Video Transcript
Given that cos of 𝜃 one is equal
to one-third, where 𝜃 one is between zero and 𝜋 over two and cos 𝜃 two is equal
to one-third, where 𝜃 two is between zero and 𝜋 over two, evaluate tan of 𝜃 one
plus 𝜃 two without using a calculator. Hint: Take tan of 𝜃 one plus 𝜃
two is equal to the tan of 𝜃 one plus the tan of 𝜃 two all over one minus the tan
of 𝜃 one times the tan of 𝜃 two.
We’ve been given the cos of 𝜃 one
and the cos of 𝜃 two, which are both one-third. In addition to that, both of the
ranges for 𝜃 one and 𝜃 two are between zero and 𝜋 over two. In the context of a coordinate
grid, if our angle falls between zero and two 𝜋, it will fall in quadrant one. And this gives us a bit more
information about this angle. To remember this, we use the CAST
diagram. The 𝐴 in quadrant one tells us
that for angles that fall in quadrant one, all three of the trig functions will be
positive. The sine, cosine, and tangent
values of a 𝜃 falling between zero and 𝜋 over two will be positive.
Since our goal is to find the tan
of 𝜃 one plus 𝜃 two and we’ve been given a trig identity to help us do that, we
need two pieces of information. We need to calculate the tan of 𝜃
one and the tan of 𝜃 two. We’ll need to know the sine,
cosine, and tangent relationships. Since we’re given a cosine
relationship, that is the adjacent angle over the hypotenuse and both 𝜃 one and 𝜃
two have the same cosine relationships. To calculate the tan of 𝜃 one and
𝜃 two, we need to know the opposite side length.
To help us find this opposite side
length, we can sketch a right-angled triangle. Here’s a right-angled triangle with
a cos of one-third. To calculate its opposite side
length, we use the Pythagorean theorem, where we would say one squared plus 𝑏
squared equals three squared. One plus 𝑏 squared equals
nine. If we subtract one from both sides,
we see that 𝑏 squared equals eight. And to get 𝑏 by itself, we take
the square root of both sides. We know the square root of eight is
equal to the square root of four times two, which is equal to the square root of
four times the square root of two. And so the most simplified form of
𝑏 is two times the square root of two.
We can add that to our right-angled
triangle. And if we want to know the tangent
relationship of this angle, it’s the opposite side length over the adjacent side
length, two times the square root of two over one. Because both of these angles have a
cos of one-third and fall in the same quadrant, the tan of 𝜃 one and the tan of 𝜃
two will be the same. Two times the square root of two
over one. At this point, we’re ready to solve
for the tan of 𝜃 one plus 𝜃 two.
We just need to plug in what we
know. The tan of 𝜃 one is two times the
square root of two, and the tan of 𝜃 two is two times the square root of two. We can add two times the square
root of two plus two times the square root of two. When we combine them, we get four
times the square root of two. In our denominator, we need to
multiply two times the square root of two times two times the square root of
two.
If we rearrange it, it could look
like this. We multiply two times two, which
gives us four. And the square root of two times
the square root of two equals two. Four times two equals eight and one
minus eight equals negative seven. So we can say the tan of 𝜃 one
plus 𝜃 two is equal to negative four times the square root of two over seven.
