- 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
Adjacency matrix of a graph.
Square matrix with aij = 1 when there is an edge from node i to node j; otherwise aij = O. A = AT when edges go both ways (undirected). Adjacency matrix of a graph. Square matrix with aij = 1 when there is an edge from node i to node j; otherwise aij = O. A = AT when edges go both ways (undirected).
A = CTC = (L.J]))(L.J]))T for positive definite A.
Remove row i and column j; multiply the determinant by (-I)i + j •
z = a - ib for any complex number z = a + ib. Then zz = Iz12.
Echelon matrix U.
The first nonzero entry (the pivot) in each row comes in a later column than the pivot in the previous row. All zero rows come last.
Elimination matrix = Elementary matrix Eij.
The identity matrix with an extra -eij in the i, j entry (i #- j). Then Eij A subtracts eij times row j of A from row i.
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 IIA-1 yll2 = Y T(AAT)-1 Y = 1 displayed by eigshow; axis lengths ad
Fourier matrix F.
Entries Fjk = e21Cijk/n give orthogonal columns FT F = nI. Then y = Fe is the (inverse) Discrete Fourier Transform Y j = L cke21Cijk/n.
Full row rank r = m.
Independent rows, at least one solution to Ax = b, column space is all of Rm. Full rank means full column rank or full row rank.
A sequence of steps intended to approach the desired solution.
Left inverse A+.
If A has full column rank n, then A+ = (AT A)-I AT has A+ A = In.
Ps = pascal(n) = the symmetric matrix with binomial entries (i1~;2). Ps = PL Pu all contain Pascal's triangle with det = 1 (see Pascal in the index).
Rank one matrix A = uvT f=. O.
Column and row spaces = lines cu and cv.
Right inverse A+.
If A has full row rank m, then A+ = AT(AAT)-l has AA+ = 1m.
R = [~ CS ] rotates the plane by () and R- 1 = RT rotates back by -(). Eigenvalues are eiO and e-iO , eigenvectors are (1, ±i). c, s = cos (), sin ().
Iv·wl < IIvll IIwll.Then IvTAwl2 < (vT Av)(wT Aw) for pos def A.
Constant down each diagonal = time-invariant (shift-invariant) filter.
Unitary matrix UH = U T = U-I.
Orthonormal columns (complex analog of Q).
Vandermonde matrix V.
V c = b gives coefficients of p(x) = Co + ... + Cn_IXn- 1 with P(Xi) = bi. Vij = (Xi)j-I and det V = product of (Xk - Xi) for k > i.
Stretch and shift the time axis to create Wjk(t) = woo(2j t - k).