CHAPTER 7 SSM - Weebly
SECTION 7-1 289CHAPTER 7Section 7–11. To verify an identity means to prove that both sides of an equation are equal when any values of thevariables for which both sides are defined are substituted into that equation.3. Values of the variables must be found for which both sides of the equation are defined, but are notequal. A single set of such values, called a counterexample, suffices to prove that the equation isnot an identity.5. sin θ sec θ sin θ 1cos sin cos Algebra tan θ7. cot u sec u sin u Quotient Identitycos u 1sin usin u cos u 19. sin xsin( x) cos xcos( x)Identities for NegativesQuotient Identitytan tan1 tan cot csc csc 1 csc 1 Reciprocal and Quotient IdentitiesAlgebra –tan x11.Reciprocal Identity1sin sin αReciprocal IdentityAlgebraReciprocal IdentityAlgebra1(cos u sin u)sin ucos usin u sin usin ucos u 1sin uAlgebra cot u 1Quotient Identity13. csc u(cos u sin u) Reciprocal IdentityAlgebra1aba 1Key Algebraic Steps: (a b) bbbb15.cos x sin xcos xsin x –sin x cos xsin x cos xsin x cos x11 –sin xcos x csc x – sec xAlgebraAlgebraReciprocal Identities
290 CHAPTER 7 TRIGONOMETRIC IDENTITIES AND CONDITIONAL EQUATIONS17.sin 2 tsin 2 tcos t cos t cos tcos t1 sin 2 tcos 2 t cos tcos tsin 2 t cos 2 tcos t1 cos t Pythagorean IdentityReciprocal Identitya2a2a2b2a 2 b2b b Key Algebraic Steps:bbbbb1cos xcos x1 sin 2 x sin 2 x cos 2 x sin 2 xcos xcos 2 x1 cos x1 – 2 sin2 x cos2 x sin2 x – 2 sin2 x cos2 x – sin2 x25. (sec t 1)(sec t – 1) sec2 t – 1 tan2 t 1 – 1 tan2 t29.AlgebraReciprocal Identity21. (1 – cos u)(1 cos u) 1 – cos2 u sin2 u cos2 u – cos2 u sin2 u27.Pythagorean IdentityAlgebra sec x23.AlgebraAlgebra sec t19.AlgebraAlgebraPythagorean IdentityAlgebraPythagorean IdentityAlgebraAlgebraPythagorean IdentityAlgebracsc2 x – cot2 x 1 cot2 x – cot2 x 1Pythagorean IdentityAlgebracos x tan xcos xtan x sin xsin xsin xAlgebra cot x sin xcos xsin xsin x cot x sin xcos xsin x1 cot x ·cos x sin x1 cot x cos x cot x sec xKey Algebraic Steps: a bcb a Quotient IdentitiesAlgebraAlgebraAlgebraReciprocal Identitiesbb 11 b a · a cc bc
SECTION 7-1 29131. Plug in x –4.22Left side: ( 4 3) ( 1) 1 1Right side: –4 3 –1The equation is not an identity.35. Plug in x .333. Plug in x –1.Left side: (–1)2 (–1) 1 – 1 0Right side: 3(–1)3 – (–1)2 –3 – 1 –4The equation is not an identity. .4 Left side: sec tan 44 Right side: cot 1437. Plug in x 3 32 11Right side: 1 – cos 1– 322Left side: sin2 1The equation is not an identity.The equation is not an identity.41.39.The two graphs appear to coincide. The equationcos x tan x sin x appears to be an identity.The two graphs do not appear to coincide. Theequation 1 sin x cos x is not an identity.45.43.Simplify the left side:x2 9( x 3)( x 3) x–3x 3x 3The equation is an identity.The two graphs do not appear to coincide. Theequation sin x – csc x –cot x csc x does notappear to be an identity.47. Simplify the left side:x2 4 x 4 ( x 2)( x 2) ( x 2) 2 x 2 The equation is not an identity: the left side mustbe positive, but the right can be negative. Forexample, if we plug in x –4, we get49. Plug in x ?: sin– cos 144411 ?– 122Obviously this is false so the equation is not an identity.?( 4) 2 4( 4) 4 –4 2?4 –2?2 –251. The equation is an identity. We can verify it as follows:cos3 x cos x cos2 xAlgebra cos x( 1 – sin2 x)Pythagorean Identity cos x – cos x sin2 x Algebra
