Question

Question #e5ad5

Answer 1

For an angle `A` in the second quadrant,
`sin(A)=4/5`
`cos(A)=-3/5`
`sin(2A)=-24/25`
`cos(2A)=-7/25`
`tan(2A)=24/7`

Explanation:

We are given that `sin(A)=4/5` and that angle `A` is in the second quadrant. Recall that the cosine value of an angle in the second quadrant is always negative.

Then, using the identity `sin^2(theta)+cos^2(theta)=1` and remembering that `cos(A)` is negative, we have
`cos(A)=-sqrt(1-sin^2(theta))`
`\ \ \ \ \ \ \ " "=-sqrt(1-(4/5)^2)`
`\ \ \ \ \ \ \ " "=-3/5`

So now we know that `sin(A)=4/5` and `cos(A)=-3/5`. Then, the problem asks us to find `sin(2A)` and `cos(2A)`. To do so, we need the double-angle identities:

`sin(2A)=2sin(A)cos(A)=2*4/5*(-3/5)=-24/25`
`cos(2A)=cos^2(A)-sin^2(A)=(-3/5)^2-(4/5)^2=-7/25`

Finally, we want to demonstrate that `tan(2A)=24/7`. Remember that `tan(theta)` is defined as `sin(theta)/cos(theta)`. Therefore,
`tan(2A)=sin(2A)/cos(2A)=(-24/25)/(-7/25)=24/7`.

Answer 2

`"see explanation"`

Explanation:

`"using the "color(blue)"trigonometric identities"`

`•color(white)(x)sin^2A+cos^2A=1`

`•color(white)(x)sin2A=2sinAcosA`

`•color(white)(x)cos2A=cos^2A-sin^2A`

`tan2A=(sin2A)/(cos2A)`

`•color(white)(x)cosA=+-sqrt(1-sin^2A)`

`"since A is in second quadrant then cos is negative"`

`rArrcosA=-sqrt(1-(4/5)^2)`

`color(white)(rArrcosA)=-sqrt(1-16/25)=-sqrt(9/25)=-3/5`

`•color(white)(x)sin2A=2xx4/5xx-3/5=-24/25`

`•color(white)(x)cos2A=(-3/5)^2-(4/5)^2`

`color(white)(xxxxxxx)=9/25-16/25=-7/25`

`•color(white)(x)tan2A=(-24/25)/(-7/25)`

`color(white)(xxxxxxx)=-24/25xx-25/7=24/7`