Solving Trig Equations

Graph of y = sin x y = sin x Max sin x = 1 when x = 90° Min sin x = -1 when x = 270°

Graph of y = cos x y = cos x Max cos x = 1 when x = 0° or 360° Min cos x = -1 when x = 180°

Graph of y = tan x These lines are called asymptotes and show that tan xo is undefined for 90o and 270o The graph of y = tan xo has no maximum or minimum values

A S T C +ve +ve +ve +ve –ve –ve –ve –ve Splitting each graph into 90o sections, shows that each has two positive and two negative sections 90o +ve +ve A S sin all 180o 0o 360o –ve –ve T C tan cos All Sinners Take Care 270o

The diagram shows where the various ratios are positive 90° 90° to 180° sin positive 0° to 90° all positive S A 180 - 180° 0°,360° 180 + C 360 - T 180° to 270° tan positive 270° to 360° cos positive 270°

Solve the following equations, giving two values for x between 0° and 360°. cos x° = 0·085 tan x° = 0·510 +ve -ve Ignore sign 90o A  S  A S 0o 180o 0o 180o 360o  360o T C  T C 270o 270o

Trigonometric Equations Solve the following equations, giving two values for x between 0° and 360°. sin x° = 0·766 cos x° = 0·565 tan x° = 4·915 cos x° = 0·906 sin x° = 0·707 tan x° = 2·050 sin x° = 0·415 tan x° = 0·193

Trigonometric Equations Solutions sin x° = 0·766 x = 50·0 or 130·0 cos x° = 0·565 x = 55·6 or 304·4 tan x° = 4·915 x = 78·5 or 258·5 cos x° = 0·906 x = 155·0 or 205·0 sin x° = 0·707 x = 225·0 or 315·0 tan x° = 2·050 x = 116·0 or 296·0 sin x° = 0·415 x = 24·5 or 155·5 tan x° = 0·193 x = 10·9 or 190·9

Solve the following equations, giving two values for x between 0° and 360°. 3sin x°  1 = 0 2tan x° + 5 = 1 2tan xo = – 4 3sin xo = 1 tan xo = – 2 sin xo = 1 ÷ 3 = 0∙333 tan-1 2 = 63∙4o A 0o 180o 270o 360o S T C 90o sin-1 0∙333 = 19∙5o    x = 180 – 63∙4o x = 19∙5o x = 116∙6o  x = 180 – 19∙5o x = 360 – 63∙4o x = 160∙5o x = 296∙6o

Trigonometric Equations Solve the following equations, giving two values for x between 0° and 360°. 4cos x°  2 = 0 5tan x°  12 = 0 5sin x° + 4 = 0 4cos x° + 3 = 0 3tan x° + 2 = 1 2tan x° - 5 = 0

4cos x°  2 = 0 cos x° = 2  4 = 0·5 x = 60 or 300 5sin x° + 4 = 0 sin x° = 4  5 = 0·8 x = 233·1 or 306·9 3tan x° + 2 = 1 tan x° = 1  3 = 0·333. x = 161·6 or 341·6 5tan x°  12 = 0 tan x° = 12  5 = 2·4 x = 67·4 or 247·4 4cos x° + 3 = 0 cos x° = 3  4 = 0·75 x = 138·6 or 221·4 2tan x° - 5 = 0 tan xo = 5 ÷ 2 = 2∙5 x = 68∙2 or 248∙2

  The diagram shows part of the graph of y = 4cos x° and y = – 3 Find the coordinates of the points A and B A B (138·6, -3) (221·4, -3) A 0o 180o 270o 360o S T C 90o  Where the graphs intersect y = y  4cos xo = – 3   cos xo = – 3/4  x = 180 - 41·4o  cos-1 ¾ = 41·4o  x = 180 + 41·4o

The maximum value of the sine of any angle is 1 and the minimum – 1 Evaluating Trig Functions The height, h km, of a satellite above the earth after x hours is given by : h(x) = 800 + 40 sin (10x)o a) What are the maximum and minimum heights of the satellite? b) When do these maximum and minimum heights occur? c) What is the period of h? d) Draw a sketch of h(x) for 0 ≤ x ≤ 180 The maximum value of the sine of any angle is 1 and the minimum – 1 Max h(x) = 800 + 40 × 1 = 840 km Min h(x) = 800 + 40 × (– 1) = 760 km

Since all values repeat every 360o Max h(x) = 800 + 40 × 1 = 840 km Occurs when angle = 90o Min h(x) = 800 + 40 × (– 1) = 760 km Note: When 0 ≤ x ≤ 180, 0 ≤ 10x ≤ 1800 Max h(x) occurs when 10x = 90, 450, 810, 1170, …….. Since all values repeat every 360o Max h(x) occurs when x = 9 hrs, 45 hrs, 81 hrs, 117 hrs, ……..