1. ## trigonometry; solve

if tan θ = 2/5 for 0< θ <360^2, find the values of sin θ

2. Originally Posted by deej813
if tan θ = 2/5 for 0< θ <360^2, find the values of sin θ
1. $\tan(\theta)=\dfrac{\sin(\theta)}{\cos(\theta)} = \dfrac25$

2. $\sin^2(\theta) + \cos^2(\theta)=1$

3. Substitute:

$\tan(\theta)=\dfrac{\sin(\theta)}{\cos(\theta)} = \dfrac{\sin(\theta)}{\sqrt{1-\sin^2(\theta)}} = \dfrac25$

4. Solve for $\sin(\theta)$ .

3. should i get to
25sin^2θ=1-sinθ
where do i go from there??

4. Originally Posted by deej813
hey ive really stuffed this up
can anyone show me what to do <<<< the only adequate answer to this question is: Yes!
Obviously solving equations isn't very familiar to you. So here we go:

$\dfrac{\sin(\theta)}{\sqrt{1-\sin^2(\theta)}} = \dfrac25$ Multiply both sides of the equation by the complete square-root:

$\sin(\theta) = \dfrac25 \cdot \sqrt{1-\sin^2(\theta)}$ Square both sides:

$\sin^2(\theta) = \dfrac4{25} \cdot (1-\sin^2(\theta))$ Expand the bracket:

$\sin^2(\theta) = \dfrac4{25}- \dfrac4{25} \cdot \sin^2(\theta)$ Collect like terms on one side:

$\dfrac{29}{25} \cdot \sin^2(\theta) = \dfrac4{25}$ Divide both sides by $\dfrac{29}{25}$ :

$\sin^2(\theta) = \dfrac4{29}$ Calculate the square root at both sides:

$| \sin(\theta) | = \sqrt{\dfrac4{29}} = \dfrac2{29} \cdot \sqrt{29}$

Due to the squaring in step #2 you'll get 2 solutions of this equation:

$\sin(\theta) = \dfrac2{29} \cdot \sqrt{29}~\vee~ \sin(\theta) = - \dfrac2{29} \cdot \sqrt{29}$

You must check if one or both are valid.

5. tangent function is positive in quadrants I and III. So, the hypotenuse is equal to square root of (5^2 + 2^2) = square root of 29. Then sin theta is equal to + 2/(sqrt of 29) and - 2/(sqrt of 29), because for sin theta, positive in quadrant I and negative in quadrant III.

6. Hello, deej813!

if $\tan \theta = \tfrac{2}{5}\;\text{ for }0< \theta <360^o$, find the values of $\sin \theta$
Since $\tan\theta$ is positive, $\theta$ is in Quadrant 1 or 3.

We have: . $\tan\theta \:=\:\frac{2}{5} \:=\:\frac{opp}{adj}$

Assume $\theta$ is an acute angle.
Then $\theta$ is in a right triangle with: . $opp = 2,\;adj = 5$
Pythagorus says: . $hyp \:=\:\sqrt{2^2 + 5^2} \:=\:\sqrt{29}$

In Quadrant 1, we have: . $\begin{Bmatrix}opp &=& 2 \\ adj &=& 5 \\ hyp &=& \sqrt{29}\end{Bmatrix}$

. . Hence: . $\sin\theta \:=\:\frac{opp}{adj} \;=\;\frac{2}{\sqrt{29}}$

In Quadrant 3, we have: . $\begin{Bmatrix}opp &=& \text{-}2 \\ adj &=& \text{-}5 \\ hyp &=& \sqrt{29} \end{Bmatrix}$

. . Hence: . $\sin\theta \:=\:\frac{opp}{hyp} \:=\:\frac{-2}{\sqrt{29}}$

Therefore: . $\boxed{\sin\theta \;=\;\pm\frac{2}{\sqrt{29}}}$