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3241004551 [841]
3 years ago
7

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?

Mathematics
2 answers:
KIM [24]3 years ago
8 0

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

Ira Lisetskai [31]3 years ago
4 0
For the answer to the question above, <span>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? </span>

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

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

<span>sin Θ = square root 2 over 2; tan Θ = −1 </span>

<span>sin Θ = negative square root 2 over 2; tan Θ = 1 </span>

<span>sin Θ = square root 2 over 2; tan Θ = negative square root 2 </span>

<span>sin Θ = negative square root 2 over 2; tan Θ = −1</span>
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