Lesson Objectives
- Learn how to work with the sum & difference identities for sine
- Learn how to work with the sum & difference identities for tangent
What are the Sum & Difference Identities for Sine and Tangent
In the last lesson, we learned about the sum & difference identities for cosine. Here, we will introduce the sum and difference identities for sine and tangent. Note: We will derive these formulas in the practice test section. A full step-by-step is provided in the practice test solution video.
Example #1: Find the exact value of each. $$\text{sin}\hspace{.1em}165°$$ Let's use our sum identity for sine: $$\text{sin}\hspace{.1em}165°$$ $$=\text{sin}(135° + 30°)$$ $$=\text{sin}135° \text{cos}30° + \text{cos}135° \text{sin}30°$$ $$=\frac{\sqrt{2}}{2}\cdot \frac{\sqrt{3}}{2}+ \frac{-\sqrt{2}}{2}\cdot \frac{1}{2}$$ $$=\frac{\sqrt{6}}{4}- \frac{\sqrt{2}}{4}$$ $$=\frac{\sqrt{6}- \sqrt{2}}{4}$$ Example #2: Find the exact value of each. $$\text{tan}\frac{π}{12}$$ Since it is easier to work with degrees, we will first convert this from radians into degrees: $$\frac{π}{12}\cdot \frac{180 °}{π}=15°$$ We can obtain 15° by subtracting 45° - 30°. We will now convert back to radians. $$45°=\frac{π}{180 °}=\frac{π}{4}$$ $$30°=\frac{π}{180 °}=\frac{π}{6}$$ For reference, let's obtain the tangent of each: $$\text{tan}\frac{π}{4}=1$$ $$\text{tan}\frac{π}{6}=\frac{\sqrt{3}}{3}$$ Now, we return to our problem. Let's use our difference identity for tangent: $$\text{tan}\frac{π}{12}$$ $$=\text{tan}\left(\frac{π}{4}- \frac{π}{6}\right)$$ $$=\frac{1 - \large{\frac{\sqrt{3}}{3}}}{1 + \large{\frac{\sqrt{3}}{3}}}$$ Simplify: $$=\frac{1 - \large{\frac{\sqrt{3}}{3}}}{1 + \large{\frac{\sqrt{3}}{3}}}\cdot \frac{1 - \large{\frac{\sqrt{3}}{3}}}{1 - \large{\frac{\sqrt{3}}{3}}}$$ $$=\frac{1 - \large{\frac{2\sqrt{3}}{3}}+ \large{\frac{1}{3}}}{1 - \large{\frac{1}{3}}}$$ $$=\frac{\Large{\frac{4 - 2 \sqrt{3}}{3}}}{\Large{\frac{2}{3}}}$$ $$=\require{cancel}\frac{4 - 2\sqrt{3}}{\cancel{3}}\cdot \frac{\cancel{3}}{2}$$ $$=\frac{\cancel{2}(2 - \sqrt{3})}{\cancel{2}}$$ $$=2 - \sqrt{3}$$ We will also see problems that ask us to simplify. Let's look at an example.
Example #3: Simplify. $$\text{sin}\hspace{.1em}3 θ \hspace{.1em}\text{cos}({-}2θ) - \text{cos}\hspace{.1em}3θ \hspace{.1em}\text{sin}(-2 θ)$$ To simplify, we use our difference identity for sine: $$\text{sin}(3θ - (-2θ))=\text{sin}\hspace{.1em}5 θ$$
Example #4: Find sin(θ + β), tan(θ + β), and the quadrant of A + B. $$\text{sin}\hspace{.1em}θ=\frac{4}{5}$$ θ is in quadrant II. $$\text{cos}\hspace{.1em}β=-\frac{5}{13}$$ β is in quadrant III.
