Question

Suppose cos(x) =1/(sqrt(5))
and sin(x) >0. what is the value of tan(2x)?

Answer 1

`tan(2\theta) = -(4)/(3)\\`

Explanation:

So cos is defined as:
`cos(\theta) = (adjacent)/(hypotenuse)`, meaning we can tell that the adjacent side is 1, and the hypotenuse is 5, from the fraction you gave.

Using this we can solve for the opposite side.

`1^2 + b^2 = √(5)^2\\1+b^2 = 5\\b^2=4\\b=2`

Now it's important to note, that b can be a negative number, so we have to use the information that sin(x) > 0, to determine the length of this side.

The sin is defined as:
`sin(\theta) = (opposite)/(hypotenuse)`, and since we we're solving for the opposite side, this means that the value +\- 2, is in the top, and since the hypotenuse is positive, this means that the opposite side is also positive.

This also tells us one more thing, since both cos(x) and sin(x) are positive, we are dealing with a angle in the first quadrant.

So we can now define sin(x), using the opposite (2) and the hypotenuse (sqrt(5))

`sin(\theta) = (2)/(√(5))`

And we can rationalize the denominator for both the cosine and sine, by multiplying by the square root in the denominator so that

`sin(\theta) = (2√(5))/(5)\\\\cos(\theta) = (√(5))/(5)`

Now we can define the value of tan(2 theta) using the double angle-identities such that:

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

And we can also define tan(theta) using the definition that:

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

So plugging in the values sin(theta) and cos(theta) we get the following:

`tan(\theta) = ((2√(5))/(5))/((√(5))/(5))\\\\tan(\theta) = (2√(5))/(5) * (5)/(√(5))\\\\tan(\theta) = 2`

Btw in the last step, I just canceled out the 5 and sqrt(5) since they were both in the denominator and numerator

So now let's plug this value, 2 as tan(theta) into the equation

`tan(2\theta) = (2\ *2)/(1-2^2)\\\\tan(2\theta) = (4)/(-3)\\tan(2\theta) = -(4)/(3)\\`