 128.1: Karla said that if cos A is positive, then , A , , , , , and cos is...
 128.2: In Example 1, can tan be found by using ? Explain why or why not.
 128.3: In 38, for each value of u, use halfangle formulas to find a. sin ...
 128.4: In 38, for each value of u, use halfangle formulas to find a. sin ...
 128.5: In 38, for each value of u, use halfangle formulas to find a. sin ...
 128.6: In 38, for each value of u, use halfangle formulas to find a. sin ...
 128.7: In 38, for each value of u, use halfangle formulas to find a. sin ...
 128.8: In 38, for each value of u, use halfangle formulas to find a. sin ...
 128.9: In 914, for each value of cos A, find a. sin b. cos c. tan . Show a...
 128.10: In 914, for each value of cos A, find a. sin b. cos c. tan . Show a...
 128.11: In 914, for each value of cos A, find a. sin b. cos c. tan . Show a...
 128.12: In 914, for each value of cos A, find a. sin b. cos c. tan . Show a...
 128.13: In 914, for each value of cos A, find a. sin b. cos c. tan . Show a...
 128.14: In 914, for each value of cos A, find a. sin b. cos c. tan . Show a...
 128.15: If sin A 5 and 90 , A , 180, find: a. sin b. cos c. tan
 128.16: If sin A 5 and 180 , A , 270, find: a. sin b. cos c. tan
 128.17: If sin A 5 and 540 , A , 630, find: a. sin b. cos c. tan
 128.18: If tan A 5 3 and 180 , A , 270, find: a. sin b. cos c. tan
 128.19: Show that tan 5 .
 128.20: Use cos A 5 cos to show that the exact value of tan 5 2 1.
 128.21: Use cos A 5 cos 30 5 to show that the exact value of tan 15 5 2 2 .
 128.22: Use cos A 5 cos 30 5 to show that the exact value of sin 15 5 .
 128.23: a. Derive an identity for in terms of cos . b. Derive an identity f...
 128.24: The top of a billboard that is mounted on a base is 60 feet above t...
Solutions for Chapter 128: FUNCTIONS OF A !10, !6B BC AC AE 508 Trigonometric Identities A 10 mi B E D C 12 mi 1 2A
Full solutions for Amsco's Algebra 2 and Trigonometry  1st Edition
ISBN: 9781567657029
Solutions for Chapter 128: FUNCTIONS OF A !10, !6B BC AC AE 508 Trigonometric Identities A 10 mi B E D C 12 mi 1 2A
Get Full SolutionsChapter 128: FUNCTIONS OF A !10, !6B BC AC AE 508 Trigonometric Identities A 10 mi B E D C 12 mi 1 2A includes 24 full stepbystep solutions. Since 24 problems in chapter 128: FUNCTIONS OF A !10, !6B BC AC AE 508 Trigonometric Identities A 10 mi B E D C 12 mi 1 2A have been answered, more than 30872 students have viewed full stepbystep solutions from this chapter. Amsco's Algebra 2 and Trigonometry was written by and is associated to the ISBN: 9781567657029. This expansive textbook survival guide covers the following chapters and their solutions. This textbook survival guide was created for the textbook: Amsco's Algebra 2 and Trigonometry, edition: 1.

Big formula for n by n determinants.
Det(A) is a sum of n! terms. For each term: Multiply one entry from each row and column of A: rows in order 1, ... , nand column order given by a permutation P. Each of the n! P 's has a + or  sign.

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).

Full column rank r = n.
Independent columns, N(A) = {O}, no free variables.

Fundamental Theorem.
The nullspace N (A) and row space C (AT) are orthogonal complements in Rn(perpendicular from Ax = 0 with dimensions rand n  r). Applied to AT, the column space C(A) is the orthogonal complement of N(AT) in Rm.

Hankel matrix H.
Constant along each antidiagonal; hij depends on i + j.

Hypercube matrix pl.
Row n + 1 counts corners, edges, faces, ... of a cube in Rn.

Length II x II.
Square root of x T x (Pythagoras in n dimensions).

Linear combination cv + d w or L C jV j.
Vector addition and scalar multiplication.

Lucas numbers
Ln = 2,J, 3, 4, ... satisfy Ln = L n l +Ln 2 = A1 +A~, with AI, A2 = (1 ± /5)/2 from the Fibonacci matrix U~]' Compare Lo = 2 with Fo = O.

Multiplication Ax
= Xl (column 1) + ... + xn(column n) = combination of columns.

Orthonormal vectors q 1 , ... , q n·
Dot products are q T q j = 0 if i =1= j and q T q i = 1. The matrix Q with these orthonormal columns has Q T Q = I. If m = n then Q T = Q 1 and q 1 ' ... , q n is an orthonormal basis for Rn : every v = L (v T q j )q j •

Pivot.
The diagonal entry (first nonzero) at the time when a row is used in elimination.

Projection matrix P onto subspace S.
Projection p = P b is the closest point to b in S, error e = b  Pb is perpendicularto S. p 2 = P = pT, eigenvalues are 1 or 0, eigenvectors are in S or S...L. If columns of A = basis for S then P = A (AT A) 1 AT.

Projection p = a(aTblaTa) onto the line through a.
P = aaT laTa has rank l.

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).

Row picture of Ax = b.
Each equation gives a plane in Rn; the planes intersect at x.

Semidefinite matrix A.
(Positive) semidefinite: all x T Ax > 0, all A > 0; A = any RT R.

Simplex method for linear programming.
The minimum cost vector x * is found by moving from comer to lower cost comer along the edges of the feasible set (where the constraints Ax = b and x > 0 are satisfied). Minimum cost at a comer!

Skewsymmetric matrix K.
The transpose is K, since Kij = Kji. Eigenvalues are pure imaginary, eigenvectors are orthogonal, eKt is an orthogonal matrix.

Sum V + W of subs paces.
Space of all (v in V) + (w in W). Direct sum: V n W = to}.