292 CHAPTER 7 TRIGONOMETRIC IDENTITIES AND CONDITIONAL EQUATIONS53.1 (sin x cos x) 2sin x 1 (sin 2 x 2sin x cos x cos 2 x)sin x1 ( 2sin x cos x sin 2 x cos 2 x)sin x1 ( 2sin x cos x 1) sin x2sin x cos x sin xAlgebra AlgebraPythagorean IdentityAlgebra 2 cos xKey Algebraic Steps: (a – b)2 a2 – 2ab b2 –2ab a2 b255.cot 1 csc cos 1sin 1sin sin cossin Reciprocal and Quotient Identities 1 sin sin Algebra1sin cos θ sin θKey Algebraic Steps:57. 59.ba 11aAlgebra a 1aba b aa a11 cos y(1 cos y ) (1 cos y ) 1 cos y(1 cos y ) (1 cos y ) Algebra1 cos 2 yAlgebra(1 cos y ) 2sin 2 y cos 2 y cos 2 yPythagorean Identity(1 cos y ) 2sin 2 yAlgebra(1 cos y ) 2tan2 x – sin2 x sin 2 xcos 2 x sin2 x– sin2 x1Quotient Identity– sin2 x · 1cos 2 x 1 sin2 x 2 1 cos x csc cot tan Reciprocal IdentityPythagorean IdentityAlgebra1sin cos sin cossin Reciprocal and Quotient Identitiescos sin cos sin AlgebraAlgebra sin2 x(sec2 x – 1) sin2 x tan2 x tan2 x sin2 x61.Algebracos sin 1sin cos sin sin cos Algebra
SECTION 7-1 293cos Algebra2cos sin 2 cos 1Pythagorean Identity cos θKey Algebraic Steps:63.1aba baab 1a ab ba ab ba bb2 a 2 sin x cos x Quotient Identityln (tan x) ln ln (sin x) – ln (cos x)65.1cos A1cos Asec A 1 sec A 1 169.Reciprocal Identity 1cos A cos1 A 1 cos AAlgebracos A cos1 A 1 cos A1 cos A1 cos AKey Algebraic Steps:67.AlgebraAlgebra1a1a 1 1 a 1a 1 aa 1a 1 a 1 a1 asin4 w – cos4 w (sin2 w)2 – (cos2 w)2 (sin2 w cos2 w)(sin2 w – cos2 w) 1(sin2 w – cos2 w) sin2 w – cos2 w 1 – cos2 w – cos2 w 1 – 2 cos2 wKey Algebraic Steps: a4 – b4 (a2)2 – (b2)2 (a2 b2)(a2 – b2)sec x –cos x1cos x –1 sin xcos x1 sin xAlgebraAlgebraPythagorean IdentityAlgebraPythagorean IdentityAlgebraReciprocal Identity 1 sin x cos 2 x(1 sin x ) cos xAlgebra sin 2 x cos 2 x sin x cos 2 x(1 sin x) cos xPythagorean Identity sin 2 x sin x(1 sin x) cos xAlgebra (1 sin x) sin x(1 sin x) cos xAlgebra sin xcos xAlgebra tan xQuotient Identity
294 CHAPTER 7 TRIGONOMETRIC IDENTITIES AND CONDITIONAL EQUATIONSKey Algebraic Steps:11 b a2a– (1 b)aa1 bbb2 b(1 b)b a(1 b)a(1 b)a71.cos 2 z 3cos z 2 sin 2 z cos 2 z 3cos z 2Algebrasin 2 z cos 2 z cos 2 zcos 2 z 3cos z 2Pythagorean Identity1 cos 2 z(cos z 1)(cos z 2) (1 cos z )(1 cos z )Algebra (cos z 2)1 cos z2 cos z 1 cos z Key Algebraic Steps:73.a 2 3a 21 a2 AlgebraAlgebra (a 2)2 a(a 1)(a 2) 1 a1 a(1 a )(1 a)cos3 sin 3 (cos sin )(cos 2 cos sin sin 2 ) cos sin cos sin cos2 θ cos θ sin θ sin2 θ 1 cos θ sin θAlgebraAlgebraPythagoreanIdentity 1 sin θ cos θKey Algebraic Steps:3322a b(a b)(a ab b ) a2 ab b2a ba b75. Graph both sides of the equation in the same viewingwindow.sin( x) –1 is not an identity, since the graphscos( x) tan( x) 4do not match. Try x – .Left side: sin 4 cos 4 tan 4 1212 1 1Right side: –1This verifies that the equation is not an identity.AlgebraCommon Error:a 3 b3 a 2 b2a b
SECTION 7-1 29577. Graph both sides of the equation in the sameviewing window.sin x –1 appears to be an identity,cos x tan( x)which we now verify:sin xsin x cos x tan( x)cos x( tan x)Identities for Negativessin xcos x tan xsin x –sin xcos x cosx –AlgebraQuotient Identitysin xsin x –Algebra –1Algebra79. Graph both sides of the equation in the sameviewing window.15cos 2 x sec x is not an identity, sincesin x the graphs do not match. Try x – .4sin x Left side: sin 4 Right side: sec 4 sin 4 cos 2 4 –12 -5 – -152121212–12 – 22This verifies that the equation is not an identity.81. Graph both sides of the equation in the sameviewing window.sin x 5cos 2 x csc x appears to be an identity,sin x15-55which we now verify:-15cos 2 xcos 2 xsin xsin x sin xsin x1 sin 2 xcos 2 x sin xsin xsin 2 x cos 2 xsin x1 sin x csc xAlgebraAlgebraAlgebraPythagorean IdentityReciprocal Identity
