Summary
This lesson opens the unit on analytic trigonometry, with an entry assessment that measures the starting point; the unit turns the trigonometric functions, studied in the preceding unit as functions of an angle, into an algebraic calculus with rules of its own: identities that hold for every angle, formulas that express the ratios of a sum through those of its summands, transformations of sums into products, equations and inequalities, and their use to measure angles between lines and to recognize the conics whose axes are oblique. It presents the six lessons of the unit in their logical order, traces the history of these ideas from the chord tables of Greek astronomy to the analytic formulation of Euler, states the problem that gave rise to them, that of calculating the chord of any arc from a few known chords, and solves it in an example with the tools that the unit will prove. It ends by pointing out where these ideas reappear in the rest of the course, in other courses on the site and in science and technology.
Learning objectives
- Describe the content of each lesson of the unit and explain how each one rests on the earlier ones.
- Place in their historical context the tables of chords, the transition from the chord to the sine and the analytic formulation of the trigonometric functions.
- Formulate the problem of calculating a chord from other known chords and recognize in its solution the difference and half-angle formulas.
- Identify applications of analytic trigonometry in calculus, in mechanics, in programming and in linear algebra.
Entry assessment
Before working through the unit, it is advisable to measure the starting point. This self-assessment does not grade: it serves to show where to slow down. The first nine questions cover the six lessons of the unit, two for each of the lessons on the identities, on the sum and difference formulas and on the equations, which support the others, and one for each of the other three; the tenth is a modeling problem, which asks for a situation in context to be set up with the double-angle formula. The same skills are measured again at the end, in the synthesis lesson, with other data. The last three check what the unit takes as known from the preceding unit and from Course I: the values on the unit circle, the inverse functions and the dot product.
Before starting the unit, measure your starting point. It is a short self-assessment: it is not graded and its result is stored only in your browser.
Question 1Which of the following equalities is an identity on its common domain?
The square root of is , not : at the left-hand side equals and the right-hand side . A single counterexample is enough to refute an identity.It holds at and at , but not at , where the left-hand side equals . No finite number of favorable checks proves an identity.The quotient has been inverted: , which at equals and not .With a common denominator, , valid at every point where , which is the common domain of the two sides.Question 2If and lies in the second quadrant, calculate .
Correct answer:
By the Pythagorean identity, , so that ; in the second quadrant the cosine is negative, and . Whoever obtains has forgotten the sign fixed by the quadrant, and whoever obtains has calculated the sine, , and given it the sign of the cosine.Question 3Calculate , writing .
This is the sum of the tangents. The tangent does not distribute over the sum of the angles: one must divide by .The sign of the denominator is reversed: has been used instead of . The angle lies in the second quadrant and its tangent is negative.This is the tangent of : the difference formula has been applied instead of the sum formula.By the formula for the tangent of a sum, , and rationalizing gives . It is negative, as befits an angle in the second quadrant.Review: Sum and difference formulas, double angle and half angle
Question 4If and , calculate .
Correct answer:
In the second quadrant the cosine is negative, and by the Pythagorean identity ; by the double-angle formula, . Whoever obtains has taken the cosine as positive, and whoever obtains has doubled the sine, which is impossible because it exceeds .Review: Sum and difference formulas, double angle and half angle
Question 5Calculate the maximum value of .
Correct answer:
By the harmonic form, , with ; its range is , and the maximum of is . Whoever obtains has added the maxima of the two terms, which are not attained at the same point, and whoever obtains has forgotten the constant.Question 6How many solutions does the equation have in ?
Correct answer:
The trinomial factors as . The equation has two solutions in one turn, and , and only one, , because . Whoever obtains has counted the solution of twice.Question 7Solving by squaring yields the candidates . Which are the solutions in ?
Squaring is not an equivalence: the candidates contain the solutions, but also those of . They must be checked.These are the solutions of . At the left-hand side of the original equation, , equals .These are the extraneous solutions, that is, those of ; the true ones have been discarded.Squaring adds the solutions of , so each candidate must be checked in the original equation: at and at the left-hand side equals , and at and at it equals .Question 8What angle do the lines and form?
This is the supplement of the angle between the lines. The angle between two lines does not exceed , and that is why the formula has an absolute value.The division by is missing: the difference of the slopes is not the tangent of the difference of the angles of inclination.Slopes of opposite signs are not enough for perpendicularity: is needed, and here .By the theorem on the angle between two lines, , and the angle between two lines lies in .Review: The angle between two lines and applications of trigonometry
Question 9What does the equation represent?
The coefficients of and of are positive, but the term changes the nature of the conic: what decides is the sign of , which here is positive.Having is not enough: a circle also requires . Here .Parabolas have zero discriminant; here .The discriminant is , and the equation is of hyperbolic type; the rotation by takes it to , which is a hyperbola and not two lines.Review: Rotation of axes and the general second-degree equation
Question 10From a log of circular cross section, centimeters in diameter, a beam of rectangular cross section is to be cut, with the four vertices of the cross section on the circle of the log. If is the angle that a diagonal of the rectangle forms with one of its sides, what is the largest possible area of the cross section, in square centimeters?
Correct answer:
The diagonals of a rectangle inscribed in a circle are diameters, so the sides measure and , and by the double-angle formula the area is . Since , with equality for , the maximum is , that of the square cross section. Whoever obtains has forgotten the factor in , and whoever obtains has taken the radius as the diagonal.Review: Sum and difference formulas, double angle and half angle
Question 11From earlier unitsWhat are the coordinates of the point on the unit circle?
This is : the arc has been traversed counterclockwise. A negative angle is traversed clockwise.These are the values of , the reference angle, without the sign of the quadrant: the point lies in the second quadrant, where the abscissa is negative.The sine and the cosine have been swapped: this is the point .The angle is coterminal with , in the second quadrant, whose cosine is and whose sine is .Question 12From earlier unitsCalculate .
The arccosine does not undo the cosine outside : its value always lies in that interval, and does not.It has the right cosine, but it does not lie in , the range of the arccosine; it is the principal value of the arcsine or of the arctangent that can be negative.This is the reference angle, whose cosine is and not .The cosine of is , and the arccosine returns the only angle in with that cosine, which is .Question 13From earlier unitsCalculate the cosine of the angle formed by the vectors and .
Correct answer:
The dot product is and each vector has norm ; the cosine is . Whoever obtains has forgotten to divide by the norms, and whoever obtains has divided by only one of them.Review: Vectors in the plane and parametric equations of a line
Loading the content…
Completed
Sign in to save your progress