To find the sine of (θ + β), we need to obtain the sine of β and the cosine of θ: $$\text{cos}^2 β + \text{sin}^2 β=1$$ $$\left(-\frac{5}{13}\right)^2 + \text{sin}^2 β=1$$ $$\frac{25}{169}+ \text{sin}^2 β=1$$ $$\text{sin}^2 β=\frac{169}{169}- \frac{25}{169}$$ $$\text{sin}^2 β=\frac{144}{169}$$ $$\text{sin}\hspace{.1em}β=\pm \sqrt{\frac{144}{169}}$$ Since β is in quadrant III, we will use the negative square root. $$\text{sin}\hspace{.1em}β=-\frac{12}{13}$$ $$\text{sin}^2 θ + \text{cos}^2 θ=1$$ $$\left(\frac{4}{5}\right)^2 + \text{cos}^2 θ=1$$ $$\frac{16}{25}+ \text{cos}^2 θ=1$$ $$\text{cos}^2 θ=\frac{25}{25}- \frac{16}{25}$$ $$\text{cos}^2 θ=\frac{9}{25}$$ $$\text{cos}\hspace{.1em}θ=\pm \sqrt{\frac{9}{25}}$$ Since θ is in quadrant II, we will use the negative square root. $$\text{cos}\hspace{.1em}θ=- \frac{3}{5}$$ Now, we can use our sum identity for sine: $$\text{sin}(θ + β)$$ $$=\frac{4}{5}\cdot -\frac{5}{13}+ \left(- \frac{3}{5}\cdot -\frac{12}{13}\right)$$ $$=-\frac{20}{65}+ \frac{36}{65}$$ $$=\frac{16}{65}$$ To find tan (θ + β), we will first find tan (θ) and the tan (β): $$\tan \hspace{.1em}θ=\frac{\text{sin}\hspace{.1em}θ}{\text{cos}\hspace{.1em}θ}$$ $$\tan \hspace{.1em}θ=\frac{4}{\cancel{5}}\cdot -\frac{\cancel{5}}{3}$$ $$=-\frac{4}{3}$$ $$\tan \hspace{.1em}β=\frac{\text{sin}\hspace{.1em}β}{\text{cos}\hspace{.1em}β}$$ $$\tan \hspace{.1em}β=-\frac{12}{\cancel{13}}\cdot -\frac{\cancel{13}}{5}$$ $$=\frac{12}{5}$$ Now we can use our tangent sum identity: $$\text{tan}(θ + β)$$ $$=\large{\frac{-\frac{4}{3}+ \frac{12}{5}}{1 - \left(-\frac{4}{3}\cdot \frac{12}{5}\right)}}$$ $$=\large{\frac{\frac{-20 + 36}{15}}{1 + \frac{48}{15}}}$$ $$=\large{\frac{\frac{16}{15}}{\frac{15}{15}+ \frac{48}{15}}}$$ $$=\large{\frac{\frac{16}{15}}{\frac{63}{15}}}$$ $$=\frac{16}{\cancel{15}}\cdot \frac{\cancel{15}}{63}$$ $$=\frac{16}{63}$$ In what quadrant is θ + β?
Since sin (θ + β) and tan (θ + β) are both positive, θ + β must be in quadrant I since it is the only quadrant where sine and tangent are positive.
