Question

what is the value of cos(x) if tan ( x)= -2 and x is in quadrant of IV?

Answer 1

Answer:

$cos(x)=\frac{\sqrt{5}}{5}$

Step-by-step explanation:

we have that

$tan(x)=-2$

The angle x is in quadrant IV

That means ---> The value of cos(x) is positive and the value of sin(x) is negative

Remember that

$cos^2(x)+sin^2(x)=1$ ----> equation A

$tan(x)=\frac{sin(x)}{cos(x)}$

so

$-2=\frac{sin(x)}{cos(x)}$

$sin(x)=-2cos(x)$ ----> equation B

substitute equation B in equation A

$cos^2(x)+(-2cos(x))^2=1$

solve for cos(x)

$cos^2(x)+4cos^2(x)=1$

$5cos^2(x)=1$

$cos^2(x)=\frac{1}{5}$

square root both sides

$cos(x)=\pm\frac{1}{\sqrt{5}}$

but remember that the value of cos(x) is positive (IV quadrant)

$cos(x)=\frac{1}{\sqrt{5}}$

simplify

$cos(x)=\frac{\sqrt{5}}{5}$