Hey everyone, I couldn't figure out how how to find the value for:

http://www.mymathforum.com/download/file.php?id=4517

I know the answer, but, I'm looking for the solution, and why.

Thanks

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- Jan 18th 2013, 07:45 AMImonarsFind the value
Hey everyone, I couldn't figure out how how to find the value for:

http://www.mymathforum.com/download/file.php?id=4517

I know the answer, but, I'm looking for the solution, and why.

Thanks - Jan 18th 2013, 08:42 AMBariothRe: Find the value
Hi you'll need your trigo circle to get this one.

If you look at the circle you'll notice that

$\displaystyle Sin(\frac{\pi}{2}-x)=cos(x)$

There is also a proof if you use the fact

$\displaystyle sin(a-b)=cos(b)sin(a)-cos(a)sin(b)$.

Try it now, let me know if you can't make it! - Jan 18th 2013, 09:00 AMImonarsRe: Find the value
Nope, sorry I have no idea how to do it. but, my professor did it like this:

1 + sin^2 79 cos^2 7

1 + 1 = 2

I think he did something wrong tho.. - Jan 18th 2013, 09:14 AMBariothRe: Find the value
- Jan 18th 2013, 09:19 AMPlatoRe: Find the value
- Jan 18th 2013, 11:17 AMImonarsRe: Find the value
I never copy wrong from the board, i always triple check my notes. This is not the first time he make these kinds of mistakes.. anyways Thanks guys I got it from Plato. and barioth, Yes I do understand that part, it's a formula

- Jan 18th 2013, 11:22 AMSorobanRe: Find the value
Hello, Imonars!

Quote:

$\displaystyle \text{Simplify: }\:1 + \sin^2(7^o) + \sin^2(83^o)$

$\displaystyle \text{Note that: }\:\sin(83^o) \;=\;\sin(90-7) \;=\;\sin(90^o)\cos)7^o - \cos(90^o)\sin(7^o)$

. . . . . . . . . . . . . . . $\displaystyle =\;1\!\cdot\!\cos(7^o) - 0\!\cdot\!\sin(7^o) \;=\;\cos(7^o)$

$\displaystyle \text{The expression becomes: }\:1 + \underbrace{\sin^2(7^o) + \cos^2(7^o)}_{\text{This is 1}} \;=\; 2$

- Jan 18th 2013, 11:25 AMImonarsRe: Find the value
That's it! Thank you all and thank you soroban that's exactly what I was looking for!