Question

Use the pythagorean identity to simplify the following expression: cos^2*50 degrees +cos^2*40 degrees.

I think it has something to do with the formula: sin(90-A)=cos A

Please help showing steps :)

Answer 1

You're quite right in saying that
sin(90 - A) = cos A
But it is also true that
cos(90 - A) = sin A

So go ahead and implement that idea, and replace one of the cosines there with a sine...

Answer 2

I'm not sure how to use the rule, would it be something along the lines of: cos^2(90-50)=sin 40^2?

Answer 3

More like this...
\[\huge \cos^2(40^o)=\cos^2(90^o-50^o)=\color{red}?\]

Answer 4

So cos^2*40=cos^2*4=1?

Answer 5

No, see, you use this identity...
\[\huge \cos(90^o - A )=\sin(A)\]

Answer 6

By that logic,
\[\huge \cos(90^o - 50^o)=\color{red}?\]

Answer 7

cos (90 degrees - 50 degrees)= sin(40)? Sorry if I'm being slow :/

Answer 8

Look more carefully at the parts I emphasize...
\[\huge \cos(90^o - \color{blue}A )=\sin(\color{blue}A)\]

Answer 9

so sin(50)?

Answer 10

Yes.
And therefore..
\[\huge \cos^2(40^o)=\cos^2(90^o-50^o)=\color{red}?\]

Answer 11

cos^2(40)=sin^2(50)?

Answer 12

Most excellent :)
Therefore...
cos^2(40) + cos^2(50) = ?

Answer 13

Do I just work out sin^2(50)?

Answer 14

No need.
It becomes
\[\huge \cos^2(50^o)+\sin^2(50^o)\]

right?

Answer 15

Ohh so then I use the rule sin^2A+cos^2A=1?

Answer 16

and \[\large \cos^2 \theta + \sin^2 \theta =...?\]

Answer 17

Yes :)

Answer 18

Yes, @thesecret20111
precisely :)

Answer 19

Meaning the final answer is 1?

Answer 20

Yup.

Answer 21

Thanks so much! You're awesome! :)

Answer 22

Well, you know, we try :D

Answer 23

Which you can also confirm yourself by calculation: http://www.wolframalpha.com/input/?i=cos%5E2%2850%29+%2Bcos%5E2%2840%29

Answer 24

Ok thanks again guys :)