Video Transcript
Find the value of two times sec of
two 𝜋 minus 𝜃 plus eight times cot of two 𝜋 minus 𝜃 given four times tan of 𝜃
equals negative three, where 𝜃 is in the open interval between 𝜋 over two and
𝜋.
Let’s begin by visualizing where 𝜃
is located as an angle in standard position. We recall that 𝜋 over two radians
equals 90 degrees and 𝜋 radians equals 180 degrees. Therefore, 𝜃 is in the second
quadrant. We are told that four times the tan
of 𝜃 equals negative three. This equation can be solved to
isolate tan of 𝜃. We do this by dividing both sides
of the equation by four. The result is tan of 𝜃 equals
negative three over four. We recall that in terms of the
coordinate point 𝑥, 𝑦 found on the terminal side of angle 𝜃 in standard position,
tangent is defined as 𝑦 over 𝑥. This will allow us to find the
exact coordinate point the terminal side of 𝜃 passes through in the second
quadrant.
We know that in the second quadrant
all 𝑥-values are negative and all 𝑦-values are positive. In our case, this means that 𝑥 is
negative four and 𝑦 is positive three. So, we plot the point negative
four, three on our 𝑥𝑦-coordinate plane. 𝜃 is our principal angle, which
means its initial side is the positive 𝑥-axis and the ray passing through negative
four, three is 𝜃’s terminal side.
Now we sketch a right triangle
whose hypotenuse is the line segment between the origin and the point negative four,
three. The hypotenuse makes an acute angle
with the negative 𝑥-axis, which we name 𝛼. 𝛼 is our reference angle, which
means it is always the acute angle created between the terminal side and the
𝑥-axis. Unlike 𝜃, 𝛼 is on the interior of
our right triangle. The side opposite angle 𝛼 has a
length of three, and the side adjacent to 𝛼 has a length of four. However, we will label the adjacent
side with a negative four because 𝑥 is always negative in the second quadrant. tan of 𝛼 should equal tan of 𝜃,
as long as we take into account what quadrant the triangle is in.
To be sure we are using the right
sign, we refer to the CAST diagram. The CAST diagram helps us remember
which trigonometric functions are positive in each quadrant. In the first quadrant, all
trigonometric functions are positive. In the second quadrant, only sine
and its reciprocal are positive. Therefore, tangent is negative. In the third quadrant, only tangent
and its reciprocal are positive. In the fourth quadrant, only cosine
and its reciprocal are positive. Therefore, tangent is negative.
In this question, we are asked to
evaluate two times sec of two 𝜋 minus 𝜃 plus eight times cot of two 𝜋 minus 𝜃
using the facts we have just illustrated in our diagram. It will be helpful at this point to
recall the coordinate definitions of secant and cotangent. We recall that secant is the
reciprocal of the cosine function and cotangent is the reciprocal of the tangent
function. Since tan equals 𝑦 over 𝑥, cot
equals 𝑥 over 𝑦. And since cos equals 𝑥 over 𝑟,
sec equals 𝑟 over 𝑥. We have the necessary 𝑥- and
𝑦-values, so we know that cot of 𝜃 equals negative four over three. But we still need to find 𝑟 before
we can evaluate cos or sec of 𝜃.
𝑟 is the hypotenuse of our
reference triangle. Since we have the other two sides,
we can use the Pythagorean theorem to solve for 𝑟. After substituting the 𝑥- and
𝑦-coordinates and simplifying, we find that 𝑟 squared equals 25. In this case, we only take the
positive square root. So, we conclude that 𝑟 equals
five. It is worth noting that we could
skip using the Pythagorean theorem here altogether by recognizing this as the
familiar three-four-five Pythagorean triple. Then, it follows that cos of 𝜃
equals negative four over five and sec of 𝜃 equals five over negative four. We will now clear some space to
make room for our next few steps.
We have found sec and cot of
𝜃. But to evaluate the given
expression, we need to find sec and cot of two 𝜋 minus 𝜃. Let’s now consider how the position
of 𝜃 in the second quadrant compares with the position of two 𝜋 minus 𝜃. We know that 𝜃 is found by
rotating counterclockwise in the positive direction. If we rotate 𝜃 clockwise in the
negative direction, we locate not only negative 𝜃, but also two 𝜋 minus 𝜃. This is because coterminal angles
in radians are calculated by adding or subtracting any multiple of two 𝜋. Another way to think about this is
that rotating two 𝜋 radians is equivalent to rotating 360 degrees, or one full
circle.
Now we can sketch the reference
triangle for the angle two 𝜋 minus 𝜃. Our new triangle is found by a
simple reflection of the original triangle in the 𝑥-axis. This new right triangle also has a
reference angle of 𝛼. However, because we are now in the
third quadrant, the 𝑦-coordinate will be negative. The 𝑦-coordinate of the point on
the terminal side of two 𝜋 minus 𝜃 is negative three. We will use the orange triangle to
determine cot and sec of two 𝜋 minus 𝜃. It follows that cot of two 𝜋 minus
𝜃 is negative four over negative three, which simplifies to positive four over
three. And sec of two 𝜋 minus 𝜃 is the
same as before, which is five over negative four.
Before incorporating these values
into the original expression, it is a good idea to verify our signs are correct by
referring to the CAST diagram. According to the CAST diagram, only
tangent and cotangent are positive in quadrant three. Therefore, it is logical that we
found cotangent to be positive and secant to be negative. We will now clear some space to
evaluate the given expression. We will substitute four-thirds for
cot of two 𝜋 minus 𝜃 and negative five-fourths for sec of two 𝜋 minus 𝜃. Then we will add the product of two
times negative five-fourths and eight times four-thirds. Then to add negative ten-fourths
and thirty-two thirds, we use a common denominator of 12, which results in a sum of
98 over 12. This fraction simplifies to our
final answer of 49 over six.
