1. ## 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

2. ## 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!

3. ## 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..

4. ## Re: Find the value

mhm maybe you prof is wrong stick with me and we shall know!

First:
Originally Posted by Barioth
$Sin(\frac{\pi}{2}-x)=cos(x)$
Do you understand this?

5. ## Re: Find the value

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 $1+\cos^2(83^o)+\sin^2(83^o)=2$

6. ## 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

7. ## Re: Find the value

Hello, Imonars!

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

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

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

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

8. ## Re: Find the value

That's it! Thank you all and thank you soroban that's exactly what I was looking for!