296 CHAPTER 7 TRIGONOMETRIC IDENTITIES AND CONDITIONAL EQUATIONS83. Graph both sides of the equation in the sameviewing window.tan x1 appears to be ansin x 2 tan xcos x 2identity, which we now verify:sin xtan xcos x sin xsin x 2 tan xsin x 2 cosx Quotient Identitysin xcos x cosxAlgebrasin xsin x cos x 2 cos x cosxsin xsin x cos x 2sin xsin x sin x(cos x 2) 85.2sin 2 x 3cos x 3sin 2 x AlgebraAlgebra1cos x 2Algebra2(1 cos 2 x ) 3cos x 3 Pythagorean Identitysin 2 x2 2 cos 2 x 3cos x 3Algebrasin 2 x 2 cos 2 x 3cos x 1Algebrasin 2 x2 2 cos x 3cos x 1Pythagorean Identity1 cos 2 x (2 cos 2 x 3cos x 1)Algebra (cos 2 x 1) (2 cos x 1)(cos x 1) (cos x 1)(cos x 1)Algebra2 cos x 1cos x 12 cos x 1 1 cos x Key Algebraic Steps:87. 2a 2 3a 11 a2 tan u sin usec u 1– tan u sin usec u 1 AlgebraAlgebra (2a 2 3a 1)2 (a 1)sin ucos usin ucos u1cos u1cos u sin u sin u– sec u 1sec u 1sin u 1 sin usin u 1 sin u cos1 u 1 sin u cos1 u 1 sin u (2a 1)(a 1) 2a 1 2a 1 1 aa 1 (a 1)(a 1)––sec u 1sec u 1sec u 1sec u 1Quotient IdentityAlgebraAlgebra
SECTION 7-1 297 1 1cos u1cos u sec u 1sec u 1–sec u 1sec u 1 1–sec u 1sec u 1AlgebraReciprocal Identity 0babaKey Algebraic Steps: b b b 1a bb b1aAlgebra 1a 1 b 1a 1 b1a1a 1 1tan cot tan cot tan cot tan cot tan cot 11 tan cot 89.AlgebraAlgebra tan α cot βReciprocal IdentitiesReason91. Statement cos x 2cot2 x 1 1 sin x cos 2 x(A) cot x 1Algebrasin 2 xcos 2 x sin 2 xAlgebrasin 2 x1 (B) cos2 x sin2 x 1sin 2 x 1 2 sin x Algebra csc2 x(C) csc x 93. Since 1 cos 2 x sin 2 x , the equation will be true whennegative. This occurs in Quadrants III, IV.95. Since 1 sin 2 x cos xsin x1sin xsin 2 x –sin x, that is, when sin x iscos 2 x cos x is an identity, this will hold in all quadrants.97. Since 1 sin 2 x sin x1 sin 2 x cos 2 x cos x if cos x is positive, this will hold whensin xsin x tan x, that is, when cos x is positive. This occurs in Quadrants I,2cos xcos xIV.99.a2 u 2 a 2 (a sin x) 2 a 2 a 2 sin 2 x a 2 (1 sin 2 x) a 2 1 sin 2 x a 2 a. cos x will be 0 if x is a quadrant I or IV angle. requires that x is such an angle.But the restriction – x 22Since a 0,Thereforecos 2 x cos xa 2 u 2 a cos xa 2 cos 2 xCommon Error:1 sin 2 x 1 sin x
298 CHAPTER 7 TRIGONOMETRIC IDENTITIES AND CONDITIONAL EQUATIONS101.a 2 (a tan x) 2 a2 u2 a 2 a 2 tan 2 x a 2 (1 tan 2 x) a 2 1 tan 2 x a 2 sec2 xa 2 a. sec x will be 0 if x is a quadrant I or IV angle. But the restriction 0 x requires that x is a quadrant I angle.2Since a 0,Thereforesec2 x sec xa 2 u 2 a sec xExercise 7–21. The six trigonometric functions can be paired as sine-cosine, tangent-cotanget, secant-cosecant.Each function in a pair is the cofunction of the other and they are related by cofunction identities ofthe type f cof(θ). 2 3. Use x y x – (–y).5. The only difference is that then 90 is used in place of7. Plug in x 1, y 1.Left side: (1 1)2 22 4Right side: 12 12 2The equation is not an identity. .29. Plug in x 2, y Left side: 2 sin .6 16 3Right side: sin 2 2 6 The equation is not an identity. ,y .66 1 Left side: cos cos 3662 Right side: cos 33 cos 6622The equation is not an identity. , y 0.2 Left side: cos 0 cos 02 2 15. Plug in x Right side: cos – cos 0 0 – 1 –12The equation is not an identity. ,y .36 1 Left side: tan tan 6363 12Right side: tan – tan 3 – 363313. Plug in x 11. Plug in x 3The equation is not an identity.17. Expand and simplify the left side:tan x tan 1 tan x tan tan x 0 1 tan x 0tan x 1tan(x – π) tan xThe equation is an identity.Difference Identitytan π sin 0 0cos 1AlgebraAlgebra
