Question

The angle,2Θ, lies in the third quadrant such that cos2Θ=-2/5. Determine an exact value for tanΘ . Show your work including any diagrams if you plan to use them. (3 marks)

Answer 1

Answer:

$tan(\theta)=\frac{\sqrt{21}}{3}$

Step-by-step explanation:

1. Approach

One is given the following information:

$cos(2\theta)=-\frac{2}{5}$

One can rewrite this as:

$cos(2\theta)=-0.4$

Also note, the problem says that the angle ($2\theta$) is found in the third quadrant.

Using the trigonometric identities ($cos(2\theta)=2(cos^2(\theta))-1$) and ($cos(2\theta)=1-2(sin^2(\theta))$) one can solve for the values of ($cos(\theta)$) and ($sin(\theta)$). After doing so one can use another trigonometric identity ($tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$).  Substitute the given information into the ratio and simplify.

2. Solve for $(cos(\theta))$

Use the following identity to solve for ($cos(\theta)$) when given the value ($cos(2\theta)$).

$cos(2\theta)=2(cos^2(\theta))-1$

Substitute the given information in and solve for ($cos(\theta)$).

$cos(2\theta)=2(cos^2(\theta))-1$

$-0.4=2(cos^2(\theta))-1$

Inverse operations,

$-0.4=2(cos^2(\theta))-1$

$0.6=2(cos^2(\theta))$

$0.3=cos^2(\theta)$

$\sqrt{0.3}=cos(\theta)$

Since this angle is found in the third quadrant its value is actually:

$cos(\theta)=-\sqrt{0.3}$

3. Solve for $(sin(\theta))$

Use the other identity to solve for the value of ($sin(\theta)$) when given the value of ($cos(2\theta)$).

$cos(2\theta)=1-2(sin^2(\theta))$

Substitute the given information in and solve for ($sin(\theta)$).

$cos(2\theta)=1-2(sin^2(\theta))$

$-0.4=1-2(sin^2(\theta))$

Inverse operations,

$-0.4=1-2(sin^2(\theta))$

$-1.4=-2(sin^2(\theta))$

$0.7=sin^2(\theta)$

$\sqrt{0.7}=sin(\theta)$

Since this angle is found in the third quadrant, its value is actually:

$sin(\theta)=-\sqrt{0.7}$

4. Solve for $(tan(\theta))$

One can use the following identity to solve for $(tan(\theta))$;

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

Substitute the values on just solved for and simplify,

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

$tan(\theta)=\frac{-\sqrt{0.7}}{-\sqrt{0.3}}$

$tan(\theta)=\frac{\sqrt{0.7}}{\sqrt{0.3}}$

$tan(\theta)=\frac{\sqrt{\frac{7}{10}}}{\sqrt{\frac{3}{10}}}$

Rationalize the denominator,

$tan(\theta)=\frac{\sqrt{\frac{7}{10}}}{\sqrt{\frac{3}{10}}}$

$tan(\theta)=\frac{\sqrt{\frac{7}{10}}}{\sqrt{\frac{3}{10}}}*\frac{\sqrt{\frac{3}{`0}}}{\sqrt{\frac{3}{10}}}$

$tan(\theta)=\frac{\sqrt{\frac{7}{10}*\frac{3}{10}}}{\sqrt{\frac{3}{10}*\frac{3}{10}}}$

$tan(\theta)=\frac{\sqrt{\frac{21}{100}}}{\frac{3}{10}}$

$tan(\theta)=\frac{\frac{\sqrt{21}}{10}}{\frac{3}{10}}$

$tan(\theta)=\frac{\sqrt{21}}{10}*\frac{10}{3}$

$tan(\theta)=\frac{\sqrt{21}}{3}$