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

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

Show that cos(sin^(-1) v + cos^(-1) v) = 0.

2. Originally Posted by magentarita
Show that cos(sin^(-1) v + cos^(-1) v) = 0.
Let $\varphi = \sin^{-1}v$ and $\vartheta=\cos^{-1}v$

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

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

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

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

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

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

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

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

--Chris

3. Originally Posted by magentarita
Show that cos(sin^(-1) v + cos^(-1) v) = 0.
$sin^{-1}(v)$ and $cos^{-1}(v)$ are complementary angles! What is cos(90 degrees)?

4. Originally Posted by HallsofIvy
$sin^{-1}(v)$ and $cos^{-1}(v)$ 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 $\sin^{-1}(v)$ and $\cos^{-1}(v)$ were complementary angles [Like I did ]

--Chris

5. ## I see......

Originally Posted by HallsofIvy
$sin^{-1}(v)$ and $cos^{-1}(v)$ are complementary angles! What is cos(90 degrees)?
Are you saying that the first reply is wrong?

6. ## this question...........

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 $\sin^{-1}(v)$ and $\cos^{-1}(v)$ were complementary angles [Like I did ]

--Chris
This question is very involved.

7. 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!

--Chris