# cos(sin^(-1) v + cos^(-1) v)

• Nov 12th 2008, 07:18 AM
magentarita
cos(sin^(-1) v + cos^(-1) v)
Show that cos(sin^(-1) v + cos^(-1) v) = 0.
• Nov 12th 2008, 07:31 AM
Chris L T521
Quote:

Originally Posted by magentarita
Show that cos(sin^(-1) v + cos^(-1) v) = 0.

Let $\displaystyle \varphi = \sin^{-1}v$ and $\displaystyle \vartheta=\cos^{-1}v$

First, we can rewrite $\displaystyle \cos(\varphi+\vartheta)$ using the identity $\displaystyle \cos(\varphi+\vartheta)=\cos\varphi\cos\vartheta-\sin\varphi\sin\vartheta$

Its good to note that $\displaystyle \cos\vartheta=v$ and $\displaystyle \sin\varphi=v$ [why do you think this is the case?]. So our expression can be written as $\displaystyle v\cos\varphi-v\sin\vartheta$

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Now let's construct two different triangles, one that has an angle $\displaystyle \varphi$ and an angle $\displaystyle \vartheta$

Starting with $\displaystyle \varphi$, we see that the side opposite of the angle has a value of $\displaystyle \text{opp}=v$ and that the hypotenuse has a value of $\displaystyle \text{hyp}=1$. Using the pythagorean theorem, we can find the value of the adjacent side:

$\displaystyle (\text{opp})^2+(\text{adj})^2=(\text{hyp})^2\impli es (\text{adj})^2=(\text{hyp})^2-(\text{opp})^2\implies (\text{adj})^2=1^2-v^2$ $\displaystyle \implies \text{adj}=\sqrt{1-v^2}$

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Now, we construct at triangle for $\displaystyle \vartheta$. We see that the side adjacent of the angle has a value of $\displaystyle \text{adj}=v$ and that the hypotenuse has a value of $\displaystyle \text{hyp}=1$. Using the pythagorean theorem, we can find the value of the opposite side:

$\displaystyle (\text{opp})^2+(\text{adj})^2=(\text{hyp})^2\impli es (\text{opp})^2=(\text{hyp})^2-(\text{adj})^2\implies (\text{opp})^2=1^2-v^2$ $\displaystyle \implies \text{opp}=\sqrt{1-v^2}$

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Now what do you think $\displaystyle v\cos\varphi-v\sin\vartheta$ equals?

--Chris
• Nov 12th 2008, 07:35 AM
HallsofIvy
Quote:

Originally Posted by magentarita
Show that cos(sin^(-1) v + cos^(-1) v) = 0.

$\displaystyle sin^{-1}(v)$ and $\displaystyle cos^{-1}(v)$ are complementary angles! What is cos(90 degrees)?
• Nov 12th 2008, 07:44 AM
Chris L T521
Quote:

Originally Posted by HallsofIvy
$\displaystyle sin^{-1}(v)$ and $\displaystyle cos^{-1}(v)$ are complementary angles! What is cos(90 degrees)?

haha...wow...I'm surprised I didn't notice that! Good one!

The way I ended up doing this is acceptable, given that they didn't realize that $\displaystyle \sin^{-1}(v)$ and $\displaystyle \cos^{-1}(v)$ were complementary angles [Like I did (Rofl)]

--Chris
• Nov 12th 2008, 04:40 PM
magentarita
I see......
Quote:

Originally Posted by HallsofIvy
$\displaystyle sin^{-1}(v)$ and $\displaystyle cos^{-1}(v)$ are complementary angles! What is cos(90 degrees)?

Are you saying that the first reply is wrong?
• Nov 12th 2008, 04:41 PM
magentarita
this question...........
Quote:

Originally Posted by Chris L T521
haha...wow...I'm surprised I didn't notice that! Good one!

The way I ended up doing this is acceptable, given that they didn't realize that $\displaystyle \sin^{-1}(v)$ and $\displaystyle \cos^{-1}(v)$ were complementary angles [Like I did (Rofl)]

--Chris

This question is very involved.
• Nov 12th 2008, 06:30 PM
Chris L T521
Quote:

Originally Posted by magentarita
Are you saying that the first reply is wrong?

He's not saying its wrong. He just happened to point out a faster way to the answer. Both ways are correct! (Nod)

--Chris