Trigonometric Equations Lec 6 0f 12 Trigonometry 2.4 Solutions Of Trigonometric Equations

Learning Outcomes: to solve equation using half-angle identity for sin θ, cos and tan in terms of Express a cos + b sin as R cos ( ) or Rsin ( ) and subsequently solve the equation a cos θ + b sin θ = c Determine the maximum and minimum value of trigonometric expressions.

The Substitution This method is very useful in solving trigonometric equations which cannot be solved by using identities. Remember the half-angle formula:

By using Phytogaras Theorem: OP= 1+ t2 2t 1- t2 P O Q q 1+ t2

By using these substitutions, the trigonometric equation is reduced to an algebraic equation in terms of t. It can be used if each term in the equation has the same angle.

Example 1 Solve the equation 5 tan + sec + 5 = 0 for using

5 tan + sec + 5 = 0 Solution: Same angle

71.560 θ/2 26.560

Example 2 Solve the equation 5 cos - 2sin = 2 for

Solution: 5 cos - 2sin = 2 Same angle

0.4049 0.7854 θ/2

Example 3 Solve the equation 2 sin2 - cos2 = -1 for

2 sin2 - cos2 = -1 Solution: Same angle

y x 63.43

Equation in the form acos + bsin = c This type of equation can be solved by using 2 different methods: Using the substitution ( as the previous examples) (ii) Express as Rcos ( ) or Rsin( ) and subsequently solve the equation

NOTES:

NOTES:

Example 4 By expressing 2 sin θ + 5 cos θ in the form Rsin ( θ + α ), hence solve the equation 2 sin θ + 5 cos θ = - 3 for 00 < θ < 3600

Solution: Let 2 sin θ + 5 cos θ ≡ R sin ( θ + α ); R > 0 2 sin θ + 5 cos θ ≡ R [sin θ cos α + cos θsin α ] 2 sin θ + 5 cos θ ≡ R sin θ cos α + R cos θsin α Equating the coefficient of sin θ and cos θ : R sin α = 5 1st quadrant R cos α = 2

5 2 R α α = 68.20 2 sin θ + 5 cos θ = sin (θ + 68.20);

2sinθ + 5 cosθ = - 3; 00 <θ < 3600 33.9 θ + 68.2 = 213.9, 326.1 θ = 145.70, 257.90

Example 5 By expressing cos 2x – sin 2x in the form R sin ( 2x + α ) where R > 0 and 0 < α < 2π, solve the equation cos 2x – sin 2x = 1 for

cos 2x - sin 2x ≡ R sin ( 2x + α ); R > 0 Solution: cos 2x - sin 2x ≡ R sin ( 2x + α ); R > 0 cos 2x - sin 2x ≡ R [sin 2x cos α + cos2x sin α] cos 2x - sin 2x ≡ Rsin 2x cos α + Rcos2x sin α Equating the coefficient of sin 2x and cos 2x : (i) (ii) R sin α = 2nd quadrant R cos α = -1

R 1 cos 2x - sin 2x = 2 sin ( 2x + )

cos 2x - sin 2x = 1 2 sin ( 2x + ) = 1 sin ( 2x + ) =

Maximum and Minimum Values of Trigonometric Expressions

Example 6 Find the maximum and minimum values ( the greatest and the least values ) for: 2 sin θ + 5 cos θ (b) cos 2x – sin 2x

Solution: (a) 2 sin θ + 5 cos θ = sin (θ + 68.20) [ Eg. 4 ] Max = 1 Min = -1 Maximum value = (1) = Minimum value = (-1) = -

(b) cos 2x - sin 2x = 2 sin ( 2x + ) [ Eg. 5] Max = 1 Min = -1 Maximum value = 2 (1) = 2 Minimum value = 2 (-1) = - 2

Example 7 Find the maximum and minimum value for 2 - cos 2x + sin 2x. Hence, find the value of x when that expression is maximum and minimum for

Solution: 2 - cos 2x + sin 2x = 2 – ( cos 2x- sin2x) [ Eg 5 ] 2 sin ( 2x + ) = 2 – Max = 1 Min = -1 Maximum value = 4 Minimum value = 0 Max = - 2 Min = 2

When the expression is maximum: 2 sin ( 2x + ) = 4 2 – Thus the value of x when the expression is maximum are: sin ( 2x + ) = - 1

2 sin ( 2x + ) = 0 2 – 2 sin ( 2x + ) = 0 2 – sin ( 2x + ) = 1 When the expression is minimum: 2 sin ( 2x + ) = 0 2 – 2 sin ( 2x + ) = 0 2 – Thus the value of x when the expression is minimum are: sin ( 2x + ) = 1

Excercise By expressing 5 cos x + 3 sin x in the form R cos ( x - α ), find the maximum value for 5 cos x + 3 sin x. Solve the equation 10 cos x + 6 sin x = 7 for values of x lying between 0 and 2л. Give your answer in two decimal places.

Solution: Let 5 cos x + 3 sin x ≡ R cos ( x - α );R > 0 5 cos x + 3 sin x ≡ R cosx cos α + Rsinx sin α Equating the coefficient of cos x and sin x : R cos α = 5 1st quadrant R sin α = 3

R 3 5 α R 3 5 α R 3 5 α

5 cos x + 3 sin x = cos ( x – 0.540 ) Maximum value for 5 cos x + 3 sin x = (1) = 0 < x < 2л 10 cos x + 6 sin x = 7; 5 cos x + 3 sin x = 3.5

cos ( x – 0.540 ) = 3.5 cos ( x – 0.540 ) = cos ( x – 0.540 ) = 0.6002 0.9270 x – 0.540 = 0.9270, 5.3562 x = 1.467, 5.8962 x = 1.47, 5.90 ( 2 dp)

Conclusion Trigonometric Equation in the form a cos θ + b sin θ = c can be solved by using: Substitution Compound angle formulae (b) To find the minimum and maximum value for trigonometris expression, express acos θ+bsinθ as Rcos ( ) or Rsin( )