SECTION 7-2 29919. Expand and simplify the left side:sin(x – π) sin x cos π – cos x sin π sin x(–1) – cos x(0) –sin xThe equation is not an identity.21.Difference Identitycos π –1; sin π 0AlgebraExpand and simplify the left side:csc(2π – x) 1sin(2 x )csc x 1sin x1sin 2 cos x cos 2 sin x1 0 cos x 1 sin x1 sin xsin 2π 0; cos 2π 1 –csc xcsc x Difference IdentityAlgebra1sin xThe equation is not an identity.23. Expand and simplify the left side: – cos x sinsin x sin x cos2 22 sin x · 0 – cos x · 1 –cos xThe equation is an identity.25. 2 cot x 2 x sin 2 x coscos 2 cos x sin 2 sin xsin 2 cos x cos 2 sin x0 cos x 1sin x1cos x 0sin xsin x cos x tan x27. 2 csc x 1sin 2 x 1sin cos x cos 2 sin x 2Difference Identitycos 0; sin 122AlgebraQuotient IdentityDifference IdentitiesKnown ValuesAlgebraQuotient IdentityReciprocal IdentityDifference Identity
300 CHAPTER 7 TRIGONOMETRIC IDENTITIES AND CONDITIONAL EQUATIONS11cos x 0sin x1 cos x Known ValuesAlgebra sec x29.31.33.Reciprocal Identitycos x cos π cos x sin π sin x (–1) cos x (0) sin x –cos xsin x y sin [ x – (–y)] sin x cos(–y) – cos x sin(–y) sin x cos y – cos x (–sin y) sin x cos y cos x sin yDifference IdentityKnown ValuesAlgebraAlgebraDifference IdentityIdentities for NegativesAlgebrasin(30 – x) sin 30 cos x – cos 30 sin xDifference Identity13cos x –sin x221(cos x – 3 sin x) 2 35.37.Algebrasin(180 – x) sin 180 cos x – cos 180 sin x 0 cos x – (–1) sin x sin x Difference IdentityKnown ValuesAlgebratan x tan 3tan x 3 1 tan x tan 3 39.Known ValuesSum Identitytan x 3Known Values1 3 tan xsec 75 sec(30 45 )41.1 cos(30 45 )sin sin1 cos 30 cos 45 sin 30 sin 45 1 3 122 122 232 2 2 212 2 2 23 143. cos 74 cos 44 sin 74 sin 44 cos(74 – 44 ) cos 30 45. 122 2 1tan 27 tan18 tan(27 18 ) tan 45 11 tan 27 tan18 7 sin 12 3 4 32 cos cossin34341 13 1· ·222232 2 3 12 212 2
SECTION 7-2 301b47.(a, 8 )3y8aaa cos x 4552 ( 3) 2 4tan x –34cos y sin(x – y) sin x cos y – cos x sin y4 8 3 1 3 4 815 – 5 3 5 332 ( 8) 2 1a 13tan(x y) 8 1tan y 8 34 8tan x tan y 1 tan x tan y 1 34 8 3 4 84 3 8 4 8 34 3 8bb49.yx-42a-1(2, -1)a-3(-4, -3)sin x –35cos x –45sin y sin(x – y) sin x cos y – cos x sin y 3 2 4 1 5 5 5 64 10 2 5 –51.53.5 5–5 5 r ( 4) 2 ( 3) 2 5r 5 5 1cos y 5tan(x y) AlgebraSum IdentityAlgebracos( x y )sin( x y )Quotient Identity cos x cos y sin x sin ysin x cos y cos x sin ySum Identities cos x cos ysin x sin y sinsin x sin yx sin ysin x cos ycos x sin y sin x sin ysin x sin yAlgebra cos x cos y 1sin x sin ycos yx cossin ysin xAlgebra cot x cot y 1cot y cot xQuotient Identitycot(x y) 525 3 1tan x tan y6 42 4 3 2 11 tan x tan y 1 4 28 3 115cos 2x cos(x x) cos x cos x – sin x sin x cos2 x – sin2 x22 ( 1) 2
302 CHAPTER 7 TRIGONOMETRIC IDENTITIES AND CONDITIONAL EQUATIONS55.tan 2x tan(x x)Algebratan x tan x 1 tan x tan x2 tan x 57.Sum IdentityAlgebra1 tan 2 xsin v cos u cos v sin usin(v u ) sin v cos u cos v sin usin(v u ) Sum and Difference Identitiessin v cos u cos v sin usin v sin usin v sin usin v cos ucos v sin u sin v sin usin v sin ucos u cos vsin usin vcos ucos v sin usin vAlgebraAlgebracot u cot vcot u cot vcos xsin y–cot x – tan y cos ysin xQuotient Identity 59.61.tan(x – y) cos x cos ysin x sin y–sin x cos ysin x cos yAlgebra cos x cos y sin x sin ysin x cos yAlgebra cos( x y )sin x cos ySum Identitytan x tan y1 tan x tan y1cot x1 cot1 xReciprocal Identity1cot y cot x cot y 1 63.Difference Identity cot1 ycot x cot y Quotient Identities1cot x cot1 y 11cot x cot ycot y cot xcot x cot y 1cos( x h) cos xcos x cos h sin x sin h cos x hhcos x cos h cos x sin x sin h hcos x(cos h 1) sin x sin h hsin h cos h 1 – sin x cos x hh AlgebraAlgebraSum IdentityAlgebraAlgebraAlgebra
