Pythagorean theorem and triangles Question

Printable View

April 13th 2010, 05:16 PM
florx

Pythagorean theorem and triangles Question

http://i40.tinypic.com/16huma0.jpg

I need help with c) and d) and also e).

For c) they are asking for 2θ however I don't know what they did to get 1 - 2y^2.
d) Can you please explain to me what we do with the (pi - θ)?
e) Cos is adjacent over hypotenuse. However what do we do with the inverse of cos and y?

Please help me with these problems. Thank you so much in advance.
April 13th 2010, 08:04 PM
Gusbob

Quote:

Originally Posted by florx

http://i40.tinypic.com/16huma0.jpg

I need help with c) and d) and also e).

For c) they are asking for 2θ however I don't know what they did to get 1 - 2y^2.
d) Can you please explain to me what we do with the (pi - θ)?
e) Cos is adjacent over hypotenuse. However what do we do with the inverse of cos and y?

Please help me with these problems. Thank you so much in advance.

c) $\cos 2\theta = cos^2 \theta - sin^2 \theta$
d) $\sin (\pi - \theta) = \sin \theta$

"Why?" you ask. Imagine the unit circle if you will. Sine is positive in the first and second quadrant (between 0 and $\pi$ .) Now $\theta < 90^{\cdot}$ so $\sin (\pi - \theta)$ lies in the second quadrant. By symmetry, it has the value of $\sin \theta$ . Draw it out to see for yourself.

e) This one is trickier. Split it up into two parts. Let $sin^2 (arccos y) = sin^2 u$ where $u = arccos y$

Now $y = \cos u$ . Looking at your triangle the only angle that satisfies this equation is $u = \frac{\pi}{2} - \theta$

So we now have $sin^2 (\frac{\pi}{2} - \theta)$ . Look at your unit circle again and you'll realise that $\sin (\frac{\pi}{2} - \theta)$ is the same thing as $\cos \theta$
April 13th 2010, 08:25 PM
florx

Thank you so much for your input. I understand what you about the sin(pi-θ). However i still don't quite understand how to get the answer.

Taken from the back of the book,

c) 1 - 2y^2
d) y
e) 1 - y^2

Thanks again for trying to help me and I hope you can help me through this and solve this problem.
April 13th 2010, 08:30 PM
harish21

Quote:

Originally Posted by florx

Thank you so much for your input. I understand what you about the sin(pi-θ). However i still don't quite understand how to get the answer.

Taken from the back of the book,

c) 1 - 2y^2
d) y
e) 1 - y^2

Thanks again for trying to help me and I hope you can help me through this and solve this problem.

You know that $sin(\pi - \theta) = sin\theta$ , right?

so, if you look at the triangle,

$sin\theta = \frac{opposite}{hypotenuse}= \frac{y}{1} = y$

Go through this link if you still have confusions:

Trigonometric functions - Wikipedia, the free encyclopedia