Which of the following is equivalent to sin ⁡ ( α + β ) \sin\left(\alpha+\beta\right) sin ( α + β ) ?

sin ⁡ α cos ⁡ β + cos ⁡ α sin ⁡ β \sin\alpha\cos\beta+\cos\alpha\sin\beta sin α cos β + cos α sin β

sin ⁡ α cos ⁡ α + sin ⁡ β cos ⁡ β \sin\alpha\cos\alpha+\sin\beta\cos\beta sin α cos α + sin β cos β

sin ⁡ α cos ⁡ α − sin ⁡ β cos ⁡ β \sin\alpha\cos\alpha-\sin\beta\cos\beta sin α cos α − sin β cos β

sin ⁡ α cos ⁡ β − cos ⁡ α sin ⁡ β \sin\alpha\cos\beta-\cos\alpha\sin\beta sin α cos β − cos α sin β

Which of the following is equivalent to cos ⁡ ( α − β ) \cos\left(\alpha-\beta\right) cos ( α − β ) ?

cos ⁡ α cos ⁡ β + sin ⁡ α sin ⁡ β \cos\alpha\cos\beta+\sin\alpha\sin\beta cos α cos β + sin α sin β

cos ⁡ α cos ⁡ β − sin ⁡ α sin ⁡ B \cos\alpha\cos\beta-\sin\alpha\sin B cos α cos β − sin α sin B

cos ⁡ α sin ⁡ β + sin ⁡ α cos ⁡ β \cos\alpha\sin\beta+\sin\alpha\cos\beta cos α sin β + sin α cos β

cos ⁡ α sin ⁡ β − sin ⁡ α cos ⁡ β \cos\alpha\sin\beta-\sin\alpha\cos\beta cos α sin β − sin α cos β

Which of the following is equivalent to tan ⁡ ( α + β ) \tan\left(\alpha+\beta\right) tan ( α + β ) ?

tan ⁡ α + tan ⁡ β 1 − tan ⁡ α tan ⁡ β \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta} 1 − tan α tan β tan α + tan β

tan ⁡ α − tan ⁡ β 1 − tan ⁡ α tan ⁡ β \frac{\tan\alpha-\tan\beta}{1-\tan\alpha\tan\beta} 1 − tan α tan β tan α − tan β

tan ⁡ α + tan ⁡ β 1 + tan ⁡ α tan ⁡ β \frac{\tan\alpha+\tan\beta}{1+\tan\alpha\tan\beta} 1 + tan α tan β tan α + tan β

tan ⁡ α − tan ⁡ β 1 + tan ⁡ α tan ⁡ β \frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta} 1 + tan α tan β tan α − tan β

Find the exact value of sin ⁡ 105 ° \sin105\degree sin 1 0 5 °

6 + 2 4 \frac{\sqrt{6}+\sqrt{2}}{4} 4 6 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> + 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

6 − 2 4 \frac{\sqrt{6}-\sqrt{2}}{4} 4 6 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> − 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

2 − 6 4 \frac{\sqrt{2}-\sqrt{6}}{4} 4 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> − 6 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

3 + 1 2 \frac{\sqrt{3}+1}{2} 2 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> + 1

cos ⁡ 105 ° \cos105\degree cos 1 0 5 °

2 − 6 4 \frac{\sqrt{2}-\sqrt{6}}{4} 4 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> − 6 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

2 + 6 4 \frac{\sqrt{2}+\sqrt{6}}{4} 4 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> + 6 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

6 − 2 4 \frac{\sqrt{6}-\sqrt{2}}{4} 4 6 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> − 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

1 − 3 2 \frac{1-\sqrt{3}}{2} 2 1 − 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

tan ⁡ 105 ° \tan105\degree tan 1 0 5 °

− 2 − 3 -2-\sqrt{3} − 2 − 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

2 − 3 2-\sqrt{3} 2 − 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

2 + 3 2+\sqrt{3} 2 + 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

− 2 + 3 -2+\sqrt{3} − 2 + 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

If sin ⁡ α = 5 13 \sin\alpha=\frac{5}{13} sin α = 1 3 5 and cos ⁡ β = 4 5 \cos\beta=\frac{4}{5} cos β = 5 4 (both in QI), then find sin ⁡ ( α − β ) \sin\left(\alpha-\beta\right) sin ( α − β )

