# Find the value

• Jan 18th 2013, 07:45 AM
Imonars
Find the value
Hey everyone, I couldn't figure out how how to find the value for:

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

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

If you look at the circle you'll notice that

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

There is also a proof if you use the fact

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

Try it now, let me know if you can't make it!
• Jan 18th 2013, 09:00 AM
Imonars
Re: 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 AM
Barioth
Re: Find the value
mhm maybe you prof is wrong stick with me and we shall know!

First:
Quote:

Originally Posted by Barioth
$Sin(\frac{\pi}{2}-x)=cos(x)$

Do you understand this?
• Jan 18th 2013, 09:19 AM
Plato
Re: Find the value
Quote:

Originally Posted by Imonars
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..

I think you may have copied incorrectly.

$\sin(7^o)=\cos(83^o)$ so $1+\cos^2(83^o)+\sin^2(83^o)=2$
• Jan 18th 2013, 11:17 AM
Imonars
Re: 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 AM
Soroban
Re: Find the value
Hello, Imonars!

Quote:

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

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

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

$\text{The expression becomes: }\:1 + \underbrace{\sin^2(7^o) + \cos^2(7^o)}_{\text{This is 1}} \;=\; 2$
• Jan 18th 2013, 11:25 AM
Imonars
Re: Find the value
That's it! Thank you all and thank you soroban that's exactly what I was looking for!