Example #5: Verify the identity. $$\text{sin}\left(\frac{π}{6}+ β\right) + \text{cos}\left(\frac{π}{3}+ β\right)=\text{cos}\hspace{.1em}β$$ Since the right-hand side is easier, let's flip the sides for formatting: $$\text{cos}\hspace{.1em}β=\text{sin}\left(\frac{π}{6}+ β\right) + \text{cos}\left(\frac{π}{3}+ β\right)$$ Use the sum identities for sine and cosine to simplify the right-hand side: $$=\text{sin}\frac{π}{6}\cdot \text{cos}\hspace{.1em}β + \text{cos}\frac{π}{6}\cdot \text{sin}\hspace{.1em}β + \text{cos}\frac{π}{3}\cdot \text{cos}\hspace{.1em}β- \text{sin}\frac{π}{3}\cdot \text{sin}\hspace{.1em}β$$ $$=\frac{1}{2}\cdot \text{cos}\hspace{.1em}β+ \frac{\sqrt{3}}{2}\cdot \text{sin}\hspace{.1em}β+ \frac{1}{2}\cdot \text{cos}\hspace{.1em}β- \frac{\sqrt{3}}{2}\cdot \text{sin}\hspace{.1em}β$$ $$=\frac{1}{2}\cdot \text{cos}\hspace{.1em}β+ \frac{1}{2}\cdot \text{cos}\hspace{.1em}β$$ $$=\text{cos}\hspace{.1em}β✓$$
Sum Identity for Sine:
$$\text{sin}(A + B)=\text{sin}\hspace{.1em}\text{A}\hspace{.1em}\text{cos}\hspace{.1em}\text{B}+ \text{cos}\hspace{.1em}\text{A}\hspace{.1em}\text{sin}\hspace{.1em}\text{B}$$Difference Identity for Sine:
$$\text{sin}(A - B)=\text{sin}\hspace{.1em}\text{A}\hspace{.1em}\text{cos}\hspace{.1em}\text{B}- \text{cos}\hspace{.1em}\text{A}\hspace{.1em}\text{sin}\hspace{.1em}\text{B}$$Sum Identity for Tangent:
$$\text{tan}(A + B)=\frac{\text{tan}\hspace{.1em}\text{A}+ \text{tan}\hspace{.1em}\text{B}}{1 - \text{tan}\hspace{.1em}\text{A}\hspace{.1em}\text{tan}\hspace{.1em}\text{B}}$$Difference Identity for Tangent:
$$\text{tan}(A - B)=\frac{\text{tan}\hspace{.1em}\text{A}- \text{tan}\hspace{.1em}\text{B}}{1 + \text{tan}\hspace{.1em}\text{A}\hspace{.1em}\text{tan}\hspace{.1em}\text{B}}$$ In some cases, we will be asked to use these identities in order to find the exact value for sine or tangent. Let's look at a couple of examples.Example #1: Find the exact value of each. $$\text{sin}\hspace{.1em}165°$$ Let's use our sum identity for sine: $$\text{sin}\hspace{.1em}165°$$ $$=\text{sin}(135° + 30°)$$ $$=\text{sin}135° \text{cos}30° + \text{cos}135° \text{sin}30°$$ $$=\frac{\sqrt{2}}{2}\cdot \frac{\sqrt{3}}{2}+ \frac{-\sqrt{2}}{2}\cdot \frac{1}{2}$$ $$=\frac{\sqrt{6}}{4}- \frac{\sqrt{2}}{4}$$ $$=\frac{\sqrt{6}- \sqrt{2}}{4}$$ Example #2: Find the exact value of each. $$\text{tan}\frac{π}{12}$$ Since it is easier to work with degrees, we will first convert this from radians into degrees: $$\frac{π}{12}\cdot \frac{180 °}{π}=15°$$ We can obtain 15° by subtracting 45° - 30°. We will now convert back to radians. $$45°=\frac{π}{180 °}=\frac{π}{4}$$ $$30°=\frac{π}{180 °}=\frac{π}{6}$$ For reference, let's obtain the tangent of each: $$\text{tan}\frac{π}{4}=1$$ $$\text{tan}\frac{π}{6}=\frac{\sqrt{3}}{3}$$ Now, we return to our problem. Let's use our difference identity for tangent: $$\text{tan}\frac{π}{12}$$ $$=\text{tan}\left(\frac{π}{4}- \frac{π}{6}\right)$$ $$=\frac{1 - \large{\frac{\sqrt{3}}{3}}}{1 + \large{\frac{\sqrt{3}}{3}}}$$ Simplify: $$=\frac{1 - \large{\frac{\sqrt{3}}{3}}}{1 + \large{\frac{\sqrt{3}}{3}}}\cdot \frac{1 - \large{\frac{\sqrt{3}}{3}}}{1 - \large{\frac{\sqrt{3}}{3}}}$$ $$=\frac{1 - \large{\frac{2\sqrt{3}}{3}}+ \large{\frac{1}{3}}}{1 - \large{\frac{1}{3}}}$$ $$=\frac{\Large{\frac{4 - 2 \sqrt{3}}{3}}}{\Large{\frac{2}{3}}}$$ $$=\require{cancel}\frac{4 - 2\sqrt{3}}{\cancel{3}}\cdot \frac{\cancel{3}}{2}$$ $$=\frac{\cancel{2}(2 - \sqrt{3})}{\cancel{2}}$$ $$=2 - \sqrt{3}$$ We will also see problems that ask us to simplify. Let's look at an example.
