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

Part A)
`sin(2\theta)=(24)/(25)`

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

Part C)
`tan(2\theta)=-(240)/(161)`

Explanation:

Part A) we have
`cos(\theta)=-(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

`(-(4)/(5))^(2)+sin^(2) (\theta)=1`

`((16)/(25))+sin^(2) (\theta)=1`

`sin^(2) (\theta)=1-(16)/(25)`

`sin^(2) (\theta)=(9)/(25)`

`sin(\theta)=-(3)/(5)` ---> remember that the sine is negative (3 quadrant)

Find
`sin(2\theta)`

we have

`cos(\theta)=-(4)/(5)`

`sin(\theta)=-(3)/(5)`

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

substitute

`sin(2\theta)=2(-(3)/(5))(-(4)/(5))`

`sin(2\theta)=(24)/(25)`

Part B) we have
`cos(\theta)=(√(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((√(2))/(2) )^(2)-1`

`cos(2\theta)=0`

Part C) we have
`sin(\theta)=(8)/(17)`

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

Find
`tan(2\theta)`

we know that

`tan(2\theta)=(2tan(\theta))/(1-tan^(2) (\theta))`

Remember that

`cos^(2) (\theta)+sin^(2) (\theta)=1`

substitute

`cos^(2) (\theta)+((8)/(17))^(2)=1`

`cos^(2) (\theta)=1-(64)/(289)`

`cos^(2) (\theta)=(225)/(289)`

`cos(\theta)=-(15)/(17)`

Find
`tan(\theta)`

`tan(\theta)=(sin(\theta))/(cos(\theta))`

substitute

`tan(\theta)=((8/17))/((-15/17))`

`tan(\theta)=-(8)/(15)`

Find
`tan(2\theta)`

`tan(2\theta)=(2tan(\theta))/(1-tan^(2) (\theta))`

substitute

`tan(2\theta)=(2(-(8)/(15)))/(1-(-(8)/(15))^(2))`

`tan(2\theta)=((-(16)/(15)))/(1-((64)/(225)))`

`tan(2\theta)=((-(16)/(15)))/(1-(64)/(225))`

`tan(2\theta)=((-(16)/(15)))/((161)/(225))`

`tan(2\theta)=-(240)/(161)`