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Lesson Explainer: Proving Trigonometric Identities Mathematics

In this explainer, we will learn how to prove trigonometric statements using known trigonometric identities.

An identity is similar to an equation in that it shows us that two expressions are the same. However, an identity is true for any value, the two expressions must be identical. In this case, we say that they are equivalent expressions and we use the symbol β€œβ‰‘β€ to show that the expressions are equivalent.

For example, we recall that the double angle formula for the sine tells us that sinsincos2πœƒβ‰‘2πœƒπœƒ.

This means that for any value of πœƒ, these two expressions are the same. This allows us to substitute this expression for sin2πœƒ into any equation or expression and we know that we will obtain an equivalent expression. We can use this to rewrite expressions and prove identities.

For instance, consider the expression sinsincos2πœƒπœƒπœƒ. We can substitute sinsincos2πœƒβ‰‘2πœƒπœƒ into this expression to obtain the equivalent expression sinsincossincossincos2πœƒπœƒπœƒβ‰‘2πœƒπœƒπœƒπœƒ.

We can then simplify to get 2πœƒπœƒπœƒπœƒβ‰‘2πœƒπœƒ.sincossincossincos

This must be equivalent to our original expression, so we have shown that the following two expressions are equivalent: sinsincossincos2πœƒπœƒπœƒβ‰‘2πœƒπœƒ.

We could verify this by substituting in any value of πœƒ to check that both expressions yield the same output. Alternatively, we could use graphing software to plot the graphs of 2πœƒπœƒsincos and sinsincos2πœƒπœƒπœƒ, and we would see that the two graphs are the same.

This highlights another key difference between equivalences and equations. We are not trying to solve an equivalence, so we are not applying the same operations to both sides of the equivalence. In fact, if we have two equivalent expressions and we do this, then we may end up with 0≑0, which is a true but not very useful result.

Instead, when working with equivalences, we often try to rewrite one or both sides of the equivalence into a useful form for the application we have in mind.

We will see an example of this in our first example, where we will prove the equivalence of two trigonometric expressions using the double-angle and Pythagorean identities.

Example 1: Proving a Trigonometric Statement Involving a Double Angle

Determine the value of π‘Ž such that sinsinsinοŠͺοŠ¨πœƒβˆ’πœƒβ‰‘π‘Ž2πœƒ.

Answer

We want to determine the value of constant π‘Ž that makes both sides of the equivalence identical, for any value of πœƒ. To do this, we can start by noting that the right-hand side of the equivalence includes a double angle, so we first recall that sinsincos2πœƒβ‰‘2πœƒπœƒ.

This allows us to rewrite the right-hand side of the equivalence to get π‘Ž2πœƒβ‰‘π‘Ž(2πœƒπœƒ)β‰‘π‘Žο€Ί4πœƒπœƒο†β‰‘4π‘Žπœƒπœƒ.sinsincossincossincos

We want to write this in terms of the sine function, so we will apply the Pythagorean identity, which tells us cossinοŠ¨οŠ¨πœƒβ‰‘1βˆ’πœƒ.

Substituting this into the equivalence and simplifying, we have 4π‘Žπœƒπœƒβ‰‘4π‘Žπœƒο€Ί1βˆ’πœƒο†β‰‘4π‘Žπœƒβˆ’4π‘Žπœƒ.sincossinsinsinsinοŠͺ

We want this to be identical to the left-hand side of the given equivalence: 4π‘Žπœƒβˆ’4π‘Žπœƒβ‰‘πœƒβˆ’πœƒ.sinsinsinsinοŠͺοŠͺ

Since both sides of the equivalence are in the same form, we can do this by equating coefficients. We have 4π‘Ž=1π‘Ž=14.

In our previous example, we wanted to prove that two trigonometric expressions were equivalent. We did this by rewriting the expression on the right into the same form as the expression on the left. This allowed us to find the value of the unknown constant.

There are many different ways of approaching a problem of this form. Often, it is a good idea to look at both expressions and consider which identities can be applied and how this will affect the expression. It is also worth noting that we can start with either expression, and it is always a good idea to consider both to determine which expression is easier to start rewriting.

In our next example, we will prove the equivalence of two trigonometric expressions by applying multiple trigonometric identities.

Example 2: Proving a Trigonometric Statement by Applying Multiple Trigonometric Identities

Determine the value of π‘Ž such that sintansintancos2πœƒβˆ’πœƒ2πœƒβ‰‘π‘Žπœƒ2πœƒοŠ¨.

