Video Transcript
In this video, we will learn how to
solve trigonometric equations involving secant, cosecant, and cotangent over
different intervals. Reciprocal trigonometric equations
have several real-world applications in fields such as physics, engineering,
architecture, robotics, and navigation. In physics, they can be used in
projectile motion, when analyzing currents, and finding the trajectory of a mass
around a massive body under the force of gravity.
Let’s begin by recalling the
trigonometric functions whose reciprocals we will examine in this video. By considering the right triangle
drawn, we know that the sin of angle 𝜃 is equal to the opposite over
hypotenuse. The cos of angle 𝜃 is the adjacent
over the hypotenuse. And the tan of angle 𝜃 is equal to
the opposite over the adjacent. This also leads us to the
trigonometric identity tan 𝜃 is equal to sin 𝜃 over cos 𝜃. We note that these ratios are
defined for acute angles between zero and 90 degrees.
In order to define these functions
for all values of 𝜃, we need to consider the unit circle. Suppose that a point moves along
the unit circle in the clockwise direction. At any point with coordinates 𝑥,
𝑦 on the unit circle with angle 𝜃, we know that 𝑥 is equal to cos 𝜃 and 𝑦 is
equal to sin 𝜃. We know that these trig functions
are periodic such that the sin of 360 degrees plus 𝜃 is equal to sin 𝜃. Likewise, the cos of 360 degrees
plus 𝜃 is equal to the cos of 𝜃. The periodicity of the tangent
function tells us that tan of 180 degrees plus 𝜃 is equal to tan 𝜃.
We also know from the symmetry of
the sine and cosine graphs that the sin of 180 degrees minus 𝜃 is equal to sin 𝜃
and the cos of 360 degrees minus 𝜃 is equal to cos 𝜃. These can be seen on a CAST diagram
as follows. This CAST diagram helps us to
remember the signs of the trig functions for each quadrant. In the first quadrant, all of the
functions are positive. In the second quadrant, the sine
function is positive, whereas the cosine and tangent functions are negative. In the third quadrant, only the
tangent function is positive. And finally, in the fourth
quadrant, the cosine function is positive, whereas the sine and tangent functions
are negative.
Let’s now consider how these relate
to the reciprocal trigonometric functions. The csc of angle 𝜃 is equal to one
over the sin of angle 𝜃. It is the reciprocal of the sine
function. The sec of angle 𝜃 is equal to one
over the cos of angle 𝜃. And finally, the cot of angle 𝜃 is
equal to one over the tan of angle 𝜃. This last function, together with
the fact that tan 𝜃 is equal to sin 𝜃 over cos 𝜃, leads us to the fact that cot
𝜃 is equal to cos 𝜃 over sin 𝜃. We will now use these reciprocal
functions together with our CAST diagram to solve trigonometric equations in given
intervals.
Find the set of values satisfying
root three cot 𝜃 equals one given 𝜃 is greater than zero and less than 360
degrees.
We can solve this problem using our
knowledge of the reciprocal trigonometric functions. We recall that the cot of angle 𝜃
is equal to one over the tan of 𝜃. We can rearrange the equation we’re
given by firstly dividing both sides by root three. The cot of 𝜃 is therefore equal to
one over root three. As tan 𝜃 is the reciprocal of
this, the tan of 𝜃 equals root three. We can find the principal angle
here using our knowledge of special angles. We know that the tan of 60 degrees
is equal to root three. This means that one solution to the
equation tan 𝜃 equals root three is 𝜃 equals 60 degrees. We are asked to find all the
solutions between zero and 360 degrees. We can do this by sketching a CAST
diagram.
As the tan of angle 𝜃 is positive,
there will be solutions in the first and third quadrants. We have already found that the
solution in the first quadrant is equal to 60 degrees. Using the periodicity of the
tangent function, we know that the tan of 180 degrees plus 𝜃 is equal to the tan of
angle 𝜃. Our second solution can therefore
be calculated by adding 60 degrees to 180 degrees. This is equal to 240 degrees. Any further solutions found by
adding or subtracting multiples of 180 degrees will be outside the required
interval. So the solution set is 60 degrees
and 240 degrees.
In our next example, we will
consider a problem where the interval is given in radians.
Find the value of 𝜃 that satisfies
the csc of 𝜃 minus root two equals zero where 𝜃 lies on the open interval from
zero to 𝜋 over two.
We begin by recalling that the csc
of angle 𝜃 is the reciprocal of sin 𝜃. In order to solve the equation
we’re given, we begin by adding root two to both sides. This means that the csc of 𝜃 is
equal to root two. The sin of angle 𝜃 is therefore
equal to one over root two. Using our knowledge of special
angles, we know that the sin of 45 degrees is equal to one over root two or root two
over two.
