Trigonometry Notes on the Sum and Difference Identities. [ ] [ ] [ ] [ ]

<strong>Trigonometry</strong><strong>Notes</strong> on the Sum and Difference Identities.The Derivations of the Sum and Difference Identities:Let’s start with cos(A-B). Look at the following Graph.Notice that the distance between P and R is the same as the distance between Q and S. Thus,d P, R = d Q,S2cos (A) cos(A) sin (A) si( ) ( )( cos(A) − cos(B) ) + ( sin(A) − sin(B) ) = ( cos(A −B) − 1) + ( sin(A −B)−0)2 2 2 22 22( cos(A) − cos(B) ) + ( sin(A) − sin(B) ) = ( cos(A −B) −1) + ( sin(A −B)−0)cos (A B) cos(A B)2 222− 2 cos( B) + cos (B) + − 2 n( A) sin(B) + sin(B) = − − 2 − + 1+2 2 2 222⎡ ⎤ ⎡ ⎤ ⎡ ⎤sin22(A − B)⎣cos (A) + sin (A) ⎦+ ⎣cos (B) + sin (B) ⎦−2cos(A) cos(B) − 2sin(A)sin(B) = ⎣cos (A −B) + sin (A −B) ⎦− 2cos(A −B)+ 11+ 1−2cos(A)cos(B) − 2sin(A) sin(B)= 1− 2cos(A−B) + 1−2cos(A)cos(B) − 2sin(A)sin( B) =−2cos(A − B)cos( A)cos(B) + sin(A) sin(B) = cos(A − B)Then, use the negative number identities to find:cos A + B = cos(A − -B) = cos(A)cos(-B) + sin(A)sin(-B) = cos(A)cos(B) − sin(A)sin(B)( )For sine, we use the cofunction identities.⎛π ⎞ ⎛π ⎞ ⎛⎛π ⎞ ⎞sin ( A + B)= cos ⎜ − (A + B) ⎟= cos⎜ −A − B⎟= cos⎜⎜ −A ⎟−B⎟⎝2 ⎠ ⎝2 ⎠ ⎝⎝2⎠ ⎠⎛π⎞ ⎛π⎞= cos⎜ − A⎟cos(B) + sin ⎜ − A⎟sin(B) = sin(A)cos(B) + cos(A)sin(B)⎝2 ⎠ ⎝2⎠Thus, using the negative number identities,sin A − B = sin(A + -B) = sin(A)cos(-B) + cos(A)sin(-B) = sin(A)cos(B) − cos(A)sin(B)( )Finally, we get to tangent.sin ( A + B)sin(A)cos(B) + cos(A)sin(B) sin(A)cos(B) + cos(A)sin(B) cos(A)cos(B)tan ( A + B)= = =cos A + B cos(A)cos(B) −sin(A)sin(B) cos(A)cos(B) −sin(A)sin(B) cos(A)cos(B)( )[ ] [ ][ ] [ ]tan(A) + tan(B)=1 − tan(A) tan(B)And,tan(A) + tan(-B) tan(A) −tan(B)tan(A − B) = tan(A + -B) = =1− tan(A) tan(-B) 1+tan(A) tan(B)SCC:Rickman <strong>Notes</strong> on the Summation and Difference Identities. Page #1 of 4

The Trigonometric Sum and Difference Identities:Thus, in summary,The Trigonometric Sum and Difference Identitiescos( A + B)= cos(A)cos(B) −sin(A)sin(B)cos( A − B)= cos(A)cos(B) + sin(A)sin(B)sin( A + B)= sin(A)cos(B) + cos(A)sin(B)sin( A − B)= sin(A)cos(B) −cos(A)sin(B)tan(A) + tan(B)tan ( A + B)=1 − tan(A) tan(B)tan(A) − tan(B)tan ( A − B)=1 + tan(A) tan(B)2Example #1: Given cos( A ) = , 3 π < A < 2 π, ( ) 13πsin B = - , and π < B < , evaluate sine, cosine and tangent exactly for a) A+B and3 2 52b) A-B. Also, what quadrants are c) A+B and d) A-B in.