SECTION 7-2 30365. x 3 ,y 44cos (x y) cos x cos y – sin x sin y 4Left side: cos 43 cos π –14 Right side: cos cos 3 3 2 2 2 2 1 1– sin sin –12 2 2 2 2 2444sin (x y) sin x cos y cos x sin y 4Left side: sin 3 3 cos cos sin 4444Right side: sin67. x 3 sin π 04 2 2 2 2 1 1 02 2 2 2 2 211 5 ,y 66cos (x y) cos x cos y – sin x sin y 11 5 cos π –1 6 6 Left side: cos Right side: cos11 11 3 3 1 1 3 1 5 5 cos – sinsin –1 662 2 2 2 4 4 6 6 sin (x y) sin x cos y cos x sin y 11 5 sin π 0 6 6 Left side: sin Right side: sin11 11 3 3 1 33 5 5 1 cos cossin 2 4 4 06666222 69. sin(x – y) sin(5.288 – 1.769) –0.3685sin x cos y – cos x sin y sin 5.288 cos 1.769 – cos 5.288 sin 1.769 –0.3685tan(x y) tan(5.288 1.769) 0.9771tan 5.288 tan1.769tan x tan y 0.97711 tan 5.288 tan1.7691 tan x tan y71. sin(x – y) sin(42.08 – 68.37 ) –0.4429sin x cos y – cos x sin y sin 42.08 cos 68.37 – cos 42.08 sin 68.37 –0.4429tan(x y) tan(42.08 68.37 ) –2.682tan 42.08 tan 68.37 tan x tan y –2.6821 tan 42.08 tan 68.37 1 tan x tan y73. Evaluate each side for a particular set of values of x and y for which each side is defined. If the leftside is not equal to the right side, then the equation is not an identity. For example, for x 0 andy 0, both sides are defined, but are not equal.
304 CHAPTER 7 TRIGONOMETRIC IDENTITIES AND CONDITIONAL EQUATIONS75. Let y1 sin x 6Then y2 sin x cos sin x · cos x sin661133 cos x · sin x cos x.2222The graphs coincide as shown at the right.77. Let y1 cos x 3 4 3 3 sin x sin44 2222 cos x cos x sin x sin x –2222 Then y2 cos x cosThe graphs coincide as shown at the right.79. Let y1 tan x Then y2 2 3 tan x tan 23 1 tan x tan2 3 tan x ( 3)1 tan x( 3) tan x 31 3 tan xThe graphs coincide as shown at the right. 4 3 81. Let u cos–1 , v sin–1 55Then we are asked to evaluate sin (u v) which is sin u cos v cos u sin v from the sum identity. 3 3 4 4We know sin v sin sin 1 – and cos u cos cos 1 – from the function–55 5 5 inverse function identities. It remains to find cos v and sin u. Note: 0 u π and –b 52 ( 4) 2 3sin u a 35 52 ( 3) 2 4cos v 4 v 2245 3 3 4 4 3 Then sin cos 1 sin 1 sin(u v) sin u cos v cos u sin v 5 5 5 5 5 5 121224 252525
SECTION 7-2 30583.We could proceed as in problem 81. Alternatively, we can shorten the process by recognizingarccos 1 and arcsin(–1) – .322Then sin[arccos 1 arcsin(–1)] sin sin cos cossin 323322 2 85. Let u sin–1 x, v cos–1 y. Then x sin u, –31 1 (0) (–1) – .22 2 u , y cos v, 0 y π.22Then cos u 1 sin 2 u (in Quadrants I, IV) 1 x 2sin v 1 cos 2 v (in Quadrants I, II) 1 y 2Hence sin (sin–1 x cos–1 y) sin (u v) sin u cos v cos u sin v xy 1 x 21 y287. cos(x y z) cos[(x y) z] cos(x y)cos z – sin(x y)sin z (cos x cos y – sin x sin y)cos z – (sin x cos y cos x sin y)sin z cos x cos y cos z – sin x sin y cos z – sin x cos y sin z – cos x sin y sin z89. tan(x – y) 93.95.97.sin x cos y cos x sin ysin x cos y cos x sin ycos x cos y 91.cos x cos y sin x sin y cos x cos y sin x sin ycos x cos ysin( x y )cos( x y )sin x cos y cos x sin ysin x sin y cos x cos y cos x cos ycos x cos y cos x cos y sin x sin ysin x sin y 1 cos x cos y cos x cos ycos x cos ytan 2 tan 11 tan 2 tan 1Difference Identity m2 m11 m2 m1Given m2 m11 m1m2Algebratan(θ2 – θ 1) Note: In the text figure we have drawn EF perpendicular to CF, thetrack of the incident ray. CDE and CEF are right triangles.Angle DCF β γ α. Hence, γ α – β.Denote EC by x. Then,M cos β sin(90 – β)xNin CEF sin γ sin(α – β)xMNTherefore x sin(90 )sin( )in CDE
