For instance, when 0β‰€πœƒβ‰€45∘ (or 0β‰€πœƒβ‰€πœ‹4), we have 0≀2πœƒβ‰€90∘ (or 0β‰€πœƒβ‰€πœ‹2). In this case, cosπœƒ, sinπœƒ, tanπœƒ, cos2πœƒ, sin2πœƒ, and tan2πœƒ are all nonnegative, and tan2πœƒ is not defined for πœƒ=45∘.

Let us check that this is consistent with the double-angle identities. The first identity is coscossin2πœƒ=πœƒβˆ’πœƒοŠ¨οŠ¨. Since |πœƒ|β‰₯|πœƒ|cossin, we have cossinοŠ¨οŠ¨πœƒβ‰₯πœƒ, and thus, cossinοŠ¨οŠ¨πœƒβˆ’πœƒβ‰₯0, which corresponds indeed to cos2πœƒ being nonnegative.

We derive from the second identity, sincossin2πœƒ=2πœƒπœƒ, that sin2πœƒ is nonnegative if cosπœƒ and sinπœƒ have the same sign or one of them is zero. This is the case for 0β‰€πœƒβ‰€45∘.

Finally, since cossinπœƒ>πœƒβ‰₯0 when 0β‰€πœƒ<45∘, we have 0β‰€πœƒ=πœƒπœƒ<1tansincos, which leads to 1βˆ’πœƒ>0tan, and thus, tantantan2πœƒ=2πœƒ1βˆ’πœƒβ‰₯0.

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 2πœƒ are then greater than 360∘ or 2πœ‹ rad) and show that the double-angle identities hold for any values of πœƒ.

We can use these two to derive a formula for tan(2πœƒ) too. In this case, purely in terms of tanπœƒ, tansincossincoscossindividingthroughoutbycostantantan(2πœƒ)=(2πœƒ)(2πœƒ)=2πœƒπœƒπœƒβˆ’πœƒο€Ήπœƒο…==2(πœƒ)(1)βˆ’=2πœƒ1βˆ’πœƒ,sincoscoscossincoscoscossincos giving us tantantan(2πœƒ)=2πœƒ1βˆ’πœƒοŠ¨ and cotcotcot(2πœƒ)=πœƒβˆ’12πœƒ.

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 1βˆ’2π‘₯1+2π‘₯.coscos

Answer

Recall that coscossin2π‘₯=π‘₯βˆ’π‘₯, which we can use to simplify the numerator and denominator: 1βˆ’2π‘₯=1βˆ’ο€Ίπ‘₯βˆ’π‘₯=1βˆ’π‘₯+π‘₯=2π‘₯,coscossincossinsin where we used the Pythagorean identity 1βˆ’π‘₯=π‘₯cossin. In the denominator, now with the identity 1βˆ’π‘₯=π‘₯sincos, 1+2π‘₯=1+ο€Ίπ‘₯βˆ’π‘₯=1+βˆ’π‘₯=2π‘₯.coscossincossincos

Therefore, 1βˆ’2π‘₯1+2π‘₯=2π‘₯2π‘₯=π‘₯.coscossincostan

Let us take a note of the two identities that we derived from coscossin2πœƒ=πœƒβˆ’πœƒοŠ¨οŠ¨ in the previous example.

Identity: Identities Derived from the Double-Angle Identity cos 2πœƒ = cos2 πœƒ - sin2 πœƒ

For any real number πœƒ, we have 1+2πœƒ=2πœƒ,1βˆ’2πœƒ=2πœƒ.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 sinsinοŠͺπ‘₯βˆ’π‘₯.

Answer

We notice that we can factor out a sinπ‘₯: sinsinsinsinοŠͺπ‘₯βˆ’π‘₯=π‘₯ο€Ί1βˆ’π‘₯. The Pythagorean identity cossinπ‘₯+π‘₯=1 gives 1βˆ’π‘₯=π‘₯sincos. Substituting this in leads to sinsinsincossincosοŠͺπ‘₯βˆ’π‘₯=π‘₯π‘₯=(π‘₯π‘₯). Using the double-angle identity sinsincos2π‘₯=2π‘₯π‘₯, we find that sincossinπ‘₯π‘₯=2π‘₯2. Substituting in gives sinsinsinsinοŠͺπ‘₯βˆ’π‘₯=ο€½2π‘₯2=14(2π‘₯).

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 sincos2𝐡22𝐡 given cos𝐡=45, where 3πœ‹2<𝐡<2πœ‹.

Answer

In order to apply the double-angle formulas, we need to know the value of sin𝐡. The Pythagorean identity sincos𝐡+𝐡=1 can be solved to give sincos𝐡=1βˆ’π΅=1βˆ’ο€Ό45=925.