It is important to note at this
point that the interval was given in radians. We know that 𝜋 radians is equal to
180 degrees. This means that we’re only
interested in solutions between zero and 90 degrees. We can therefore conclude that the
value of 𝜃 that satisfies the csc of 𝜃 minus root two equals zero between zero and
90 degrees is 45 degrees.
In our next example, we will solve
a trigonometric equation by changing the interval over which solutions exist.
Find the solution set of 𝜃 that
satisfies root three multiplied by the csc of 90 degrees minus 𝜃 minus two equals
zero, where 𝜃 lies on the closed interval from zero degrees to 180 degrees.
In this question, we begin by
considering the reciprocal trigonometric functions. We know that the csc of any angle
𝛼 is equal to one over the sin of angle 𝛼. Before using this identity, we will
let 𝛼 equal 90 degrees minus 𝜃. This enables us to solve the
simpler equation root three csc of 𝛼 minus two equals zero. Adding two to both sides of this
equation and then dividing through by root three, we have the csc of 𝛼 equals two
over root three. Using the reciprocal identity, we
have sin 𝛼 is equal to one over two over root three, which is equal to root three
over two.
From our knowledge of special
angles, we know that the sin of 60 degrees is equal to root three over two. This means that a solution to the
equation sin 𝛼 equals root three over two is 𝛼 equals 60 degrees. We are looking for solutions of 𝜃
between zero and 180 degrees inclusive. To write this interval in terms of
𝛼, we subtract 90 degrees from the inequality such that 𝛼 is greater than or equal
to negative 90 degrees and less than or equal to 90 degrees. Using a CAST diagram, we can see
we’re only looking for solutions in the first or fourth quadrants. As the sin of angle 𝛼 is positive,
there will therefore only be a solution in the first quadrant. This is the solution we have
already found: 𝛼 is equal to 60 degrees.
As 𝛼 is equal to 90 degrees minus
𝜃, then 𝜃 must be equal to 90 degrees minus 𝛼. Substituting in our value of 𝛼, we
have 𝜃 is equal to 90 degrees minus 60 degrees, which gives us an answer of 30
degrees. The solution set that satisfies the
equation root three multiplied by the csc of 90 degrees minus 𝜃 minus two equals
zero is 30 degrees.
In our final example, we will look
at a more complicated trigonometric equation.
Find the set of values satisfying
two sin 𝜃 csc 𝜃 plus sec 𝜃 cot 𝜃 equals zero given that 𝜃 is greater than or
equal to zero degrees and less than or equal to 360 degrees.
We will begin this question by
recalling the definition of the cosecant, secant, and cotangent functions. csc of
angle 𝜃 is equal to one over sin 𝜃. The sec of angle 𝜃 is equal to one
over cos 𝜃. And the cot of 𝜃 is equal to one
over the tan of 𝜃. We also recall that since tan 𝜃 is
equal to sin 𝜃 over cos 𝜃, the cot of 𝜃 is equal to cos 𝜃 over sin 𝜃. Substituting these identities into
our equation, we have two sin 𝜃 multiplied by one over sin 𝜃 plus one over cos 𝜃
multiplied by cos 𝜃 over sin 𝜃 is equal to zero. The first term on the left-hand
side simplifies to two, and the second term simplifies to one over sin 𝜃.
Subtracting two from both sides, we
have one over sin 𝜃 equals negative two. This means that the csc of 𝜃
equals negative two, which in turn tells us that sin of angle 𝜃 is equal to
negative one-half. We can solve this equation by
firstly sketching a CAST diagram. As the sign of angle 𝜃 is
negative, there will be solutions in the third and fourth quadrants.
From our knowledge of special
angles, we know that the sin of 30 degrees is equal to a half. We can then use our knowledge of
the symmetry of the sine function to calculate the solutions in the third and fourth
quadrants. In the third quadrant, we have 𝜃
is equal to 180 degrees plus 30 degrees. This is equal to 210 degrees. In the fourth quadrant, we have 𝜃
is equal to 360 degrees minus 30 degrees. This is equal to 330 degrees. The set of values that satisfy two
sin 𝜃 csc 𝜃 plus sec 𝜃 cot 𝜃 equals zero where 𝜃 lies between zero and 360
degrees inclusive are 210 and 330 degrees.
We will now summarize the key
points from this video. In order to solve reciprocal
trigonometric equations, we use the definition of the cosecant, secant, and
cotangent functions. The csc of angle 𝜃 is equal to one
over sin 𝜃. The sec of angle 𝜃 is equal to one
over the cos of angle 𝜃. And the cot of angle 𝜃 is equal to
one over the tan of angle 𝜃. This is also equal to the cos of
angle 𝜃 over the sin of angle 𝜃.
After finding the principal value
solution, using our knowledge of special angles or the inverse trig functions, we
can find all solutions in a given interval using the CAST diagram and the
periodicity of the trigonometric functions. Using the fact that the
trigonometric functions are periodic, we can develop this further to find a general
solution to trigonometric equations. However, this is outside of the
scope of this video.