Example #3: Simplify. $$\text{sin}\hspace{.1em}3 θ \hspace{.1em}\text{cos}({-}2θ) - \text{cos}\hspace{.1em}3θ \hspace{.1em}\text{sin}(-2 θ)$$ To simplify, we use our difference identity for sine: $$\text{sin}(3θ - (-2θ))=\text{sin}\hspace{.1em}5 θ$$
Finding Function Values and the Quadrant of A + B
Another type of problem involves finding trigonometric function values using the sum and difference identities. Let's look at some examples.Example #4: Find sin(θ + β), tan(θ + β), and the quadrant of A + B. $$\text{sin}\hspace{.1em}θ=\frac{4}{5}$$ θ is in quadrant II. $$\text{cos}\hspace{.1em}β=-\frac{5}{13}$$ β is in quadrant III.
To find the sine of (θ + β), we need to obtain the sine of β and the cosine of θ: $$\text{cos}^2 β + \text{sin}^2 β=1$$ $$\left(-\frac{5}{13}\right)^2 + \text{sin}^2 β=1$$ $$\frac{25}{169}+ \text{sin}^2 β=1$$ $$\text{sin}^2 β=\frac{169}{169}- \frac{25}{169}$$ $$\text{sin}^2 β=\frac{144}{169}$$ $$\text{sin}\hspace{.1em}β=\pm \sqrt{\frac{144}{169}}$$ Since β is in quadrant III, we will use the negative square root. $$\text{sin}\hspace{.1em}β=-\frac{12}{13}$$ $$\text{sin}^2 θ + \text{cos}^2 θ=1$$ $$\left(\frac{4}{5}\right)^2 + \text{cos}^2 θ=1$$ $$\frac{16}{25}+ \text{cos}^2 θ=1$$ $$\text{cos}^2 θ=\frac{25}{25}- \frac{16}{25}$$ $$\text{cos}^2 θ=\frac{9}{25}$$ $$\text{cos}\hspace{.1em}θ=\pm \sqrt{\frac{9}{25}}$$ Since θ is in quadrant II, we will use the negative square root. $$\text{cos}\hspace{.1em}θ=- \frac{3}{5}$$ Now, we can use our sum identity for sine: $$\text{sin}(θ + β)$$ $$=\frac{4}{5}\cdot -\frac{5}{13}+ \left(- \frac{3}{5}\cdot -\frac{12}{13}\right)$$ $$=-\frac{20}{65}+ \frac{36}{65}$$ $$=\frac{16}{65}$$ To find tan (θ + β), we will first find tan (θ) and the tan (β): $$\tan \hspace{.1em}θ=\frac{\text{sin}\hspace{.1em}θ}{\text{cos}\hspace{.1em}θ}$$ $$\tan \hspace{.1em}θ=\frac{4}{\cancel{5}}\cdot -\frac{\cancel{5}}{3}$$ $$=-\frac{4}{3}$$ $$\tan \hspace{.1em}β=\frac{\text{sin}\hspace{.1em}β}{\text{cos}\hspace{.1em}β}$$ $$\tan \hspace{.1em}β=-\frac{12}{\cancel{13}}\cdot -\frac{\cancel{13}}{5}$$ $$=\frac{12}{5}$$ Now we can use our tangent sum identity: $$\text{tan}(θ + β)$$ $$=\large{\frac{-\frac{4}{3}+ \frac{12}{5}}{1 - \left(-\frac{4}{3}\cdot \frac{12}{5}\right)}}$$ $$=\large{\frac{\frac{-20 + 36}{15}}{1 + \frac{48}{15}}}$$ $$=\large{\frac{\frac{16}{15}}{\frac{15}{15}+ \frac{48}{15}}}$$ $$=\large{\frac{\frac{16}{15}}{\frac{63}{15}}}$$ $$=\frac{16}{\cancel{15}}\cdot \frac{\cancel{15}}{63}$$ $$=\frac{16}{63}$$ In what quadrant is θ + β?
