Question

a. If cosθ=−45 where θ is in quadrant 3, find sin2θ. b. If cosθ=2√2 where θ is in quadrant 1, find cos2θ. c. If sinθ=817 where θ is in quadrant 2, find tan2θ.

Answer 1

Answer:

Part A) $sin(2\theta)=\frac{24}{25}$

Part B) $cos(2\theta)=0$

Part C) $tan(2\theta)=-\frac{240}{161}$

Step-by-step explanation:

Part A) we have $cos(\theta)=-\frac{4}{5}$

θ is in quadrant 3 ----> the sine is negative

Find $sin(2\theta)$

we know that

$sin(2\theta)=2sin(\theta)cos(\theta)$

Remember that

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

substitute

$(-\frac{4}{5})^{2}+sin^{2} (\theta)=1$

$(\frac{16}{25})+sin^{2} (\theta)=1$

$sin^{2} (\theta)=1-\frac{16}{25}$

$sin^{2} (\theta)=\frac{9}{25}$

$sin(\theta)=-\frac{3}{5}$ ---> remember that the sine is negative (3 quadrant)

Find $sin(2\theta)$

we have

$cos(\theta)=-\frac{4}{5}$

$sin(\theta)=-\frac{3}{5}$

$sin(2\theta)=2sin(\theta)cos(\theta)$

substitute

$sin(2\theta)=2(-\frac{3}{5})(-\frac{4}{5})$

$sin(2\theta)=\frac{24}{25}$

Part B) we have $cos(\theta)=\frac{\sqrt{2}}{2}$

θ is in quadrant 1

Find $cos(2\theta)$      

we know that

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

substitute

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

$cos(2\theta)=0$

Part C) we have $sin(\theta)=\frac{8}{17}$

θ is in quadrant 2 ----> the cosine is negative

Find $tan(2\theta)$  

we know that

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

Remember that

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

substitute

$cos^{2} (\theta)+(\frac{8}{17})^{2}=1$

$cos^{2} (\theta)=1-\frac{64}{289}$

$cos^{2} (\theta)=\frac{225}{289}$

$cos(\theta)=-\frac{15}{17}$

Find $tan(\theta)$  

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

substitute

$tan(\theta)=\frac{(8/17)}{(-15/17)}$

$tan(\theta)=-\frac{8}{15}$

Find $tan(2\theta)$  

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

substitute

$tan(2\theta)=\frac{2(-\frac{8}{15})}{1-(-\frac{8}{15})^{2}}$

$tan(2\theta)=\frac{(-\frac{16}{15})}{1-(\frac{64}{225})}$

$tan(2\theta)=\frac{(-\frac{16}{15})}{1-\frac{64}{225}}$

$tan(2\theta)=\frac{(-\frac{16}{15})}{\frac{161}{225}}$

$tan(2\theta)=-\frac{240}{161}$