Solutions for Sum-to-Product and Product-to-Sum Formulas
Solutions to Try Its
1. 1 2 ( cos 6 θ + cos 2 θ ) \frac{1}{2}\left(\cos 6\theta +\cos 2\theta \right)\\ 2 1 ( cos 6 θ + cos 2 θ ) 1 2 ( sin 2 x + sin 2 y ) \frac{1}{2}\left(\sin 2x+\sin 2y\right)\\ 2 1 ( sin 2 x + sin 2 y ) − 2 − 3 4 \frac{-2-\sqrt{3}}{4}\\ 4 − 2 − 3 2 sin ( 2 θ ) cos ( θ ) 2\sin \left(2\theta \right)\cos \left(\theta \right) 2 sin ( 2 θ ) cos ( θ ) tan θ cot θ − cos 2 θ = ( sin θ cos θ ) ( cos θ sin θ ) − cos 2 θ = 1 − cos 2 θ = sin 2 θ \begin{array}{l}\tan \theta \cot \theta -{\cos }^{2}\theta =\left(\frac{\sin \theta }{\cos \theta }\right)\left(\frac{\cos \theta }{\sin \theta }\right)-{\cos }^{2}\theta \hfill \\ =1-{\cos }^{2}\theta \hfill \\ ={\sin }^{2}\theta \hfill \end{array}\\ tan θ cot θ − cos 2 θ = ( c o s θ s i n θ ) ( s i n θ c o s θ ) − cos 2 θ = 1 − cos 2 θ = sin 2 θ Solutions to Odd-Numbered Exercises
1. Substitute α \alpha α β \beta β sin ( 3 x ) + sin x cos x = 1 \frac{\sin \left(3x\right)+\sin x}{\cos x}=1 c o s x s i n ( 3 x ) + s i n x = 1 2 sin ( 2 x ) cos x cos x = 1 \frac{2\sin \left(2x\right)\cos x}{\cos x}=1\\ c o s x 2 s i n ( 2 x ) c o s x = 1 8 ( cos ( 5 x ) − cos ( 27 x ) ) 8\left(\cos \left(5x\right)-\cos \left(27x\right)\right) 8 ( cos ( 5 x ) − cos ( 27 x ) ) sin ( 2 x ) + sin ( 8 x ) \sin \left(2x\right)+\sin \left(8x\right) sin ( 2 x ) + sin ( 8 x ) 1 2 ( cos ( 6 x ) − cos ( 4 x ) ) \frac{1}{2}\left(\cos \left(6x\right)-\cos \left(4x\right)\right) 2 1 ( cos ( 6 x ) − cos ( 4 x ) ) 2 cos ( 5 t ) cos t 2\cos \left(5t\right)\cos t 2 cos ( 5 t ) cos t 2 cos ( 7 x ) 2\cos \left(7x\right) 2 cos ( 7 x ) 2 cos ( 6 x ) cos ( 3 x ) 2\cos \left(6x\right)\cos \left(3x\right) 2 cos ( 6 x ) cos ( 3 x ) 1 4 ( 1 + 3 ) \frac{1}{4}\left(1+\sqrt{3}\right) 4 1 ( 1 + 3 ) 1 4 ( 3 − 2 ) \frac{1}{4}\left(\sqrt{3}-2\right) 4 1 ( 3 − 2 ) 1 4 ( 3 − 1 ) \frac{1}{4}\left(\sqrt{3}-1\right) 4 1 ( 3 − 1 ) cos ( 8 0 ∘ ) − cos ( 12 0 ∘ ) \cos \left(80^\circ \right)-\cos \left(120^\circ \right) cos ( 8 0 ∘ ) − cos ( 12 0 ∘ ) 1 2 ( sin ( 22 1 ∘ ) + sin ( 20 5 ∘ ) ) \frac{1}{2}\left(\sin \left(221^\circ \right)+\sin \left(205^\circ \right)\right) 2 1 ( sin ( 22 1 ∘ ) + sin ( 20 5 ∘ ) ) 2 cos ( 3 1 ∘ ) \sqrt{2}\cos \left(31^\circ \right) 2 cos ( 3 1 ∘ ) 2 cos ( 66. 5 ∘ ) sin ( 34. 5 ∘ ) 2\cos \left(66.5^\circ \right)\sin \left(34.5^\circ \right) 2 cos ( 66. 5 ∘ ) sin ( 34. 5 ∘ ) 2 sin ( − 1. 5 ∘ ) cos ( 0. 5 ∘ ) 2\sin \left(-1.5^\circ \right)\cos \left(0.5^\circ \right) 2 sin ( − 1. 5 ∘ ) cos ( 0. 