So, sin𝐡 is either 35 or βˆ’35. The information about the location of 𝐡 tells us it is in the 4th quadrant, where the sine is negative. Therefore, sin𝐡=βˆ’35.

Using the double-angle identities sinsincos2𝐡=2𝐡𝐡 and coscossin2𝐡=π΅βˆ’π΅οŠ¨οŠ¨, we find that sincossincoscossin2𝐡22𝐡=2𝐡𝐡2ο€Ίπ΅βˆ’π΅ο†=2ο€»βˆ’ο‡ο€»ο‡2ο€»βˆ’ο‡=βˆ’=βˆ’127.οŠͺοŠͺ

Of course an alternative is to observe that we are being asked to evaluate tan2𝐡2 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 2πœƒ=1+2πœƒcoscos that we derived from the double-angle identity coscossin2πœƒ=πœƒβˆ’πœƒοŠ¨οŠ¨ and substituting in 2πœƒ=πœ‘ and πœƒ=πœ‘2, we get 2πœ‘2=1+πœ‘πœ‘2=Β±ο„ž1+πœ‘2.coscoscoscos

Starting from the other identity derived from the double-angle identity coscossin2πœƒ=πœƒβˆ’πœƒοŠ¨οŠ¨ (i.e., 2πœƒ=1βˆ’2πœƒsincos), we find that 2πœ‘2=1βˆ’πœ‘πœ‘2=Β±ο„ž1βˆ’πœ‘2.sincossincos

Finally, to find tanπœ‘2, we use the expressions we just established for cosπœ‘2 and sinπœ‘2 by writing tansincoscoscoscoscosπœ‘2==Β±ο„ž1βˆ’πœ‘2β‹…οΒ±ο„Ÿ21+πœ‘ο=Β±ο„Ÿ1βˆ’πœ‘1+πœ‘.

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 β€œΒ±β€ 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 πœ‘=140∘, πœ‘=220∘, πœ‘=500∘, and πœ‘=580οŠͺ∘. While we have coscoscoscosπœ‘=πœ‘=πœ‘=πœ‘οŠ§οŠ¨οŠ©οŠͺ, we see that coscoscoscosπœ‘2=πœ‘2=βˆ’πœ‘2=βˆ’πœ‘2οŠͺ and sinsinsinsinπœ‘2=πœ‘2=βˆ’πœ‘2=βˆ’πœ‘2οŠͺ, 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 tansincosπœ‘2=, we multiply the numerator and denominator of the fraction on the right-hand side by 2πœ‘2cos, which gives tansincoscosπœ‘2=22.

The numerator can be rewritten as sinπœ‘ using the double-angle identities 2πœƒπœƒ=2πœƒsincossin (with 2πœƒ=πœ‘ and πœƒ=πœ‘2) and the denominator as 1+πœ‘cos using the identity 1+2πœƒ=2πœƒcoscos (with 2πœƒ=πœ‘ and πœƒ=πœ‘2) derived from the double-angle identity coscossin2πœƒ=πœƒβˆ’πœƒοŠ¨οŠ¨.

Hence, we have tansincosπœ‘2=πœ‘1+πœ‘.

Starting again from tansincosπœ‘2=, we now multiply the numerator and denominator of the fraction on the right-hand side by 2πœ‘2sin. We get tansinsincosπœ‘2=22.

Using here the identity 1βˆ’2πœƒ=2πœƒcossin, we can rewrite 2πœ‘2sin as 1βˆ’πœ‘cos and, as before, 2πœ‘2πœ‘2sincos as sinπœ‘. Hence, we find that tancossinπœ‘2=1βˆ’πœ‘πœ‘.

It is worth noting that rearranging tancoscosπœ‘2=Β±ο„Ÿ1βˆ’πœ‘1+πœ‘ by multiplying the right-hand side of the equation by either ο„Ÿ1+πœ‘1+πœ‘coscos or ο„Ÿ1βˆ’πœ‘1βˆ’πœ‘coscos leads to tansincosπœ‘2=Β±|πœ‘|1+πœ‘ and tancossinπœ‘2=Β±1βˆ’πœ‘|πœ‘|. Studying the respective signs of tanπœ‘2 and sinπœ‘ as shown in the following table, we conclude that tanπœ‘2 and sinπœ‘ always have the same sign. This allows us to establish the identities tansincosπœ‘2=πœ‘1+πœ‘ and tancossinπœ‘2=1βˆ’πœ‘πœ‘.

