Video Transcript
Find the set of values satisfying
four times the sin of 𝜃 squared minus one equals zero given that 𝜃 is greater than
zero degrees and less than 180 degrees.
Let’s begin by solving for sin of
𝜃. We add one to each side of the
equation. Now, four times the sin of 𝜃
squared equals one. Then, we divide each side by
four. So the sin of 𝜃 squared equals
one-fourth. It follows that sin of 𝜃 equals
the positive and negative square roots of one-fourth. To simplify the resulting square
root of a quotient, we square root both the numerator and the denominator. Thus, sin of 𝜃 equals positive and
negative one-half. Because of the periodic nature of
the sine function, this equation would have infinitely many solutions if we did not
have restrictions on 𝜃. We are told that 𝜃 is greater than
zero degrees and less than 180 degrees.
We now recall the familiar graph of
sine of an angle plotted against that angle. We see that sin of 𝜃 equals
one-half two times in the open interval from zero to 180 degrees. We have marked these places with
blue crosses. However, since the sin of 𝜃 never
equals negative one-half between zero and 180 degrees, we have no solutions from the
equation sin of 𝜃 equals negative one-half.
Now we must find the two values of
𝜃 for which the sin of 𝜃 equals one-half. The exact solutions are not obvious
from the graph. However, we see that one solution
is less than 90 degrees and the other solution is closer to 180 degrees. However, by applying the inverse
sine function to our equation, we obtain that the equation 𝜃 equals sin inverse of
one-half. Then, we can use a calculator in
degree mode to find the first answer of 30 degrees. Unfortunately, by applying inverse
sine, we are only able to find one of the two solutions between zero and 180
degrees. Inverse sine is only defined
between negative 90 and 90 degrees. That is why it only returns one
value of 𝜃.
To find the second value of 𝜃, we
will need to recall the relationship between the sides of a 30-60-90 special right
triangle. Let’s clear some space to show our
reasoning. For this demonstration, we sketch a
30-60-90 reference triangle on the 𝑥𝑦-coordinate plane, where the hypotenuse is
the terminal side of a 30-degree angle in standard position. According to coordinate definitions
of trigonometric functions, if we let 𝑟, the hypotenuse, equal one, then sin equals
𝑥 and cos equals 𝑦. Within a 30-60-90 triangle, the
side opposite the 30-degree angle is always half the length of the hypotenuse. And the side adjacent to the
30-degree angle is always the square root of three times the length of the opposite
side. Therefore, 𝑦 equals one-half and
𝑥 equals the square root of three over two. This reference triangle we have
drawn confirms that sin of 30 degrees equals one-half.
Now we will look in the second
quadrant for our second solution, where 𝜃 is between 90 and 180 degrees. We sketch another 30-60-90
reference triangle in the second quadrant along the negative 𝑥-axis, with the same
measures as our first right triangle. Since 𝑥-values in the second
quadrant are always negative, we now have 𝑥 equal to negative square root of three
over two. But more importantly, the value of
𝑦 remains the same. Therefore, sine still equals
one-half.
The reference angle in the second
quadrant is still 30 degrees. But we need to identify the measure
of this angle in standard position. We know that an angle terminating
on the negative side of the 𝑥-axis equals 180 degrees. Our angle is a 30-degree clockwise
turn from the negative 𝑥-axis. This reminds us of one of the
cofunction identities of sine, which tells us the sin of 𝜃 equals sin of 180
degrees minus 𝜃. The hypotenuse of our first
triangle is the terminal side of angle 𝜃, and the hypotenuse of our second triangle
is the terminal side of the 180 degrees minus 𝜃 angle.
We found 𝜃 in the first quadrant
to equal 30 degrees. So, according to the cofunction
identify we just recalled, sin of 30 degrees equals sin of 180 degrees minus 30
degrees, which is 150 degrees. We have demonstrated that both sin
of 30 degrees and sin of 150 degrees equal one-half. In conclusion, the set containing
30 degrees and 150 degrees satisfies the equation four times the sin of 𝜃 squared
minus one equals zero, given that 𝜃 is greater than zero degrees and less than 180
degrees.