Answer

We want to determine the value of constant π‘Ž that makes both sides of the equivalence identical for any value of πœƒ. To do this, we can try and rewrite each side of the equivalence to be in the same form. One way we could start is to notice that the left-hand side of the equivalence has two terms with a factor of sin2πœƒ. Taking out this shared factor gives us sintansinsintan2πœƒβˆ’πœƒ2πœƒβ‰‘2πœƒο€Ή1βˆ’πœƒο….

We now apply the double-angle formula for the sine function, which tells us sinsincos2πœƒβ‰‘2πœƒπœƒ.

Substituting this into the expression gives us sintansinsincostan2πœƒβˆ’πœƒ2πœƒβ‰‘2πœƒπœƒο€Ή1βˆ’πœƒο….

We can simplify this further by recalling that tangent is the quotient of the sine and cosine function. We have 2πœƒπœƒο€Ή1βˆ’πœƒο…β‰‘2πœƒπœƒο€Ώ1βˆ’πœƒπœƒο‹β‰‘2πœƒπœƒβˆ’2πœƒπœƒβ‰‘2πœƒπœƒβˆ’2πœƒπœƒ.sincostansincossincossincossincossincossintan

We can rewrite the right-hand side of the given equivalence in this form by applying the double-angle identity. We need to be careful here and use the version involving both sine and cosine since this will allow simplification. We recall that coscossin2πœƒβ‰‘πœƒβˆ’πœƒ.

Substituting this into the right-hand side of the given equivalence gives us π‘Žπœƒ2πœƒβ‰‘π‘Žπœƒο€Ίπœƒβˆ’πœƒο†.tancostancossin

We can distribute over the parentheses and simplify using the definition of the tangent function: π‘Žπœƒο€Ίπœƒβˆ’πœƒο†β‰‘π‘Žπœƒπœƒβˆ’π‘Žπœƒπœƒβ‰‘π‘Žο€½πœƒπœƒο‰πœƒβˆ’π‘Žπœƒπœƒβ‰‘π‘Žπœƒπœƒβˆ’π‘Žπœƒπœƒ.tancossintancostansinsincoscostansinsincossintan

We can compare this to our equivalent expression for the left-hand side of the given equivalence: 2πœƒπœƒβˆ’2πœƒπœƒsincossintan.

We see that they are identical expressions when π‘Ž=2.

In our next example, we will prove the equivalence of two trigonometric expressions by applying an angle sum identity.

Example 3: Proving a Trigonometric Statement Using an Angle Sum Identity

Determine the values of π‘Ž and 𝑏 such that tantantanο€»πœƒ+πœ‹4ο‡β‰‘π‘Ž+π‘πœƒ1βˆ’πœƒ.

Answer

We want to determine the values of constants π‘Ž and 𝑏 that make both sides of the equivalence identical, for any value of πœƒ. To do this, we can try and rewrite each side of the equivalence to be in the same form.

We note that the left-hand side of the given equivalence includes a sum of angles, so we can rewrite this using the angle sum identity: tantantantantan(π‘₯+𝑦)≑π‘₯+𝑦1βˆ’π‘₯𝑦.

This allows us to rewrite the left-hand side of the given equivalence as tantantantantanο€»πœƒ+πœ‹4ο‡β‰‘πœƒ+1βˆ’πœƒο€»ο‡.οŽ„οŠͺοŽ„οŠͺ

We then recall that tanο€»πœ‹4=1, so tantantantantantantantanπœƒ+1βˆ’πœƒο€»ο‡β‰‘πœƒ+11βˆ’πœƒβ‰‘1+πœƒ1βˆ’πœƒ.οŽ„οŠͺοŽ„οŠͺ

We can rewrite the right-hand side of the given equivalence in this form by adding the terms together by rewriting them to have the same denominator. We have π‘Ž+π‘πœƒ1βˆ’πœƒβ‰‘π‘Žο€½1βˆ’πœƒ1βˆ’πœƒο‰+π‘πœƒ1βˆ’πœƒβ‰‘π‘Žβˆ’π‘Žπœƒ1βˆ’πœƒ+π‘πœƒ1βˆ’πœƒβ‰‘π‘Žβˆ’π‘Žπœƒ+π‘πœƒ1βˆ’πœƒβ‰‘π‘Ž+(π‘βˆ’π‘Ž)πœƒ1βˆ’πœƒ.tantantantantantantantantantantantantantantan

We can now see that each of our expressions for either side of the equivalence is in the same form. In particular, for the expressions to be equivalent, their numerators must be equivalent since the denominators are identical.

