# Thread:

1. ## Pythagorean theorem and triangles Question

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?

2. Originally Posted by florx

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?

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 $\cos \theta$

3. 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.

4. 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
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