Since sin (θ + β) and tan (θ + β) are both positive, θ + β must be in quadrant I since it is the only quadrant where sine and tangent are positive.
Verifying an Identity Using the Sum and Difference Identities for Sine
We previously learned how to verify an identity. Let's look at an example that uses the sum and difference identities for sine and cosine.Example #5: Verify the identity. $$\text{sin}\left(\frac{π}{6}+ β\right) + \text{cos}\left(\frac{π}{3}+ β\right)=\text{cos}\hspace{.1em}β$$ Since the right-hand side is easier, let's flip the sides for formatting: $$\text{cos}\hspace{.1em}β=\text{sin}\left(\frac{π}{6}+ β\right) + \text{cos}\left(\frac{π}{3}+ β\right)$$ Use the sum identities for sine and cosine to simplify the right-hand side: $$=\text{sin}\frac{π}{6}\cdot \text{cos}\hspace{.1em}β + \text{cos}\frac{π}{6}\cdot \text{sin}\hspace{.1em}β + \text{cos}\frac{π}{3}\cdot \text{cos}\hspace{.1em}β- \text{sin}\frac{π}{3}\cdot \text{sin}\hspace{.1em}β$$ $$=\frac{1}{2}\cdot \text{cos}\hspace{.1em}β+ \frac{\sqrt{3}}{2}\cdot \text{sin}\hspace{.1em}β+ \frac{1}{2}\cdot \text{cos}\hspace{.1em}β- \frac{\sqrt{3}}{2}\cdot \text{sin}\hspace{.1em}β$$ $$=\frac{1}{2}\cdot \text{cos}\hspace{.1em}β+ \frac{1}{2}\cdot \text{cos}\hspace{.1em}β$$ $$=\text{cos}\hspace{.1em}β✓$$
Skills Check:
Example #1
Find the exact value of each. $$\text{tan}\hspace{.1em}195°$$
Please choose the best answer.
A
$$\frac{-\sqrt{6}- \sqrt{2}}{4}$$
B
$$2 + \sqrt{3}$$
C
$$-2 - \sqrt{3}$$
D
$$2 - \sqrt{3}$$
E
$$1 + \sqrt{2}$$
Example #2
Find the exact value. $$\text{sin}\frac{π}{18}\text{cos}\frac{13π}{9}+ \text{cos}\frac{π}{18}\text{sin}\frac{13π}{9}$$
Please choose the best answer.
A
$$0$$
B
$$\frac{\sqrt{3}}{3}$$
C
$$\frac{\sqrt{3}}{2}$$
D
$$1$$
E
$$-1$$
Example #3
Complete the identity. $$\text{tan}(45° + θ)=$$
Please choose the best answer.
A
$$\frac{1 - \text{tan}\hspace{.1em}θ}{1 + \text{tan}\hspace{.1em}θ}$$
B
$$\frac{1 + \text{tan}\hspace{.1em}θ}{1 - \text{tan}\hspace{.1em}θ}$$
C
$$\text{sin}θ - 1$$
D
$$\text{cos}θ - 1$$
E
$$\frac{1 + \text{sin}\hspace{.1em}θ}{2}$$
Congrats, Your Score is 100%
Better Luck Next Time, Your Score is %
Try again?
Ready for more?
Watch the Step by Step Video Lesson Take the Practice Test
