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.
$\endgroup$ 24 Answers
$\begingroup$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.
$\endgroup$ $\begingroup$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
$$\tan(x+y)=\frac{\tan x + \tan y}{1-\tan x \tan y}$$
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$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.
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