πœ‘ (∘)(0–90)(90–180)(180–270)(270–360)(360–450)(450–540)(540–630)(630–720)
cosπœ‘+βˆ’βˆ’++βˆ’βˆ’+
sinπœ‘++βˆ’βˆ’++βˆ’βˆ’
tanπœ‘+βˆ’+βˆ’+βˆ’+βˆ’
πœ‘2 (∘)(0–45)(45–90)(90–135)(135–180)(180–225)(225–270)(270–315)(315–360)
cosπœ‘2++βˆ’βˆ’βˆ’βˆ’++
sinπœ‘2++++βˆ’βˆ’βˆ’βˆ’
tanπœ‘2++βˆ’βˆ’++βˆ’βˆ’

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

Identity: Half-Angle Identities

For any real number πœƒ, we have coscossincosπœƒ2=Β±ο„ž1+πœƒ2,πœƒ2=Β±ο„ž1βˆ’πœƒ2.

For πœƒβ‰ 180+𝑛360(πœƒβ‰ πœ‹+2π‘›πœ‹)∘∘, where 𝑛 is an integer, we have tancoscostansincosπœƒ2=Β±ο„ž1βˆ’πœƒ1+πœƒ,πœƒ2=πœƒ1+πœƒ.

For πœƒβ‰ (0,180)+𝑛360(πœƒβ‰ (0,πœ‹)+2π‘›πœ‹)∘∘, we have tancossinπœƒ2=1βˆ’πœƒπœƒ.

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 cosπœƒ2 given cosπœƒ=1517, where 0<πœƒ<90∘∘, without using a calculator.

Answer

Recall the half-angle identity coscosπœƒ2=Β±ο„ž1+πœƒ2.

We are told that 0<πœƒ<90∘∘. Hence, 0<πœƒ2<45∘∘ and cosπœƒ2>0. This means that we have coscosπœƒ2=ο„ž1+πœƒ2 since βˆ’ο„ž1+πœƒ2≀0.cos

Let us work out the value of cosπœƒ2 by substituting the value of cosπœƒ into the above equation: cosπœƒ2=ο„‘ο„£ο„ 1+2=ο„ž322Γ—17=ο„ž1617=4√17.

We can rationalize this fraction by multiplying it by √17√17, which gives cosπœƒ2=4√1717.

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 tanπœ‹8.

Answer

The angle πœ‹8 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 πœ‹4 radians. Using the half-angle identity tansincosπœƒ2=πœƒ1+πœƒ, we find that tansincosπœ‹8=1+.οŽ„οŠͺοŽ„οŠͺ

Since sincosπœ‹4=πœ‹4=1√2, we have tanπœ‹8=1+=1√2+1.√√

Multiplying the right-hand side by √2βˆ’1√2βˆ’1 gives tanπœ‹8=√2βˆ’12βˆ’1=√2βˆ’1.√2βˆ’1 or βˆ’1+√2 is the exact value of tanπœ‹8.

We would of course find the same result using tancossinπœƒ2=1βˆ’πœƒπœƒ.

We could also use tancoscosπœƒ2=Β±ο„ž1βˆ’πœƒ1+πœƒ, which gives tancoscosπœ‹8=ο„‘ο„£ο„£ο„£ο„ 1βˆ’1+οŽ„οŠͺοŽ„οŠͺ since we know that tanπœ‹8 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 πœƒ, we have coscossinsincossin2πœƒ=πœƒβˆ’πœƒ,2πœƒ=2πœƒπœƒ, and for πœƒβ‰ (45,135,225,315)+𝑛360ο€Όπœƒβ‰ ο€Όπœ‹4,3πœ‹4,5πœ‹4,7πœ‹4+2π‘›πœ‹οˆβˆ˜βˆ˜βˆ˜βˆ˜βˆ˜, we have tantantan2πœƒ=2πœƒ1βˆ’πœƒ.
  • From the above identity coscossin2πœƒ=πœƒβˆ’πœƒοŠ¨οŠ¨, we can derive the following two identities for any real number πœƒ: 1+2πœƒ=2πœƒ,1βˆ’2πœƒ=2πœƒ.coscoscossin
  • The half-angle identities state that, for any real number πœƒ, we have coscossincosπœƒ2=Β±ο„ž1+πœƒ2,πœƒ2=Β±ο„ž1βˆ’πœƒ2. For πœƒβ‰ 180+𝑛360(πœƒβ‰ πœ‹+2π‘›πœ‹)∘∘, we have tancoscostansincosπœƒ2=Β±ο„ž1βˆ’πœƒ1+πœƒ,πœƒ2=πœƒ1+πœƒ, and for πœƒβ‰ (0,180)+𝑛360(πœƒβ‰ (0,πœ‹)+2π‘›πœ‹)∘∘, we have tancossinπœƒ2=1βˆ’πœƒπœƒ.

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