 1.5.Problem 1: Write sec in terms of sin . P
 1.5.a: What do we mean when we want tan written in terms of sin ?
 1.5.1: One way to simplify an expression containing trigonometric function...
 1.5.Problem 2: Write tan
 1.5.b: Write tan in terms of sin .
 1.5.2: To prove, or verify, an identity, we can start with the _____ _____...
 1.5.Problem 3: Add sin . 1 cos P
 1.5.c: What is the least common denominator for the expression ?
 1.5.3: cos
 1.5.Problem 4: Multiply (2 cos 1)(3 cos 2).
 1.5.d: How would you prove the identity cos tan sin ?
 1.5.4: csc
 1.5.Problem 5: Simplify the expression
 1.5.5: cot
 1.5.Problem 6: Show that sin cos cot csc is true by transforming the left side int...
 1.5.6: sec
 1.5.Problem 7: Prove the identity
 1.5.7: sec
 1.5.8: sin
 1.5.9: csc
 1.5.10: tan
 1.5.11: 1 a 1 b
 1.5.12: co 1 s a
 1.5.13: csc cot
 1.5.14: sec cot
 1.5.15: csc tan
 1.5.16: sec tan csc
 1.5.17: sec csc
 1.5.18: csc sec
 1.5.19: sec tan
 1.5.20: csc cot
 1.5.21: tan cot
 1.5.22: cot tan
 1.5.23: sin csc
 1.5.24: cos sec
 1.5.25: tan sec 2
 1.5.26: cot csc
 1.5.27: cos tan sin
 1.5.28: sec tan sin
 1.5.29: csc cot cos
 1.5.30: a. b. 1 cos
 1.5.31: a. a b. sin 1 sin
 1.5.32: a. b. 1 cos
 1.5.33: a. b. cos sin
 1.5.34: sin 3
 1.5.35: cos
 1.5.36: 1
 1.5.37: sin
 1.5.38: c
 1.5.39: a. (a b) 2
 1.5.40: a. (a 1)(a 1)
 1.5.41: a. (a 2)(a 3)
 1.5.42: a. (a 1)(a 4)
 1.5.43: (sin 4)(sin 3) 4
 1.5.44: (cos 2)(cos 5)
 1.5.45: (2 cos 3)(4 cos 5)
 1.5.46: (3 sin 2)(5 cos 4)
 1.5.47: (1 sin )(1 sin )
 1.5.48: (1 cos )(1 cos )
 1.5.49: (1 tan )(1 tan )
 1.5.50: (1 cot )(1 cot )
 1.5.51: (sin cos ) 2
 1.5.52: (cos sin ) 2
 1.5.53: (sin 4)2
 1.5.54: (cos 2)2
 1.5.55: Simplify the expression x2 4 as much as possible after substituting...
 1.5.56: Simplify the expression x2 1 as much as possible after substituting...
 1.5.57: Simplify the expression 9 x 2 as much as possible after substitutin...
 1.5.58: Simplify the expression 25 x2 as much as possible after substitutin...
 1.5.59: Simplify the expression x2 36as much as possible after substituting...
 1.5.60: Simplify the expression x2 64as much as possible after substituting...
 1.5.61: Simplify the expression 64 4x2 as much as possible after substituti...
 1.5.62: Simplify the expression 4x2 100 as much as possible after substitut...
 1.5.63: cos tan sin
 1.5.64: sin cot cos
 1.5.65: sin sec cot 1
 1.5.66: cos csc tan 1
 1.5.67: sin
 1.5.68: cos
 1.5.69: csc
 1.5.70: csc s
 1.5.71: csc
 1.5.72: sec
 1.5.73: sec cot
 1.5.74: csc tan
 1.5.75: sin tan cos sec 7
 1.5.76: cos cot sin csc
 1.5.77: tan cot sec csc 8
 1.5.78: tan2 1 sec2
 1.5.79: csc sin
 1.5.80: sec cos
 1.5.81: csc tan cos
 1.5.82: sec cot sin
 1.5.83: (1 cos )(1 cos ) sin2 8
 1.5.84: (1 sin )(1 sin ) cos2
 1.5.85: (sin 1)(sin 1) cos2 8
 1.5.86: (cos 1)(cos 1) sin2
 1.5.87: cos sec
 1.5.88: 1 c
 1.5.89: (cos sin ) 2 1 2 sin cos
 1.5.90: (sin cos ) 2 1 2 sin cos
 1.5.91: sin (sec csc ) tan 1
 1.5.92: cos (csc tan ) cot sin
 1.5.93: sin (csc sin ) cos2
 1.5.94: cos (sec cos ) sin2
 1.5.95: Express tan in terms of sin only.
 1.5.96: Simplify . a. cos b. sin c. sec d. tan
 1.5.97: Simplify x2 16as much as possible after substituting 4 tan for x. a...
 1.5.98: Which of the following is a valid step in proving the identity sin ...
Solutions for Chapter 1.5: More on Identities
Full solutions for Trigonometry  7th Edition
ISBN: 9781111826857
Solutions for Chapter 1.5: More on Identities
Get Full SolutionsThis textbook survival guide was created for the textbook: Trigonometry, edition: 7. This expansive textbook survival guide covers the following chapters and their solutions. Chapter 1.5: More on Identities includes 109 full stepbystep solutions. Trigonometry was written by and is associated to the ISBN: 9781111826857. Since 109 problems in chapter 1.5: More on Identities have been answered, more than 16958 students have viewed full stepbystep solutions from this chapter.

