Question

Find tan x if cos x = 4/9 and 0 < x < pi/2 (posting in equation editor)

Answer 1

\[\tan x = \frac{ \sin x }{ \cos x }\]

\[\cos x = \frac{ 4 }{ 9 }\]\[\Rightarrow \sin x = \sqrt{1-\frac{ 4 }{ 9 }}=\frac{ \sqrt{5} }{ 3 }\]

\[\Rightarrow \tan x = \frac{ 3*\sqrt{5} }{ 4 }\]

Is this correct or how does the condition 0 <a < pi/2 affect it?

Answer 2

The interval 0 < x < pi/2 must simply tell the sign of the value of tan x right? Since tan > 0 for all values in 0 < x < pi / 2 the sign should be positive?

Answer 3

looks good, except you just have to square the 4/9 under the radical

Answer 4

Also, since your answer should be given in the range given, divide by pi (since tan x is periodic in period pi) to find how many periods back you need to subtract to get to that range.

Answer 5

To get to the range
$$
0\le x \le \pi/2
$$

Use this formula:

\(\bf \Large x - \pi \times \lfloor\frac{x}{\pi}\rfloor\\
\text{ Where } \lfloor*\rfloor \text{ is the floor function}\\
\)
I get \( \tan x = .82 \)

Answer 6

Another way to check if you answer is correct. Get a calculator and type tan(32.24) (use radians), then take the arctan of whatever you get. You should get .82.

Answer 7

Does this make sense?

Answer 8

When calculating I get\[\tan x = \frac{\sqrt{65}}{4}, x = \arctan{\frac{\sqrt{65}}{4}} \approx 1.1102\] which is in the range [0,pi/2]