We can find the values of π‘Ž and 𝑏 that cause the numerators to be identical by comparing coefficients. First, we want the constant term in the numerator to be equal to 1, so we have π‘Ž=1. Next, the coefficient of tanπœƒ in the numerator should be 1, so we have π‘βˆ’π‘Ž=1.

We know that π‘Ž=1, so π‘βˆ’1=1𝑏=2.

Hence, the expressions are equivalent when π‘Ž=1 and 𝑏=2.

In our next example, we will prove the equivalence of two trigonometric expressions by applying double-angle identities.

Example 4: Proving a Trigonometric Statement Using Double-Angle Identities

Determine the values of π‘Ž and 𝑏 such that tancoscosοŠ¨πœƒβ‰‘1+π‘Ž2πœƒ1+𝑏2πœƒ.

Answer

We want to determine the values of constants π‘Ž and 𝑏 that make both sides of the equivalence identical, for any value of πœƒ. To do this, we can try and rewrite each side of the equivalence to be in the same form.

We can start by noting that the left-hand side of the equivalence can be written as a fraction using the fact that tansincosπœƒβ‰‘πœƒπœƒ. Applying this yields tansincosοŠ¨οŠ¨οŠ¨πœƒβ‰‘πœƒπœƒ.

We can then rewrite the right-hand side of the given equivalence in this form by applying double-angle identities for cosine.

We have cossincoscos2πœƒβ‰‘1βˆ’2πœƒ,2πœƒβ‰‘2πœƒβˆ’1.

We want the numerator in terms of the sine function and the denominator in terms of the cosine function, so we rewrite the right-hand side of the given equivalence as follows: 1+π‘Ž2πœƒ1+𝑏2πœƒβ‰‘1+π‘Žο€Ί1βˆ’2πœƒο†1+𝑏(2πœƒβˆ’1)≑1+π‘Žβˆ’2π‘Žπœƒ1+2π‘πœƒβˆ’π‘.coscossincossincos

We want this expression to be identical to sincosοŠ¨οŠ¨πœƒπœƒ. We note that there are no constant terms in the numerator or denominator, so we set the constant terms in the numerator and denominator to be equal to zero. We have π‘Ž=βˆ’1 and 𝑏=1.

Substituting these values into the right-hand side of the identity yields 1+(βˆ’1)βˆ’2(βˆ’1)πœƒ1+2(1)πœƒβˆ’1≑0+2πœƒ0+2(1)πœƒβ‰‘πœƒπœƒ.sincossincossincos

Hence, the expressions are equivalent when π‘Ž=βˆ’1 and 𝑏=1.

In our final example, we will prove the equivalence of two trigonometric expressions by applying an angle difference identity.

Example 5: Proving a Trigonometric Statement Using an Angle Difference Identity

Determine the values of π‘Ž and 𝑏 such that 2ο€»πœ‹3βˆ’2πœƒο‡β‰‘π‘Ž2πœƒ+𝑏2πœƒsinsincos.

Answer

We want to determine the values of the constants π‘Ž and 𝑏 that make both sides of the equivalence identical, for any value of πœƒ. To do this, we can try and rewrite each side of the equivalence to be in the same form.

We can note that the left-hand side of the given equivalence includes a difference of angles in the argument of the sine function. We can rewrite this by applying the angle difference identity for sine, which tells us sinsincoscossin(π‘₯βˆ’π‘¦)=π‘₯π‘¦βˆ’π‘₯𝑦.

We have 2ο€»πœ‹3βˆ’2πœƒο‡β‰‘2ο€»ο€»πœ‹32πœƒβˆ’ο€»πœ‹32πœƒο‡.sinsincoscossin

We recall that sinο€»πœ‹3=√32 and cosο€»πœ‹3=12, so we have 2ο€»ο€»πœ‹32πœƒβˆ’ο€»πœ‹32πœƒο‡β‰‘2ο€Ώβˆš322πœƒβˆ’122πœƒο‹β‰‘βˆš32πœƒβˆ’2πœƒ.sincoscossincossincossin

We can see that this is identical to the right-hand side of the given equivalence when π‘Ž=βˆ’1 and 𝑏=√3.

Let’s finish by recapping some of the important points from this explainer.

Key Points

  • We say that two expressions are equivalent if they are equal for any value of the variable.
  • We can rewrite trigonometric expressions into equivalent expressions using any trigonometric identity. These include the double-angle and half-angle identities, the Pythagorean identity, and the angle sum and difference identities.
  • When attempting to show that two trigonometric expressions are equivalent, we want to rewrite both expressions to be in the same form. It is a good idea to consider which identities can be applied to each expression to find the easiest way of writing both expressions in the same form.

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