►First we need to find ( )2( )2( )2cos B , sin ( A ) , tan ( A ), and ( )cos A + sin A = 1⎛2⎞ 2⎜ ⎟ + sin ( A)= 1⎝3⎠4 2 9+ sin ( A)=9 92 5sin ( A)=95sin ( A)=±35( ) =3tan B .thsin A - ,since Aisin the 4 Quad.( ) ( )2 2cos B + sin B = 122 ⎛ 1 ⎞( )cos B + ⎜- ⎟ = 1⎝ 5 ⎠2 1 25cos ( B)+ =25 252 24cos ( B)=252 6cos( B)=±52 6( ) =5rdcos B - ,since Bisin the3 Quad.a)- 5 3 5-1 5 1 6tan ( A)= = -tan ( B)= = =23 2-2 6 5 2 6 12⎛2⎞⎛ 2 6 ⎞ ⎛ 5 ⎞ ⎛ 1⎞4 6 5 4 6+5cos( A + B)= cos(A)cos(B) − sin(A)sin(B) = ⎜ ⎟ - − - ⎜- ⎟= - − = -⎝3 ⎠⎜ 5 ⎟ ⎜ 3 ⎟⎝ ⎠ ⎝ ⎠⎝ 5 ⎠ 15 15 15⎛ 5 ⎞⎛ 2 6 ⎞ ⎛2 ⎞⎛ 1 ⎞ 2 30 2 2 30 − 2sin( A + B)= sin(A)cos(B) + cos(A)sin(B) = ⎜- - -3 ⎟⎜ + = − =5 ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ⎝3 ⎠⎝ 5 ⎠ 15 15 15⎛ 5⎞ ⎛ 6 ⎞ ⎡⎛ 5⎞ ⎛ 6 ⎞⎤-- + ( 24)tan(A) + tan(B) ⎜ 2 ⎟+ ⎜ 12 ⎟⎢ ⎜ 2 ⎟ ⎜ 12 ⎟⎥⎢⎝ ⎠ ⎝ ⎠⎥ -12 5 + 2 6tan ( A + B)==⎝ ⎠ ⎝ ⎠=⎣ ⎦=1− tan(A) tan(B) ⎛ 5⎞⎛ 6 ⎞ ⎡ ⎛ 5⎞⎛ 6 ⎞⎤ 24 + 301−⎜- 1−- ( 24)2 ⎟⎜ ⎢ ⎥12 ⎟ ⎜ 2 ⎟⎜ 12 ⎟⎝ ⎠⎝ ⎠ ⎣⎢⎝ ⎠⎝ ⎠⎦⎥( )( )( 24 + 30 )( 24 − 30 )-12 5 + 2 6 24 − 30 -288 5 + 12 150 + 48 6 −2 180 -288 5 + 60 6 + 48 6 −12 5= = =576 − 30546108 6 −300 5 54 6 −150 5= =546 273(continued on next page)SCC:Rickman <strong>Notes</strong> on the Summation and Difference Identities. Page #2 of 4

Example #1(cont):b)⎛2⎞⎛ 2 6 ⎞ ⎛ 5 ⎞⎛ 1⎞4 6 5 5−4 6cos( A − B)= cos(A)cos(B) + sin(A)sin(B) = ⎜ ⎟ - + - ⎜- ⎟= - + =⎝3 ⎠⎜ 5 ⎟ ⎜ 3 ⎟⎝ ⎠ ⎝ ⎠⎝ 5 ⎠ 15 15 15⎛ 5 ⎞⎛ 2 6 ⎞ ⎛2 ⎞⎛ 1 ⎞ 2 30 2 2 30 + 2sin( A − B)= sin(A)cos(B) − cos(A)sin(B) = ⎜- - -3 ⎟⎜ − = + =5 ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ⎝3 ⎠⎝ 5 ⎠ 15 15 15⎛ 5⎞ ⎛ 6 ⎞ ⎡⎛ 5⎞ ⎛ 6 ⎞⎤-- − ( 24)tan(A) − tan(B) ⎜ 2 ⎟− ⎜ 12 ⎟⎢ ⎜ 2 ⎟ ⎜ 12 ⎟⎥⎢⎝ ⎠ ⎝ ⎠⎥ -12 5 − 2 6tan ( A − B)==⎝ ⎠ ⎝ ⎠=⎣ ⎦=1+ tan(A) tan(B) ⎛ 5⎞⎛ 6 ⎞ ⎡ ⎛ 5⎞⎛ 6 ⎞⎤ 24 − 301+ ⎜- 1 - ( 24)2 ⎟⎜ 12 ⎟ ⎢ +⎜ 2 ⎟⎜ 12 ⎟⎥⎝ ⎠⎝ ⎠ ⎣⎢⎝ ⎠⎝ ⎠⎦⎥(-12 5 − 2 6 )( 24 + 30 ) -288 5 −12 150 −48 6 −2 180 -288 5−60 6−48 6−12 5= = =24 − 30 24 + 30576 − 30546( )( )-108 6 − 300 5 54 6 + 150 5= = -546 2734 6+515cos A B5−4 6150.5041 0c) Since cos( A + B)= - ≈ − 0.8023 < 0 and ( )d) Because ( − ) = ≈− < and ( )□2 30−2sin A + B = ≈ 0.5970 > 0 , A+B would be in the 2 nd quadrant.152 30+2sin A − B = ≈ 0.8636 > 0 , A-B would also be in the 2 nd quadrant.15Note, that we could have also used the results from sine and cosine to get tangent in the above problem. I just wanted to usethe identities.Example #2: Evaluate tan ( 15° ) exactly.