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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) $\displaystyle \cos 2\theta = cos^2 \theta - sin^2 \theta $Quote:
Originally Posted by florx
d) $\displaystyle \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 $\displaystyle \pi $.) Now $\displaystyle \theta < 90^{\cdot} $ so $\displaystyle \sin (\pi - \theta)$ lies in the second quadrant. By symmetry, it has the value of $\displaystyle \sin \theta $. Draw it out to see for yourself.
e) This one is trickier. Split it up into two parts. Let $\displaystyle sin^2 (arccos y) = sin^2 u $ where $\displaystyle u = arccos y $
Now $\displaystyle y = \cos u $. Looking at your triangle the only angle that satisfies this equation is $\displaystyle u = \frac{\pi}{2} - \theta $
So we now have $\displaystyle sin^2 (\frac{\pi}{2} - \theta) $. Look at your unit circle again and you'll realise that $\displaystyle \sin (\frac{\pi}{2} - \theta) $ is the same thing as $\displaystyle \cos \theta $
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 $\displaystyle sin(\pi - \theta) = sin\theta$, right?Quote:
Originally Posted by florx
so, if you look at the triangle,
$\displaystyle sin\theta = \frac{opposite}{hypotenuse}= \frac{y}{1} = y$
Go through this link if you still have confusions:
Trigonometric functions - Wikipedia, the free encyclopedia
