Evaluate tan(x +y), and determine the quadrant it is located in.

Question

If $x$ is in Q1 and $y$ is in Q2, $\sin x = \frac{24}{25}$ and $\sin y = \frac45$, evaluate $\tan(x + y)$, and determine the quadrant in which $x + y$ is located.

Answer 1

HINT

We have that by $\sin^2 \theta + \cos^2 \theta =1$

• $\sin x =\frac{24}{25} \implies \cos x =\frac{7}{25}$

• $\sin y =\frac{4}{5} \implies \cos y =-\frac{3}{5}$

then use

$$\tan(x+y)=\frac{\tan x + \tan y}{1-\tan x \tan y}$$

Answer 2

Hint: use $$\tan (x+y) = \frac{\tan x+\tan y}{1-\tan x\tan y}.$$ Since $0\leq x \leq \pi/2$, then $\cos x = \sqrt{1-\sin^2x}$ and you have $\cos x = \sqrt{1-576/625} = 7/25$. Then $\tan x = \frac{24/25}{7/25} = 24/7$. Try to find $\tan y$ with a similar method.

Answer 3

Hint: One has $$\tan^2 \theta=\frac{\sin^2 \theta}{1-\sin^2 \theta},$$ from which you can deduce $\tan x$ and $\tan y$, determining their signs from the quadrant $x$ and $y$ belong to. Then apply the addition formula for the tangent.

Answer 4

Looks like a homework problem.

Recognize the Pythagorean triples and use properties of $\sin$ and $\cos$ to get $\cos x = \frac{7}{25}$ and $\cos y = -\frac{3}{5}$. Thus, $\tan x = \frac{24}{7}$ and $\tan y = -\frac{4}{3}$. Now use the trigonometric identity

$\tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}$.

This tells us that

$\tan(x+y) = \frac{\frac{24}{7} - \frac{4}{3}}{1 + \frac{24}{7}\frac{4}{3}} = \frac{44}{117}$.

Angle $x+y$ lies in quadrant $3$ by use of inverse trigonometric functions.