Trigonometric Equations and Identities

Question 1

Find the solutions of the equation sin( 3x – 15^o ) = 1 / 2 for which 0 <= x <= 180^o.

Answer 1

• Sin( 3x – 15^o ) = 1 / 2
• 3x - 15 = Sin^-1 ( 1 / 2 )
• 3x - 15 = 30 ( Sin is positive )
• • 15 <= 3x - 15 <= 525 ( Region is altered by substituting the region values into x )
• ( Since sin is positive, we look at the second quadrant ( 180 - theta )
• 180 - 30 = 150
• ( As the region has been altered, we can add 360 to the existing thetas to get values in other quadrants )
• 30 + 360 = 390
• 150 + 360 = 510
• 3x - 15 = 30, 150, 390, 510
• ( To find x we add 15 and divide by 3 for each theta value )
• x = 15, 55, 135 and 175

Question 2

Show that the equation

tan 2x = 5 sin 2x

can be written in the form

(1 – 5 cos 2x) sin 2x = 0.

Answer 2

• ( We know that tan x = sin x / cos x )
• sin 2x / cos 2x = 5 sin 2x
• sin 2x = 5 sin 2x cos 2x ( Rearrange by multiplication )
• sin 2x - 5 sin 2x cos 2x = 0
• sin 2x ( 1 - 5 cos 2x ) = 0 ( Factorise using sin 2x as the multiplier )

Question 3

Hence solve, for 0 <= x <= 180^o,

tan 2x = 5 sin 2x,

giving your answers to 1 decimal place where appropriate.

( (1 – 5 cos 2x) sin 2x = 0 )

Answer 3

• ( Separate the equations )
• sin 2x = 0
• 2x = 0 ( This equation cannot be solved )
• 1 - 5 cos 2x = 0
• 5 cos 2x = 1
• cos 2x = 1 / 5
• 2x = cos^-1 ( 1 / 5 )
• 2x = 78.46304097 ( Cos is positive )
• ( Cos is in the fourth quadrant )
• 0<= 2x <= 360
• ( Adjust degrees )
• 360 - 78.4… = 281.536959
• 2x = 78.4…., 281….
• x = 39.2, 140.8

Question 4

Show that the equation

sin x tan x = 3 cos x + 2

can be written in the form

4 cos 2x + 2 cos x – 1 = 0.

Answer 4

• ( tan x = sin x / cos x )
• sin x ( sin x / cos x ) = 3 cos x + 2
• sin^2 x = 3 cos^2 x + 2 cos x ( Multiplied by cos x )
• ( sin^2 x = 1 - cos^2 x )
• 1 - cos^2 x = 3 cos^2 x + 2 cos x
• 4 cos^2 x + 2 cos x - 1 = 0

Question 5

Find all the values of x, in the interval 0 <= x < 360^o , for which

9 cos^2 x – 11 cos x + 3 sin^2 x = 0

giving your answers to 1 decimal place.

You must show clearly how you obtained your answers.

Answer 5

• ( sin^2 x = 1 - cos^2 x )
• 9 cos^2 x - 11 cos x + 3 sin^2 x = 0
• 9 cos^2 x - 11 cos x + 3 ( 1 - cos^2 x ) = 0
• 6 cos^2 x - 11 cos x + 3 = 0
• let cos x = x
• 6x^2 - 11x + 3 = 0
• a = 6
• b = -11
• c = 3
• ( Use quadratic formula to work out x values )
• x = 1 / 3
• x = 3 / 2
• cos x = 1 / 3
• x = 70.52877937 ( cos is positive )
• cos x = 3 / 2
• x = no value
• ( cos is in the fourth quadrant )
• 360 - x
• 360 - 70.52877937 = 289.4712206
• x = 70.5 and 289.5

Question 6

Given that

4 sin^2 x + cos x = 4 − k, 0 <= k <= 3,

find cos x in terms of k.

Answer 6

• 4 sin^2 x + cos x = 4 − k
• ( sin^2 x = 1 - cos^2 x )
• 4 ( 1 - cos^2 x ) + cos x = 4 - k
• 4 - 4 cos ^2 x + cos x = 4 - k
• • 4 cos^2 x + cos x = - k
• 4 cos^2 x - cos x - k = 0
• a = 4
• b = -1
• c = k
• ( Substitute values into quadratic formula )
• • b +- ( root b^2 - 4ac ) / 2a
• cos x = - ( - 1 ) +- ( root ( - 1 )^2 - 4 ( 4 ) ( k ) ) / 2 ( 4 ) ( Substitute values into quadratic equation and simplify them )
• cos x = 1 +- ( root 1 - 16k ) / 8

Question 7

Show that the equation

cos^2 x = 8 sin^2 x – 6 sin x

can be written in the form

( 3 sin x – 1 )^2 = 2

Answer 7

• cos^2 x = 8 sin^2 x – 6 sin x
• ( cos^2 x = 1 - sin^2 x )
• 1 - sin^2 x = 8 sin^2 x – 6 sin x
• 9 sin^2 x - 6 sin x - 1 = 0
• ( Expand the end equation )
• ( 3 sin x – 1 )^2 = 9 sin^2 x - 6 sin x + 1
• ( 1 - 2 = - 1 ) ( from the first equation )
• ( Using the expanded end equation )
• 9 sin^2 x - 6 sin x +1 - 2 = 0
• 9 sin^2 x - 6 sin x + 1 = 2
• ( 3 sin x – 1 )^2 = 2