Question

If sine theta equals three over four, what are the values of cos θ and tan θ?

Answer 1

Answer:

Part 1) $cos(\theta)=(+/-)\frac{\sqrt{7}}{4}$

cosine theta equals plus or minus square root of seven over 4

Part 2) $tan(\theta)=(+/-)\frac{3}{\sqrt{7}}$

tangent theta equals plus or minus three over square root of seven

or

$tan(\theta)=(+/-)3\frac{\sqrt{7}}{7}$

tangent theta equals plus or minus three times square root of seven over seven

Step-by-step explanation:

we have that

The sine of angle theta is equal to

$sin(\theta)=\frac{3}{4}$

Is positive

therefore

The angle theta lie on the I Quadrant or in the II Quadrant

Part 1) Find the value of the cosine of angle theta

Remember that

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

we have

$sin(\theta)=\frac{3}{4}$

substitute and solve for cosine of angle theta

$(\frac{3}{4})^{2}+cos^{2} (\theta)=1$

$cos^{2} (\theta)=1-(\frac{3}{4})^{2}$

$cos^{2} (\theta)=1-\frac{9}{16}$

$cos^{2} (\theta)=\frac{7}{16}$

$cos(\theta)=(+/-)\frac{\sqrt{7}}{4}$

cosine theta equals plus or minus square root of seven over 4

Part 2) Find the value of tangent of angle theta

we know that

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

we have

$sin(\theta)=\frac{3}{4}$

$cos(\theta)=(+/-)\frac{\sqrt{7}}{4}$

substitute

$tan(\theta)=\frac{\frac{3}{4}}{(+/-)\frac{\sqrt{7}}{4}}$

$tan(\theta)=(+/-)\frac{3}{\sqrt{7}}$

tangent theta equals plus or minus three over square root of seven

Simplify

$tan(\theta)=(+/-)3\frac{\sqrt{7}}{7}$

tangent theta equals plus or minus three times square root of seven over seven