Change of basis matrix M.
The old basis vectors v j are combinations L mij Wi of the new basis vectors. The coordinates of CI VI + ... + cnvn = dl wI + ... + dn Wn are related by d = M c. (For n = 2 set VI = mll WI +m21 W2, V2 = m12WI +m22w2.)

Commuting matrices AB = BA.
If diagonalizable, they share n eigenvectors.

Companion matrix.
Put CI, ... ,Cn in row n and put n  1 ones just above the main diagonal. Then det(A  AI) = ±(CI + c2A + C3A 2 + .•. + cnA nl  An).

Diagonal matrix D.
dij = 0 if i # j. Blockdiagonal: zero outside square blocks Du.

Ellipse (or ellipsoid) x T Ax = 1.
A must be positive definite; the axes of the ellipse are eigenvectors of A, with lengths 1/.JI. (For IIx II = 1 the vectors y = Ax lie on the ellipse IIA1 yll2 = Y T(AAT)1 Y = 1 displayed by eigshow; axis lengths ad

Exponential eAt = I + At + (At)2 12! + ...
has derivative AeAt; eAt u(O) solves u' = Au.

Fibonacci numbers
0,1,1,2,3,5, ... satisfy Fn = Fnl + Fn 2 = (A7 A~)I()q A2). Growth rate Al = (1 + .J5) 12 is the largest eigenvalue of the Fibonacci matrix [ } A].

Kirchhoff's Laws.
Current Law: net current (in minus out) is zero at each node. Voltage Law: Potential differences (voltage drops) add to zero around any closed loop.

Left nullspace N (AT).
Nullspace of AT = "left nullspace" of A because y T A = OT.

Pivot columns of A.
Columns that contain pivots after row reduction. These are not combinations of earlier columns. The pivot columns are a basis for the column space.

Polar decomposition A = Q H.
Orthogonal Q times positive (semi)definite H.

Pseudoinverse A+ (MoorePenrose inverse).
The n by m matrix that "inverts" A from column space back to row space, with N(A+) = N(AT). A+ A and AA+ are the projection matrices onto the row space and column space. Rank(A +) = rank(A).

Rank r (A)
= number of pivots = dimension of column space = dimension of row space.

Right inverse A+.
If A has full row rank m, then A+ = AT(AAT)l has AA+ = 1m.

Schur complement S, D  C A } B.
Appears in block elimination on [~ g ].

Schwarz inequality
Iv·wl < IIvll IIwll.Then IvTAwl2 < (vT Av)(wT Aw) for pos def A.

Singular matrix A.
A square matrix that has no inverse: det(A) = o.

Singular Value Decomposition
(SVD) A = U:E VT = (orthogonal) ( diag)( orthogonal) First r columns of U and V are orthonormal bases of C (A) and C (AT), AVi = O'iUi with singular value O'i > O. Last columns are orthonormal bases of nullspaces.

Symmetric matrix A.
The transpose is AT = A, and aU = a ji. AI is also symmetric.

Transpose matrix AT.
Entries AL = Ajj. AT is n by In, AT A is square, symmetric, positive semidefinite. The transposes of AB and AI are BT AT and (AT)I.