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Trigonometry. After completing this chapter you should know the functions of secant ϴ , cosecant ϴ and cotangent ϴ the graphs of sec ϴ , cosec ϴ and cot ϴ how to solve equations and prove identities involving sec ϴ , cosec ϴ and cot ϴ how to prove and use the identities
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After completing this chapter you should know • the functions of secant ϴ, cosecant ϴ and cotangent ϴ • the graphs of sec ϴ, cosec ϴ and cot ϴ • how to solve equations and prove identities involving sec ϴ, cosec ϴ and cot ϴ • how to prove and use the identities 1 + tan² ϴ ≡ sec² ϴ 1 + cot² ϴ≡ cosec² ϴ • how to sketch and use the inverse trigonometric functions arcsinx, arccosxand arctanx
The functions of secant ϴ, cosecant ϴ and cotangent ϴ these are more commonly known as sec, cosec and cot. sec ϴ = undefined for values of cosϴ = 0 cosec ϴ = undefined for values of sinϴ = 0 cot ϴ = undefined for values of tanϴ = 0
Example 2 looks at finding exact values for sec and cot, lets see what they do Exercise 6A goes over finding these values on your calculator and giving exact values!
the graphs of sec ϴ, cosec ϴ and cot ϴ These look so pretty in the book that we’ll look at them there ( page 87)
Simplify Expressions • sinϴcotϴsecϴ apply what you know ( or ½ know) = Cancelling gives sinϴcotϴsecϴ = 1
How about sinϴcosϴ(secϴ + cosecϴ)? Lets look inside the bracket first secϴ + cosecϴ = = This gives us sinϴcosϴ(secϴ + cosecϴ) = sinϴcosϴ X = sinϴ + cosϴ
Prove Identities Show that Simplify the numerator cotϴcosecϴ = x = Simplify the denominator sec²ϴ + cosec²ϴ = + = =
Solve Equations Solve the equation sec ϴ = -2.5 in the interval 0≤ϴ≤ 360° secϴ = -5/4 so cosϴ = -0.4 cos is negative in the second and third quadrant ϴ = 113.6°, 246.4° = 114°, 246° (3s.f.) S A 66.4° 66.4° T C
Solve the equation cot2ϴ = 0.6 So tan2ϴ = 5/3 Let x = 2ϴ then we are solving tanx = 5/3 in the interval 0≤ x ≤ 720° S A x = 59.0°, 239.0°, 419.0, 599.0° So ϴ = 29.5°, 120°, 210°, 300° (3 s.f) 59.0° 59.0° T C
Exercise 6C page 92 to use and apply this new found knowledge
prove and use these identities 1 + tan² ϴ ≡sec² ϴ 1 + cot² ϴ ≡cosec² ϴ
So here goes with the first proof 1 + tan² ϴ ≡sec²ϴ You know sin²ϴ + cos²ϴ≡ 1 Divide through by cos²ϴgives us tan²ϴ + 1 ≡ 1 + tan²ϴ≡ sec²ϴ
And for your second proof 1 + cot² ϴ ≡ cosec²ϴ Take sin²ϴ + cos²ϴ≡1 and divide through by sin²ϴ No I haven’t done this one it’s your turn
Now we have these identities we get to use them in some very exciting (for mathematicians) ways. Example 12 tells us that tan A = , that A is obtuse, and asks us to find a) sec A b) sin A there are two methods for part a Method 1 uses the identity 1 + tan²A≡ sec²A sec²A = 1 + sec A = because A is obtuse sec A = (cos A is negative in the second quadrant)
Method 2 this uses Pythagoras to find the hypotenuse, when tanφ = 13 5 φ 12 so secφ =
Part b) asked us to find sin A from the diagram in method 2 we can see sin A = (sin is positive in the second quadrant so there is no need to adjust this answer) using identities we have tanA this gives us sinA ≡ tanAcosA ≡ sinA ≡
you may be asked to use these identities to prove other identities example 13 on page 95 deals with this or solve equations by substituting in the identities example 14 on page 96 shows you what may happen
and finally for chapter 6 The inverse trig functions arcsin x (sin-1 x) arccos x (cos-1 x) and arctan x (tan-1 x) and their graphs.
Lets start with their graphs (because I know how much you like graph work!) sinx, cosx and tanx only have inverse functions if their domains are restricted so that they are one-to-one functions. You can draw the graph of arcsin x with the restricted domain of • ≤ x ≤ by drawing the graph of sin x and reflecting it in the line y = x Example 15 shows this with pictures Example 16 shows arccos x Example 17 shows arctan x