Question

If cos Θ = square root 2 over 2 and 3 pi over 2 < Θ < 2π, what are the values of sin Θ and tan Θ?

Answer 1

For the answer to the question above, If cos Θ = square root 2 over 2 and 3 pi over 2 < Θ < 2π, what are the values of sin Θ and tan Θ? 

If 3 pi over 2 < Θ < 2π Then the signs of sinx and tanx are both negative. 

cos(x) = sqrt(2)/2. So sin(x) = -sqrt(2)/2 and tan(x) = -1. 

sin Θ = square root 2 over 2; tan Θ = −1 

sin Θ = negative square root 2 over 2; tan Θ = 1 

sin Θ = square root 2 over 2; tan Θ = negative square root 2 

sin Θ = negative square root 2 over 2; tan Θ = −1

Answer 2

Answer:

The answer is

$sin(\theta)=-\frac{\sqrt{2}}{2}$

$tan(\theta)=-1$

Step-by-step explanation:

we know that

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

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

In this problem we have

$cos(\theta)=\frac{\sqrt{2}}{2}$

$\frac{3\pi}{2} <\theta < 2\pi$

so

The angle $\theta$ belong to the third or fourth quadrant

The value of $sin(\theta)$ is negative

Step 1

Find the value of  $sin(\theta)$

Remember

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

we have

$cos(\theta)=\frac{\sqrt{2}}{2}$

substitute

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

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

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

$sin(\theta)=-\frac{\sqrt{2}}{2}$ ------> remember that the value is negative

Step 2

Find the value of $tan(\theta)$

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

we have

$sin(\theta)=-\frac{\sqrt{2}}{2}$

$cos(\theta)=\frac{\sqrt{2}}{2}$

substitute

$tan(\theta)=\frac{-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}$

$tan(\theta)=-1$