►⎛ 3 ⎞31 1−( 3)tan ( 45° ) − tan ( 30° −)⎜ 3 ⎟3 3 ( 3 3)( 3 3)9 6 3 3 12 6 3tan ( 15 ) tan ( 45 30 )3− − − − + −° = °− ° = = =⎝ ⎠= = = =1+ tan( 45° ) tan( 30° ) ⎛ 3⎞ ⎛ 3⎞ 3+ 3()( ) ( 3+ 3)( 3−3)9−3 61+ 1 ⎜ 1+33 ⎟ ⎜ 3 ⎟⎝ ⎠ ⎝ ⎠= 2−3□⎛7π⎞Example #3: Evaluate sin ⎜ ⎟⎝ 12 ⎠ exactly.►⎛7π⎞ ⎛3π 4π⎞ ⎛π π⎞ ⎛π⎞ ⎛π⎞ ⎛π⎞ ⎛π⎞ ⎛ 2 ⎞⎛ 3⎞ ⎛ 2 ⎞⎛1⎞sin ⎜ ⎟= sin ⎜ + ⎟= sin ⎜ + ⎟= sin ⎜ ⎟cos⎜ ⎟+ cos⎜ ⎟sin⎜ ⎟= +⎜ ⎟⎝12 ⎠ ⎝12 12 ⎠ ⎝4 3 ⎠ ⎝4 ⎠ ⎝3 ⎠ ⎝4 ⎠ ⎝ 3 ⎠⎜ 2 ⎟⎜ 2 ⎟ ⎜ 2 ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠⎝2⎠6+2=4□SCC:Rickman <strong>Notes</strong> on the Summation and Difference Identities. Page #3 of 4

Proving Identities:⎛ π ⎞ cos( θ ) + sin ( θ)Example #4: Prove tan ⎜θ+ ⎟=.⎝ 4⎠cos( θ) −sin( θ)►⎛π⎞tan ( θ ) + tansin( θ⎜ ⎟)⎛ π ⎞ 4 tan ( ) 1 cos( )1 sin costan⎝ ⎠ θ + +θ θ + θ⎜θ+ ⎟= = = =sin4 1 tan( θ)⎝ ⎠ ⎛π ⎞ −( )( θ)1 cos θ −sinθ1−tan θ tan−cos( θ⎜ ⎟)⎝ 4 ⎠cos=cos□Solving Equations:Example #5: Find all real solutions for►Either set of work would be fine.□( θ ) + sin ( θ)( θ) −sin( θ)( θ) ( θ)+ = -1 for in [ 0 ,360 )cos sin 32 2( θ) ( θ)cos sin 3+ = -12 2⎛1⎞⎛ 3⎞cos( θ ) ⎜ ⎟+ sin ( θ )= -1⎝2⎠ ⎜ 2 ⎟⎝ ⎠cos( θ ) cos( 60° ) + sin ( θ ) sin ( 60° ) = -1cos( θ− 60° ) = -1θ− 60°= 180°θ = 240°orθ ° ° .( ) ( )( ) ( )( θ) ( θ)cos sin 3+ = -12 2⎛1⎞⎛ 3⎞cos( θ ) ⎜ ⎟+ sin ( θ )= -1⎝2⎠ ⎜ 2 ⎟⎝ ⎠cos( θ ) sin ( 30° ) + sin ( θ ) cos( 30° ) = -1sin ( θ+ 30° ) = -1θ+ 30°= 270°θ= 240°Applications:Example #6: It can be shown that the formula t = 2V 0⎡g ⎣sin ( θ+φ) −tan ( θ) cos( θ+φ)⎤⎦ will give the timeof flight, t, for an object thrown up an inclined ground at an initial speed of V0. g is the acceleration dueto gravity, 32 ft m2= 9.8 2. θ is the incline of the ground from horizontal, and φ is the angle that the objects s2V0sin ( φ)is thrown from the ground. See figure to the right. a) Show that the formula simplifies to t = .gcos( θ)mb) Find the time of flight for θ= 45 °φ= , 30 ° ,and V0 = 20 . Round answer to the tenths place.s►a)b)2V0220mt = ⎡sin ( θ+φ) −tan ( θ) cos( θ+φ)⎤( ) sin30 ( °s)g ⎣⎦t =( 9.8 m) ( )2V0sin( θ)t = ⎡sin g( θ) cos( φ ) + cos( θ) sin ( φ)− ⎡cos cos( )( θ) cos( φ) −sin ( θ) sinθ( φ)⎤⎤2 cos 45°s⎣⎣⎦⎦40s 1( 2 ) 40st = =222V0sin ( θ) sin( φ)9.8t = ⎡sin ( θ) cos( φ ) + cos( θ) sin ( φ) −sin ( θ) cos( φ ) + ⎤( 2 )9.8 2□g ⎢⎣cos( θ)2 22V0 sin ( θ) sin( φ)2V0sin ( θ)t = ⎡cos g ( ) sin ( ) ⎤ sincos( ) g ( ) ⎡cosθ( ) ⎤⎢θ φ + = φ θ +⎣ ⎥⎦ ⎢⎣ cos( θ)⎥⎦2 22V0 cos ( θ+ ) sin ( θ)2V01t = sin ( φ ) ⎡ ⎤ = singcos( ) g( φ) ⎡ ⎤⎣⎢ θ ⎥⎦⎣cos( θ)⎦2V0sin ( φ)t =gcos θ( )⎥⎦t ≈ 2.9sφθSCC:Rickman <strong>Notes</strong> on the Summation and Difference Identities. Page #4 of 4