# Math Help - Values of Trig Functions

1. ## Values of Trig Functions

Hello, 1st post.
A little help needed.
We're not allowed to use calculators, so if you could provide answers with working, great!

Sinθ = - 2/5 (negative two fifths) Range: π ≤ θ ≤ 3π/2 (π=pi)

Find Values for:
Cosθ
Tanθ

Im told i am to use the triangles, but i dont get how they work.
Help is appreciated.

2. Originally Posted by Privatealpha
Hello, 1st post.
A little help needed.
We're not allowed to use calculators, so if you could provide answers with working, great!

Sinθ = - 2/5 (negative two fifths) Range: π ≤ θ ≤ 3π/2 (π=pi)

Find Values for:
Cosθ
Tanθ

Im told i am to use the triangles, but i dont get how they work.
Help is appreciated.
The sine function is negative in Quadrants III and IV

but we only want the Values in Quadrant III becuase of the range given.

Using the pythagorean theorem we can solve for the other leg of the triangle

$(-2)^2+x^2=5^2 \iff 4+x^2=25 \iff x=\sqrt{21}$

Note that since we are in quadrant III the x coordinate is negative.

so we get $-\sqrt{21}$

$\tan(\theta)=\frac{opposite}{adjacent}=\frac{-2}{-\sqrt{21}}=\frac{2\sqrt{21}}{21}$

$\cos(\theta)=\frac{adjacent}{hypotenuse}=\frac{-\sqrt{21}}{5}

$

I hope this helps.

Good luck.

3. Thanks for that.
Working through my assignment, it goes on to ask:
Evaluate Sin2 and Cos2 with the values above.

Ive come up with:
Sin2 = 2sin cos
=> 4√21 / 25

Cos2 = Cos² - Sin²
=> 1

Could someone confirm or correct my answers?

4. Hello,

Originally Posted by Privatealpha
Thanks for that.
Working through my assignment, it goes on to ask:
Evaluate Sin2 and Cos2 with the values above.

Ive come up with:
Sin2 = 2sin cos
=> 4√21 / 25

Cos2 = Cos² - Sin²
=> 1

Could someone confirm or correct my answers?
Ok for $\sin 2x$

$\cos 2x=\cos^2 x-\sin^2 x=\left(\frac{-\sqrt{21}}{5}\right)^2-\left(\frac{-2}{5}\right)^2=\frac{21}{25}-\frac{4}{25} \neq 1$

Also, you can remember the two other formulas for $\cos 2x$ : $2 \cos^2 x-1=1-2 \sin^2 x$