− 16 65 -\frac{16}{65} − 6 5 1 6

− 65 16 -\frac{65}{16} − 1 6 6 5

If sin ⁡ α = 5 13 \sin\alpha=\frac{5}{13} sin α = 1 3 5 and cos ⁡ β = 4 5 \cos\beta=\frac{4}{5} cos β = 5 4 (both in QI), find cos ⁡ ( α + β ) \cos\left(\alpha+\beta\right) cos ( α + β )

− 33 65 -\frac{33}{65} − 6 5 3 3

− 63 65 -\frac{63}{65} − 6 5 6 3

If sin ⁡ α = 5 13 \sin\alpha=\frac{5}{13} sin α = 1 3 5 and cos ⁡ β = 4 5 \cos\beta=\frac{4}{5} cos β = 5 4 (both in QI), find tan ⁡ ( α − β ) \tan\left(\alpha-\beta\right) tan ( α − β )

− 16 63 -\frac{16}{63} − 6 3 1 6

− 63 16 -\frac{63}{16} − 1 6 6 3

Trig - Sum and Difference

Quiz

•

Mathematics

•

9th - 12th Grade

•

Practice Problem

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Medium

•

CCSS

HSF.TF.C.9, 8.G.B.7

Standards-aligned

Scott Mocabee

Used 263+ times

FREE Resource

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CCSS.HSF.TF.C.9

10.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

If the 2 legs of a right triangle are 7 in. and 24 in., how long is the hypotenuse?

31 in.

17 in.

$\sqrt{527}\$ in.

25 in.

Tags

CCSS.8.G.B.7

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Trig - Sum and Difference

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Which of the following is equivalent to sin⁡(α+β)\sin\left(\alpha+\beta\right)sin(α+β) ?

Which of the following is equivalent to cos⁡(α−β)\cos\left(\alpha-\beta\right)cos(α−β) ?

Which of the following is equivalent to tan⁡(α+β)\tan\left(\alpha+\beta\right)tan(α+β) ?

Find the exact value of sin⁡105°\sin105\degreesin105°

cos⁡105°\cos105\degreecos105°

tan⁡105°\tan105\degreetan105°

If sin⁡α=513\sin\alpha=\frac{5}{13}sinα=135​ and cos⁡β=45\cos\beta=\frac{4}{5}cosβ=54​ (both in QI), then find sin⁡(α−β)\sin\left(\alpha-\beta\right)sin(α−β)

If sin⁡α=513\sin\alpha=\frac{5}{13}sinα=135​ and cos⁡β=45\cos\beta=\frac{4}{5}cosβ=54​ (both in QI), find cos⁡(α+β)\cos\left(\alpha+\beta\right)cos(α+β)

If sin⁡α=513\sin\alpha=\frac{5}{13}sinα=135​ and cos⁡β=45\cos\beta=\frac{4}{5}cosβ=54​ (both in QI), find tan⁡(α−β)\tan\left(\alpha-\beta\right)tan(α−β)

If the 2 legs of a right triangle are 7 in. and 24 in., how long is the hypotenuse?

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Which of the following is equivalent to $\sin\left(\alpha+\beta\right)$ ?

Which of the following is equivalent to $\cos\left(\alpha-\beta\right)$ ?

Which of the following is equivalent to $\tan\left(\alpha+\beta\right)$ ?

Find the exact value of $\sin105\degree$

$\cos105\degree$

$\tan105\degree$

If $\sin\alpha=\frac{5}{13}$ and $\cos\beta=\frac{4}{5}$ (both in QI), then find $\sin\left(\alpha-\beta\right)$

If $\sin\alpha=\frac{5}{13}$ and $\cos\beta=\frac{4}{5}$ (both in QI), find $\cos\left(\alpha+\beta\right)$

If $\sin\alpha=\frac{5}{13}$ and $\cos\beta=\frac{4}{5}$ (both in QI), find $\tan\left(\alpha-\beta\right)$