Question

Find exact values for sin θ and tan θ if cos θ = -4/9 and tan θ > 0.

Answer 1

Answer:

Part 1) $sin(\theta)=-\frac{\sqrt{65}}{9}$

Part 2) $tan(\theta)=\frac{\sqrt{65}}{4}$

Step-by-step explanation:

we have that

The cosine of angle theta is negative and the tangent of angle theta is positive

That means that the sine of angle theta is negative

step 1

Find $sin(\theta)$

we know that

$sin^{2}(\theta) +cos^{2}(\theta)=1$

we have

$cos(\theta)=-\frac{4}{9}$

substitute

$sin^{2}(\theta) +(-\frac{4}{9})^{2}=1$

$sin^{2}(\theta) +\frac{16}{81}=1$

$sin^{2}(\theta)=1-\frac{16}{81}$

$sin^{2}(\theta)=\frac{65}{81}$

square root both sides

$sin(\theta)=\pm\frac{\sqrt{65}}{9}$

Remember that

In this problem the sine of angle theta is negative

so

$sin(\theta)=-\frac{\sqrt{65}}{9}$

step 2

Find $tan(\theta)$

we know that

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

we have

$sin(\theta)=-\frac{\sqrt{65}}{9}$

$cos(\theta)=-\frac{4}{9}$

substitute the given values

$tan(\theta)=-\frac{\sqrt{65}}{9}:-\frac{4}{9}=\frac{\sqrt{65}}{4}$