if tan θ = 2/5 for 0< θ <360^2, find the values of sin θ
1. $\displaystyle \tan(\theta)=\dfrac{\sin(\theta)}{\cos(\theta)} = \dfrac25$
2. $\displaystyle \sin^2(\theta) + \cos^2(\theta)=1$
3. Substitute:
$\displaystyle \tan(\theta)=\dfrac{\sin(\theta)}{\cos(\theta)} = \dfrac{\sin(\theta)}{\sqrt{1-\sin^2(\theta)}} = \dfrac25$
4. Solve for $\displaystyle \sin(\theta)$ .
Obviously solving equations isn't very familiar to you. So here we go:
$\displaystyle \dfrac{\sin(\theta)}{\sqrt{1-\sin^2(\theta)}} = \dfrac25$ Multiply both sides of the equation by the complete square-root:
$\displaystyle \sin(\theta) = \dfrac25 \cdot \sqrt{1-\sin^2(\theta)}$ Square both sides:
$\displaystyle \sin^2(\theta) = \dfrac4{25} \cdot (1-\sin^2(\theta))$ Expand the bracket:
$\displaystyle \sin^2(\theta) = \dfrac4{25}- \dfrac4{25} \cdot \sin^2(\theta)$ Collect like terms on one side:
$\displaystyle \dfrac{29}{25} \cdot \sin^2(\theta) = \dfrac4{25}$ Divide both sides by $\displaystyle \dfrac{29}{25}$ :
$\displaystyle \sin^2(\theta) = \dfrac4{29}$ Calculate the square root at both sides:
$\displaystyle | \sin(\theta) | = \sqrt{\dfrac4{29}} = \dfrac2{29} \cdot \sqrt{29}$
Due to the squaring in step #2 you'll get 2 solutions of this equation:
$\displaystyle \sin(\theta) = \dfrac2{29} \cdot \sqrt{29}~\vee~ \sin(\theta) = - \dfrac2{29} \cdot \sqrt{29}$
You must check if one or both are valid.
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.
Hello, deej813!
Since $\displaystyle \tan\theta$ is positive, $\displaystyle \theta$ is in Quadrant 1 or 3.if $\displaystyle \tan \theta = \tfrac{2}{5}\;\text{ for }0< \theta <360^o$, find the values of $\displaystyle \sin \theta$
We have: .$\displaystyle \tan\theta \:=\:\frac{2}{5} \:=\:\frac{opp}{adj}$
Assume $\displaystyle \theta$ is an acute angle.
Then $\displaystyle \theta$ is in a right triangle with: .$\displaystyle opp = 2,\;adj = 5$
Pythagorus says: .$\displaystyle hyp \:=\:\sqrt{2^2 + 5^2} \:=\:\sqrt{29}$
In Quadrant 1, we have: .$\displaystyle \begin{Bmatrix}opp &=& 2 \\ adj &=& 5 \\ hyp &=& \sqrt{29}\end{Bmatrix}$
. . Hence: .$\displaystyle \sin\theta \:=\:\frac{opp}{adj} \;=\;\frac{2}{\sqrt{29}}$
In Quadrant 3, we have: .$\displaystyle \begin{Bmatrix}opp &=& \text{-}2 \\ adj &=& \text{-}5 \\ hyp &=& \sqrt{29} \end{Bmatrix}$
. . Hence: .$\displaystyle \sin\theta \:=\:\frac{opp}{hyp} \:=\:\frac{-2}{\sqrt{29}}$
Therefore: .$\displaystyle \boxed{\sin\theta \;=\;\pm\frac{2}{\sqrt{29}}} $