5 ∘ ) 2 sin ( 7 x ) − 2 sin x = 2 sin ( 4 x + 3 x ) − 2 sin ( 4 x − 3 x ) = 2 ( sin ( 4 x ) cos ( 3 x ) + sin ( 3 x ) cos ( 4 x ) ) − 2 ( sin ( 4 x ) cos ( 3 x ) − sin ( 3 x ) cos ( 4 x ) ) = 2 sin ( 4 x ) cos ( 3 x ) + 2 sin ( 3 x ) cos ( 4 x ) − 2 sin ( 4 x ) cos ( 3 x ) + 2 sin ( 3 x ) cos ( 4 x ) = 4 sin ( 3 x ) cos ( 4 x ) {2}\sin \left({7x}\right){-2}\sin{ x}={ 2}\sin \left({4x}+{ 3x }\right)-{ 2 }\sin\left({4x } - { 3x }\right)=\\ {2}\left(\sin\left({ 4x }\right)\cos\left({ 3x }\right)+\sin\left({ 3x }\right)\cos\left({ 4x }\right)\right)-{ 2 }\left(\sin\left({ 4x }\right)\cos\left({ 3x }\right)-\sin \left({ 3x }\right)\cos\left({ 4x }\right)\right)=\\{2}\sin\left({ 4x }\right)\cos\left({ 3x }\right)+{2}\sin\left({ 3x }\right)\cos\left({ 4x }\right)-{ 2 }\sin\left({ 4x }\right)\cos\left({ 3x }\right)+{ 2 }\sin\left({ 3x }\right)\cos\left({ 4x }\right)=\\{ 4 }\sin\left({ 3x }\right)\cos\left({ 4x }\right)\\ 2 sin ( 7 x ) − 2 sin x = 2 sin ( 4 x + 3 x ) − 2 sin ( 4 x − 3 x ) = 2 ( sin ( 4 x ) cos ( 3 x ) + sin ( 3 x ) cos ( 4 x ) ) − 2 ( sin ( 4 x ) cos ( 3 x ) − sin ( 3 x ) cos ( 4 x ) ) = 2 sin ( 4 x ) cos ( 3 x ) + 2 sin ( 3 x ) cos ( 4 x ) − 2 sin ( 4 x ) cos ( 3 x ) + 2 sin ( 3 x ) cos ( 4 x ) = 4 sin ( 3 x ) cos ( 4 x ) sin x + sin ( 3 x ) = 2 sin ( 4 x 2 ) cos ( − 2 x 2 ) = \sin x+\sin \left(3x\right)=2\sin \left(\frac{4x}{2}\right)\cos \left(\frac{-2x}{2}\right)= sin x + sin ( 3 x ) = 2 sin ( 2 4 x ) cos ( 2 − 2 x ) = 2 sin ( 2 x ) cos x = 2 ( 2 sin x cos x ) cos x = 2\sin \left(2x\right)\cos x=2\left(2\sin x\cos x\right)\cos x= 2 sin ( 2 x ) cos x = 2 ( 2 sin x cos x ) cos x = 4 sin x cos 2 x 4\sin x{\cos }^{2}x 4 sin x cos 2 x 2 tan x cos ( 3 x ) = 2 sin x cos ( 3 x ) cos x = 2 ( . 5 ( sin ( 4 x ) − sin ( 2 x ) ) ) cos x 2\tan x\cos \left(3x\right)=\frac{2\sin x\cos \left(3x\right)}{\cos x}=\frac{2\left(.5\left(\sin \left(4x\right)-\sin \left(2x\right)\right)\right)}{\cos x} 2 tan x cos ( 3 x ) = c o s x 2 s i n x c o s ( 3 x ) = c o s x 2 ( .5 ( s i n ( 4 x ) − s i n ( 2 x ) ) ) = 1 cos x ( sin ( 4 x ) − sin ( 2 x ) ) = sec x ( sin ( 4 x ) − sin ( 2 x ) ) =\frac{1}{\cos x}\left(\sin \left(4x\right)-\sin \left(2x\right)\right)=\sec x\left(\sin \left(4x\right)-\sin \left(2x\right)\right) = cos x 1 ( sin ( 4 x ) − sin ( 2 x ) ) = sec x ( sin ( 4 x ) − sin ( 2 x ) ) 2 cos ( 35 ∘ ) cos ( 23 ∘ ) , 1.5081 2\cos \left({35}^{\circ }\right)\cos \left({23}^{\circ }\right),\text{ 1}\text{.5081} 2 cos ( 35 ∘ ) cos ( 23 ∘ ) , 1 .5081 − 2 sin ( 33 ∘ ) sin ( 11 ∘ ) , − 0.2078 -2\sin \left({33}^{\circ }\right)\sin \left({11}^{\circ }\right),\text{ }-0.2078 − 2 sin ( 33 ∘ ) sin ( 11 ∘ ) , − 0.2078 1 2 ( cos ( 99 ∘ ) − cos ( 71 ∘ ) ) , − 0.2410 \frac{1}{2}\left(\cos \left({99}^{\circ }\right)-\cos \left({71}^{\circ }\right)\right),\text{ }-0.2410 2 1 ( cos ( 99 ∘ ) − cos ( 71 ∘ ) ) , − 0.2410 2 cos 3 x 2{\cos }^{3}x 2 cos 3 x tan ( 3 t ) \tan \left(3t\right) tan ( 3 t ) 2 cos ( 2 x ) 2\cos \left(2x\right) 2 cos ( 2 x ) − sin ( 14 x ) -\sin \left(14x\right) − sin ( 14 x ) cos x + cos y \cos x+\cos y cos x + cos y x = α + β x=\alpha +\beta x = α + β y = α − β y=\alpha -\beta y = α − β cos x + cos y \cos x+\cos y cos x + cos y cos ( α + β ) + cos ( α − β ) = cos α cos β − sin α sin β + cos α cos β + sin α sin β = 2 cos α cos β \cos \left(\alpha +\beta \right)+\cos \left(\alpha -\beta \right)=\cos \alpha \cos \beta -\sin \alpha \sin \beta +\cos \alpha \cos \beta +\sin \alpha \sin \beta =2\cos \alpha \cos \beta cos ( α + β ) + cos ( α − β ) = cos α cos β − sin α sin β + cos α cos β + sin α sin β = 2 cos α cos β
Since x = α + β x=\alpha +\beta x = α + β y = α − β y=\alpha -\beta y = α − β α \alpha α β \beta β x and y and substitute in for 2 cos α cos β 2\cos \alpha \cos \beta 2 cos α cos β 2 cos ( x + y 2 ) cos ( x − y 2 ) 2\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right) 2 cos ( 2 x + y ) cos ( 2 x − y ) cos ( 3 x ) + cos x cos ( 3 x ) − cos x = 2 cos ( 2 x ) cos x − 2 sin ( 2 x ) sin x = − cot ( 2 x ) cot x \frac{\cos \left(3x\right)+\cos x}{\cos \left(3x\right)-\cos x}=\frac{2\cos \left(2x\right)\cos x}{-2\sin \left(2x\right)\sin x}=-\cot \left(2x\right)\cot x c o s ( 3 x ) − c o s x c o s ( 3 x ) + c o s x = − 2 s i n ( 2 x ) s i n x 2 c o s ( 2 x ) c o s x = − cot ( 2 x ) cot x cos ( 2 y ) − cos ( 4 y ) sin ( 2 y ) + sin ( 4 y ) = − 2 sin ( 3 y ) sin ( − y ) 2 sin ( 3 y ) cos y = 2 sin ( 3 y ) sin ( y ) 2 sin ( 3 y ) cos y = tan y \begin{array}{l}\frac{\cos \left(2y\right)-\cos \left(4y\right)}{\sin \left(2y\right)+\sin \left(4y\right)}=\frac{-2\sin \left(3y\right)\sin \left(-y\right)}{2\sin \left(3y\right)\cos y}=\\ \frac{2\sin \left(3y\right)\sin \left(y\right)}{2\sin \left(3y\right)\cos y}=\tan y\end{array} s i n ( 2 y ) + s i n ( 4 y ) c o s ( 2 y ) − c o s ( 4 y ) = 2 s i n ( 3 y ) c o s y − 2 s i n ( 3 y ) s i n ( − y ) = 2 s i n ( 3 y ) c o s y 2 s i n ( 3 y ) s i n ( y ) = tan y cos x − cos ( 3 x ) = − 2 sin ( 2 x ) sin ( − x ) = 2 ( 2 sin x cos x ) sin x = 4 sin 2 x cos x \begin{array}{l}\cos x-\cos \left(3x\right)=-2\sin \left(2x\right)\sin \left(-x\right)=\\ 2\left(2\sin x\cos x\right)\sin x=4{\sin }^{2}x\cos x\end{array} cos x − cos ( 3 x ) = − 2 sin ( 2 x ) sin ( − x ) = 2 ( 2 sin x cos x ) sin x = 4 sin 2 x cos x tan ( π 4 − t ) = tan ( π 4 ) − tan t 1 + tan ( π 4 ) tan ( t ) = 1 − tan t 1 + tan t \tan \left(\frac{\pi }{4}-t\right)=\frac{\tan \left(\frac{\pi }{4}\right)-\tan t}{1+\tan \left(\frac{\pi }{4}\right)\tan \left(t\right)}=\frac{1-\tan t}{1+\tan t} tan ( 4 π − t ) = 1 + t a n ( 4 π ) t a n ( t ) t a n ( 4 π ) − t a n t = 1 + t a n t 1 − t a n t Licenses & Attributions CC licensed content, Specific attribution Precalculus. Provided by: OpenStax Authored by: OpenStax College. Located at: https://cnx.org/contents/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1/Preface. License: CC BY: Attribution .
