MHB Simple Trigonometric Identities

[courtbits]

If (sinΘ) = 2/3 with Θ in quadrant 1, find (secΘ)

Θ = theta

Completely new at trigonometric identities, would be a great help!

 

[mathmari]

We have that $\sec (\theta)=\frac{1}{\cos (\theta)}$ and $\sin^2 (\theta)+\cos^2 (\theta)=1$.

Using these two identities we get:

$$\sec^2 (\theta)=\frac{1}{\cos^2 (\theta)}=\frac{1}{1-\sin^2 (\theta)}$$

 

[courtbits]

So what exactly is the answer, is the answer the sec^2 Θ =? to all those other identities or is the rest to solve by me? o.o" I'm a dunce I know.

 

[mathmari]

We are given that $\sin (\theta)=\frac{2}{3}$, so replacing this at the relation of post #2 we have:

$$\sec^2 (\theta)=\frac{1}{1-\sin^2 (\theta)}=\frac{1}{1-\left ( \frac{2}{3}\right )^2}=\frac{1}{1-\frac{4}{9}}=\frac{1}{\frac{9}{9}-\frac{4}{9}}=\frac{1}{\frac{5}{9}}=\frac{9}{5}$$

So, we have $\sec^2 (\theta)=\frac{9}{5}$.

To get the desired result, $\sec (\theta)$, you have to take the square root of the above equation.

 

[courtbits]

so does that mean i square root sec^2(Θ) so that they eliminate and then i square root 9/5?
OR am I completely off? >~>"

 

[mathmari]

Yes, you do the following:

$$\sec^2 (\theta)=\frac{9}{5} \Rightarrow \sqrt{\sec^2 (\theta)}=\sqrt{\frac{9}{5}} \Rightarrow \sec (\theta)=\pm \frac{3}{\sqrt{5}}$$

 

[courtbits]

Yeah, I'm still completely confused. However, I will read this over and over again until I understand. Thank you very much on helping my empty brain. e.e"

 

[MarkFL]

Yes, you do the following:

$$\sec^2 (\theta)=\frac{9}{5} \Rightarrow \sqrt{\sec^2 (\theta)}=\sqrt{\frac{9}{5}} \Rightarrow \sec (\theta)=\pm \frac{3}{\sqrt{5}}$$

Just a very minor quibble...since we are given that $\theta$ is a first quadrant angle, we then know to take the positive root. :)

 

[courtbits]

Just a very minor quibble...since we are give that $\theta$ is a first quadrant angle, we then know to take the positive root. :)

Wow, that did leave a bit of confusion because I had be precise with my answer. Thank you!

 

[Euge]

If (sinΘ) = 2/3 with Θ in quadrant 1, find (secΘ)

Θ = theta

Completely new at trigonometric identities, would be a great help!

Hi courtbits,

Since $\Theta$ is a first quadrant angle and $\sin \Theta = 2/3$, $\Theta$ is an acute angle, so we can draw a right triangle with one of the angles $\Theta$ such that the side opposite $\Theta$ is $2$ and the hypotenuse is $3$ (recall sine = opposite/hypotenuse). By the Pythagorean theorem, the side adjacent $\Theta$ is $\sqrt{3^2 - 2^2} = \sqrt{9- 4} = \sqrt{5}$. So then $$\sec \Theta = \frac{\mathrm{hypotenuse}}{\mathrm{adjacent}} = \frac{3}{\sqrt{5}}$$

Make sure you draw the picture yourself to see what's going on. :D