Question

Question #fe8b2

Answer 1

`sin 2x = 10/26 = 5/13`

Explanation:

Use trig identity:
sin 2x = 2sin x.cos x.
First, find cos x by the trig identity:
`cos^2 x = 1/(1 + tan^2 x)`
In this case:
`cos^2 x = 1/(1 + 1/25) = 25/26`
`cos x = +- 5/sqrt26`
There are 2 values of cos x because `tan x = 1/5` --> x is either in Quadrant 1 or in Quadrant 3.
Next find sin x.
`sin^2 x = 1 - cos^2 x = 1 - 25/26 = 1/26`
`sin x = +- 1/sqrt26`
Finally:
`sin 2x = +- 2(5/sqrt26)(1/sqrt26) = 10/26 = 5/13`
Remark.
If x is in Q. 1 --> 2x is in Q. 2 --> sin 2x is positive.
If x is in Q. 3 --> 2x is in Q. 1 or Q. 22 --> sin 2x is also positive

Answer 2

Given:

`tan(x) = -1/5`

Because the tangent is negative, we know that `x` is in the 2nd or 4th quadrant and one but not both of the `sin(x)` and `cos(x)` must be negative.

We know that `sin(2x) = 2sin(x)cos(x)`, therefore, `x` must be in the 2nd quadrant and `2x` must be in the 4th quadrant.

We can start with the identity `1+tan^2(x) = sec^2(x)` derive and equation for `cos(x)`:

`1+tan^2(x) = sec^2(x)`

`1+tan^2(x) = 1/cos^2(x)`

`cos^2(x)= 1/(1+tan^2(x))`

`cos(x)= +-1/sqrt(1+tan^2(x))`

We deduced that `x` is in the 2nd quadrant, therefore, we choose the negative value for the cosine function:

`cos(x)= -1/sqrt(1+tan^2(x))`

Substitute the given value for the tangent function;

`cos(x)= -1/sqrt(1+(-1/5)^2)`

`cos(x)= -1/sqrt(25/25+1/25)`

`cos(x)= -1/sqrt(26/25)`

`cos(x)= (-5sqrt26)/26`

Use the identity, `tan(x) = sin(x)/cos(x)`, to find the value of `sin(x)`

`sin(x) = (-1/5)(-5sqrt26)/26`

`sin(x) = sqrt26/26`

`sin(2x) = 2sin(x)cos(x)`

`sin(2x) = 2(sqrt26/26)((-5sqrt26)/26)`

`sin(2x) =-10/26`

`sin(2x) =-5/13`