Question

Question #b2446

Answer 1

See below

Explanation:

`sinx=7/11`

Since `x` is in quadrant `"II"`, the angle is obtuse. This means that both `cosx` and `tanx` are negative.

`cosx=-sqrt(1-sin^2x)`

`tanx=sinx/cosx`

`cosx=-sqrt(1-(7/11)^2)=-6/11sqrt2`

`tanx=(7/11)/(-6/11sqrt2)=-7/12sqrt2`

In order to find `sin2x`, `cos2x` and `tan2x`, we need to use the double angle identities. These will be given below:

`sin2x=2sinxcosx=2(7/11)(-6/11sqrt2)=-84/121sqrt2`

`cos2x=2cos^2x-1=2(-6/11sqrt2)^2-1=23/121`

`tan2x=(2tanx)/(1-tan^2x)=(2(-7/12sqrt2))/(1-(-7/12sqrt2)^2)=-84/23sqrt2`

Answer 2

There are formulas for `sin2x, cos2x` and `tan2x` (they're called "Double-Angle Formulas")

`sin2x=2sinxcosx`
`cos2x=cos^2x-sin^2x=2cos^2x-1=1-sin^2x`
`tan2x=(2*tanx)/(1-tan^2x)`

If we are given `sinx`, we know two sides of the triangle: the hypotenuse and one leg. From there, using Pythagorean's Theorem `(a^2+b^2=c^2)`, we could find the remaining side, and thus find the ratios for both `tan` and `cos`. That will allow us to solve the Double Angle Formulas without knowing all the angles