306 CHAPTER 7 TRIGONOMETRIC IDENTITIES AND CONDITIONAL EQUATIONSM sin(α – β) N sin(90 – β)M(sin α cos β – cos α sin β) N(sin 90 cos β – cos 90 sin β)M(sin α cos β – cos α sin β) N(1 cos β – 0 sin β)M(sin α cos β – cos α sin β) N cos βM sin α cos β – M cos α sin β N cos βM sin α cos β – N cos β M cos α sin βcos β (M sin α – N) M cos α sin βM sin Nsin cos M cos M sin Ntan β M cos NM sin tan β –M cos M cos sin N 1tan β –M cos cos Nsec αtan β tan α –MAlgebraDifference IdentitiesKnown ValuesAlgebraAlgebraAlgebraAlgebraAlgebraQuotient IdentityAlgebraAlgebraQuotient and Reciprocal Identities99. (A) From the text figure:ABAB hAEBCBCIn right triangle BCD, we have (2) cot α CDHH hH hE'DIn right triangle EE'D, we have (3) tan β ACAB BCEE 'In right triangle ABE, we have (1) cot α From (3), H – h (AB BC)tan βFrom (1) and (2), AB h cot α and BC H cot αHence, substituting, we have (4) H – h (h cot α H cot α)tan β, or (h H)cot α tan βSolving for H in terms of h yields:H – h h cot α tan β H cot α tan βH – H cot α tan β h cot α tan β hH(1 – cot α tan β) h(cot α tan β 1)H h(B) H hsin cos cos sin sin cos 1 cossin 1 1 H h 1 cos sin sin cos cos sin sin cos sin cos sin cos sin cos cos sin sin cos cos sin sin( )H hsin( )H h1 cot tan 1 cot tan Quotient IdentitiesAlgebraAlgebraSum and Difference Identities
SECTION 7-2 307(C) Substitute the given values to obtainH 4.90sin(46.23 46.15 )sin(92.38 ) 4.90 3510 ft (to three significant digits)sin(46.23 46.15 )sin(0.08 )Section 7-31. Substitute x for y in the equation for sin(x y).3. Substitute x for y in the equation for cos(x y).5. Solve the formulas cos 2x 1 – 2sin2 x for sin x and cos 2x 2 cos2 x – 1 for cos x, then replace xwith1x.27. cos 2(30 ) cos 60 12 3 2 1 2 3121– – 4442 2 2 cos2 30 – sin2 30 2 2 cot 3 tan 3 – 39. tan 2 tan3 3 11. sin1 cos 2 12 1 2 13. sin 22.5 sin 45 1 cos 45 22 31331 ( 1) 22 21 2 1122 2 2 2 313 3 2 32 3 – 31 3 21 12( 2 1) 2 2 2 2 4(We use since 22.5 is a I quadrant angle)1 cos135 1 15. cos 67.5 cos 135 2 2 1 22221 222 2 4 2 22(We use since 67.5 is a I quadrant angle) 1 17. tan tan 8 2 4 19. cos 1 12 4 1 sin421 cos5 1 5 cos 12 2 6 23. 3 5 1 2 6 221 cos(We use since21.2 1or12 12 3 45 is a I quadrant angle)12(sin x cos x)2 sin2 x 2 sin x cos x cos2 x 1 2 sin x cos x 1 sin 2x11(1 – cos 2x) [1 – (1 – 2 sin2 x)]221 [1 – 1 2 sin2 x)]21 (2 sin2 x)2 sin2 x2 32AlgebraPythagorean IdentityDouble–angle IdentityDouble–angle IdentityAlgebraAlgebraAlgebra2 22
308 CHAPTER 7 TRIGONOMETRIC IDENTITIES AND CONDITIONAL EQUATIONStan x sin 2x 25.sin x2 sin x cos xcos xQuotient and Double–angle Identities 2 sin2 x 2 sin2 x – 1 1 1 – (1 – 2 sin2 x) 1 – cos 2x27.sin2 1 cos xx 22 1 cos x2 AlgebraAlgebraAlgebraDouble–angle Identity 2 Half–angle Identity 2 Algebra1 cos x21cot 2x tan 2 x1 29. 31.cot 2 2Double–angle Identity1 tan 2 x2 tan xcos 2AlgebraQuotient Identitysin 2 1 cos 2 1 cos 2Half–angle Identities1 cos 1 cos Algebra 1 cos 1 cos Algebra 1 cos 1 cos 1 cos 1 cos Algebra1 cos 2 sin 2 sin 1 cos obtain2Pythagorean Identity(1 cos ) 2Since 1 – cos θ 0 and sin θ has the same sign as cot Algebra(1 cos ) 2 cotReciprocal Identity2 tan x1 tan 2 x cotAlgebra sin 1 cos Algebra 2, we may drop the absolute value signs to
SECTION 7-3 309 [To show that sin θ has the same sign as cot , we note the following cases:2 If 0 θ π, sin θ 0 then 0 , cot 0.222sin sin 0 If π θ, cot cot 0, 01 cos 1 cos 1 ( 1)22 If π θ 2π, sin θ 0, then π, cot 0.222The truth of the statement for other values of θ follows since sin( 2k ) sin and cot 2k cot . 2Key Algebraic Steps:1 a, then y 1 aIf y 1 a 1 a(1 a)(1 a ) (1 a )(1 a)1 a2(1 a ) 2An alternative proof can be given that avoids dealing with the sign ambiguity:cos 2 cot Quotient Identity sin 22 2sin2sin 2sin 2 2 22sin cos2 sin2 cos22 2 sin 2 2 2sin 22sin 1 1 2sin 2 2 sin 1 cos 2 2 sin 1 cos 33.cos 2u cos2 u – sin2 u 22cos u sin u1cos 2 u sin 2 ucos 2 u sin 2 u2cos 2 u sin 2 u2cos ucos 2 ucos 2 u Algebracos usin 2 ucos 2 uAlgebraDouble-angle IdentityAlgebraDouble-angle IdentityAlgebraDouble–angle IdentityAlgebraPythagorean IdentityAlgebra2
