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?

Asked by maham237 @ in Mathematics viewed by 402 People

Answered by maham237 @


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}

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


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