Exercise 5.5.5
cos ( 3 x ) = 1 / 2
Draw the unit circle and draw the vertical line crossing the x-axis at 1 / 2. The points where the vertical line cross the unit circle are the points where the terminal sides of solutions for 3 x cross the unit circle. The angles π / 3 and 5 π / 3 are the only two values of 3 x between 0 and 2 π satisfying cos ( 3 x ) = 1 / 2. These correspond to solutions for x of π / 9 and 5 π / 9. But there are more solutions for x in the interval [ 0, 2 π ). To find all the solutions for for x in the interval [ 0, 2 π ), we must find all solutions for 3 x in the interval [ 0, 6 π ). They are π / 3 , π / 3 + 2 π, π / 3 + 4 π, and 5 π / 3, 5 π / 3 + 2 π, 5 π / 3 + 4 π.. Divide these values by 3 to get six solutions for x in the interval [ 0, 2 π ), namely
x = π / 9, 5π / 9, 7π / 9, 11π / 9, 13π / 9, 17π / 9
Exercise 5.5.4
sin2 A – sin A = 0
sin A ( sin A – 1 ) = 0
So either sin A = 0 or sin A = 1.
sin A = 0 for every multiple of π. sin A = 1 for π / 2 plus every even multiple of π.
So the solution is A = n π, π / 2 + 2 n π.
Exercise 5.5.3
sin A = 0.4
Draw the unit circle and draw a horizontal line through the number 0.4 on the y-axis. The two points where this horizontal line crosses the unit circle represent the angles whose sines are 0.4, since their terminal sides must pass through one of these two points. Now all the even multiples of π pass through the point (1,0) and all the odd multiples of π pass through the point (-1,0). Thus the solutions lying in quadrant I are all the even multiples of π plus arcsin (0.4) and all the solutions lying in quadrant II are all the odd multiples of π minus arcsin (0.4). So the solutions are
A = 2 n π + arcsin(0.4), (2 n + 1) π – arcsin(0.4)
Exercise 5.5.2
sin A = - 1 / 2
Draw the unit circle and draw a horizontal line crossing the y-axis at – 1 / 2. The two points where the horizontal line crosses the unit circle represent the angles whose sines are equal to - 1 / 2 since their terminal sides cross the unit circle at those two points. These two points lie in quadrants III and IV. Now the even multiples of π cross the unit circle at (1,0) and the odd multiples of π cross the unit circle at (-1,0). The solutions will be the even multiples of π minus arcsin(1/2) or the odd multiples of π plus arcsin(1/2). But arcsin(1/2) = π / 6, so the solutions are
A = 2 n π – π / 6, (2 n + 1) π + π / 6.
Exercise 5.5.1
cos x = 0
Draw the unit circle and draw a vertical line through 0 on the horizontal axis. The two points where this vertical lines crosses the unit circle represent the solutions to the equation, since every angle whose cosine is zero will have a terminal side which passes through one of these two points. The two points are (0,1) and (0,-1). Now all the even multiples of π pass through the point (1,0) and all the odd multiples of π pass through the point (-1,0). So the angles passing through the point (0,1) are the even multiples of π plus π / 2, and the angles passing through the point (0,-1) are the odd multiples of π plus π / 2. So every multiple of π, whether even or odd, plus π / 2 is a solution to this equation. That is,
x = n π + π / 2.
Exercise 5.4.5
Since A is in quadrant III, y must be negative, hence the – 3 rather than + 3.
Exercise 5.4.4
Since x is in quadrant II, we know its sine is positive and its cosine is negative, thus
Exercise 5.4.3
Exercise 5.4.2
Exercise 5.4.1
In quadrant II, the x coordinate is negative, hence the negative square root of five.
Since cos ( 2A ) is positive and sin ( 2A ) is negative, 2A must lie in quadrant IV.
Exercise 5.3.5
Factor out 
to get
where φ is an angle in quadrant II.
Then
where 
. Then the phase
shift is – φ = - 2.159 radians.
The amplitude is 
and the period is 2
π.
Exercise 5.3.4
Exercise 5.3.3
sin17ºcos13º + cos17ºsin13º = sin ( 17º + 13º ) = sin30º = 1 / 2.
Exercise 5.3.2
cos ( 90º - A ) = cos90ºcosA + sin90ºsinA = 0 + sinA = sinA
Exercise 5.3.1
sin75º= sin ( 45º + 30º ) = sin45ºcos30º + cos45ºsin30º = 
cos75º = cos ( 45º + 30º ) =cos45ºcos30º - sin45ºsin30º = 
tan75º = sin75º / cos75º = 
csc75º = 1 / sin75º = 
sec75º = 1 / cos75º = 
cot75º = 1 / tan75º = 
Since 15º is the complementary angle to 75º, we know that
sin15º = cos75º = 
cos15º = sin75º = 
tan15º = cot75º = 
csc15º = sec75º = 
sec15º = csc75º = 
cot15º = tan75º = 
Exercise 5.2.4
Exercise 5.2.3
Exercise 5.2.2
Exercise 5.2.1
sin2A = 1 – cos2A = ( 1 – cosA )( 1 + cosA )