310 CHAPTER 7 TRIGONOMETRIC IDENTITIES AND CONDITIONAL EQUATIONS1 1 Quotient Identity1 tan 2 u1sin 2 x2 sin 2 x2 2sin x cos x1 sin x cos xReciprocal Identity2 csc 2x 2 ·Double–angle IdentityAlgebracos 2 x sin 2 xsin x cos xPythagorean Identity 2cos 2 x sin 2 xcos 2 xcos xsin x cos xcos 2 xAlgebra 1 tan 2Algebra 37.Algebra1 tan u 35.sin 2 ucos 2 usin 2 ucos 2 u2 22 1 tan 2sin 2 xcos 2 xsin xcos x21 Algebra1 tan xtan x 1 1 Quotient Identity1 cos 1 cos 1 cos 1 cos 2Half–angle Identity2 1 11 coscos Algebra 1 11 coscos 1 cos (1 cos ) 1 1 cos (1 cos ) 1 11 coscos Algebra1 cos (1 cos )1 cos 1 cos 1 cos 1 cos 1 cos 1 cos 2 cos 2 cos α Key Algebraic Steps:1 1 1 x1 x1 x1 x AlgebraAlgebraAlgebra22 1 11 xx1 1 x1 x 1 x(1 x) 1 1 x (1 x) 1 11 xx 1 x (1 x) 1 x 1 x2x x1 x 1 x1 x 1 x2
SECTION 7-3 31139.Plug in x .6 .3 31 Left side: sin sin 226Plug in x 41. Left side: tan 2 tan3 6 3 2 63Right side: 2 tanRight side:The equation is not an identity.The equation is not an identity.43.4 .32 4 31Left side: cos cos –322Plug in x 1 cos 4 3 2Right side:1 12 24 .32 4 3Left side: tan tan – 332Plug in x 45.12Plug in x 1 cos 43 Right side:The equation is not an identity.47.1 3sin 4231 cos 4 3 .249. The equation is not an identity. Plug in x ? sin 4 4 sin cos44 4 Right side: 2 sin 2·1 221?sin π 4 ·The equation is not an identity.2·1210 4·2?This is false.51. Expand the left side using a double–angle formula: 1tan(2 x)cot x 11tan xDouble–angle formula2 tan x1 tan 2 x1 tan 2 x2 tan xAlgebraNow simplify the right side: tan x 12 1tan x(cot 2 x 1)tan x 221 tan xtan x 2tan x·tan x 1 tan 2 x2 tan xcot2 x 1tan 2 xAlgebra tan1 x tan x 3The equation is not an identity.Left side: sin 2 sin π 02cot 2x 1 12 1 122AlgebraAlgebraBoth sides can be simplified to the same expression, so the equation is an identity. :4
312 CHAPTER 7 TRIGONOMETRIC IDENTITIES AND CONDITIONAL EQUATIONS53. The equation is not an identity. Plug in x π.?cos(2π) 1 – 2 cos2 π?1 1 – 2(–1)2?1 1–2This is false.55.a – 52 32 –4cos x –45tan x –34 3 4 24sin 2x 2 sin x cos x 2 –2555 3 2187 cos 2x 1 – 2 sin2 x 1 – 2 1 –2525 5 tan 2x sin 2 x24 2524 24 7 –· –cos 2 x2577 25 25 r 122 ( 5) 2 1357.sin x –513cos x 1213 1205 12 – 169 13 13 sin 2x 2 sin x cos x 2 288119 12 2–1 –1 169169 13 cos 2x 2 cos2 x – 1 2 sin 2 x120 169120 120 119 –· – cos 2 x169 119119 169 169 3 x3 , ,59. Since π x 2 242xxxsin will be positive, cos , tanwill be negative.222tan 2x 2bx2a – 3 ( 1) – 8cos x 831 cos x 2 1 83sin1x 2cos1 1 cos x1x – –222tan1 11 cos xx – –1 cos x21 2 83 8383-13(a, -1)1 83 2 – –1 8323 83 83 8 6 – –3 2 263 83 2 2 –66(3 8)(3 8)(3 8)(3 8) – (3 8) 2 –(3 8 ) –3 – 8 –3 – 2 2a
SECTION 7-3 313 x xxx,– – , coswill be positive, sin , tan will be negative.222422243b22( 3) ( 4) 5 sin x –cos x –5561. Since –π x –r 81 531 531 cos x – – – 5222212sin x –tan1x21x2 -4x 1 5325 22 15 55 2 552 55 –· –25555 xr(-3, -4)1 cos x 2sin 12cos 12ax2 54 –55 –cos-363. (A) 2θ is a second quadrant angle, since θ is a first quadrant angleand tan 2θ is negative for 2θ in the second quadrant and not for 2θ inthe first.(B) Construct a reference triangle for 2θ in the second quadrant with(a, b) (–3, 4). Use the Pythagorean theorem to find r 5.43Thus, sin 2θ and cos 2θ – .55b(-3, 4)r42 -3(C) The double angle identities cos 2θ 1 – 2 sin2 θ andcos 2θ 2 cos2 θ – 1.(D) Use the identities in part (C) in the formsin θ 1 cos 2 and cos θ 21 cos 2 2The positive radicals are used because θ is in quadrant one.(E) sin θ 1 532 1 53cos θ 2 5 3 105 3 108 102 1065. (A) tan[2(252.06 )] –0.723352 tan x21 tan x 2 tan(252.06 )1 tan 2 (252.06 ) –0.7233567. (A) tan[2(0.93457)] –3.25182 tan(0.93457)1 tan 2 (0.93457) –3.251842 52 555151 555252.06 –0.5882121 cos 252.06 – –0.588212(B) cos0.93457 0.8927921 cos 0.93457 0.892792(B) cosa
314 CHAPTER 7 TRIGONOMETRIC IDENTITIES AND CONDITIONAL EQUATIONS71.69.The graphs appear to coincide on the interval[–π, π].The graphs appear to coincide on the interval[–2π, 0].73.cos 3x cos(2x x) cos 2x cos x – sin 2x sin x (2 cos2 x – 1)cos x – 2 sin x cos x sin x 2 cos3 x – cos x – 2 sin2 x cos x 2 cos3 x – cos x – 2(1 – cos2 x)cos x 2 cos3 x – cos x – 2 cos x 2 cos3 x 4 cos3 x – 3 cos x75.cos 4x cos 2(2x) 2 cos2 2x – 1 2(2 cos2 x – 1)2 – 1 2(4 cos4 x – 4 cos2 x 1) – 1 8 cos4 x – 8 cos2 x 2 – 1 8 cos4 x – 8 cos2 x 1AlgebraSum IdentityDouble–angle IdentitiesAlgebraPythagorean IdentityAlgebraAlgebraAlgebraDouble–angle IdentityDouble–angle IdentityAlgebraAlgebraAlgebra33. Then cos u , 0 u π.553 7 3 2 cos 2 cos 1 cos 2u 2 cos2 u – 1 2 – 1 –255 5 77. Let u cos–1 4 4b79. Let u cos–1 . Then cos u – , 0 u π.55 52 ( 4) 2 3b tan u – 4
296 CHAPTER 7 TRIGONOMETRIC IDENTITIES AND CONDITIONAL EQUATIONS 83. Graph both sides of the equation in the same viewing window. tan sin 2tan x x x 1 cos 2x appears to be an identity, which we now verify: tan sin 2tan x x x sin cos sin cos sin 2 x x x x x Quotient Identity sin cos sin
SSM-30 refers to Bit 30 of the ARINC 429 word SSM- Dont care means ignore SSM-31 and SSM-30 settings. Don [t care means that there will not be any filtering of the respective SDI and/or SSM fields. 2.4 ARINC 429 Label and data word format A typical ARINC 429 data word (Label 164 - Rad. Alt.) is shown below. Starting at Bit-32 is PARITY
Part One: Heir of Ash Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26 Chapter 27 Chapter 28 Chapter 29 Chapter 30 .
TO KILL A MOCKINGBIRD. Contents Dedication Epigraph Part One Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Part Two Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18. Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26
DEDICATION PART ONE Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 PART TWO Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 .
About the husband’s secret. Dedication Epigraph Pandora Monday Chapter One Chapter Two Chapter Three Chapter Four Chapter Five Tuesday Chapter Six Chapter Seven. Chapter Eight Chapter Nine Chapter Ten Chapter Eleven Chapter Twelve Chapter Thirteen Chapter Fourteen Chapter Fifteen Chapter Sixteen Chapter Seventeen Chapter Eighteen
18.4 35 18.5 35 I Solutions to Applying the Concepts Questions II Answers to End-of-chapter Conceptual Questions Chapter 1 37 Chapter 2 38 Chapter 3 39 Chapter 4 40 Chapter 5 43 Chapter 6 45 Chapter 7 46 Chapter 8 47 Chapter 9 50 Chapter 10 52 Chapter 11 55 Chapter 12 56 Chapter 13 57 Chapter 14 61 Chapter 15 62 Chapter 16 63 Chapter 17 65 .
HUNTER. Special thanks to Kate Cary. Contents Cover Title Page Prologue Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter
Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 . Within was a room as familiar to her as her home back in Oparium. A large desk was situated i
