# Complementary Angle Theorem

• Aug 6th 2006, 09:29 AM
spiritualfields
Complementary Angle Theorem
There is an effect of the complementary angle theorom that I don't understand. For example:

sin squared(40) + sin squared(50) = sin squared(40) + cos squared(40) = 1

This previous statement's purpose was to show how sin(50) = cos(40), but what got me was the answer being equal to 1. Doing some experimenting, to see if this is a rule:

sin squared(10) + sin squared(80) = sin squared(10) + cos squared(10) = 1

and so on:

cos squared(10) + cos squared(80) = 1
sin squared(30) + sin squared(60) = 1

Why is it, when these functions are squared, that the answer = 1?
• Aug 6th 2006, 10:03 AM
CaptainBlack
Quote:

Originally Posted by spiritualfields
There is an effect of the complementary angle theorom that I don't understand. For example:

sin squared(40) + sin squared(50) = sin squared(40) + cos squared(40) = 1

This previous statement's purpose was to show how sin(50) = cos(40), but what got me was the answer being equal to 1. Doing some experimenting, to see if this is a rule:

sin squared(10) + sin squared(80) = sin squared(10) + cos squared(10) = 1

and so on:

cos squared(10) + cos squared(80) = 1
sin squared(30) + sin squared(60) = 1

Why is it, when these functions are squared, that the answer = 1?

Pythagoras' theorem in terms of trig functions is:

$\displaystyle \sin^2(x)+\cos^2(x)=1$

RonL
• Aug 6th 2006, 10:11 AM
spiritualfields
Okay, I see it now. Thanks.
• Aug 6th 2006, 10:23 AM
earboth
Quote:

Originally Posted by spiritualfields
There is an effect of the complementary angle theorom that I don't understand. For example:

sin squared(40) + sin squared(50) = sin squared(40) + cos squared(40) = 1
...

Hello,

I've attached a diagram to show you what I've calculated.

If the sum of the 2 angles is 90° you are dealing with a right triangle.

$\displaystyle \alpha+\beta=90^\circ \Longrightarrow \beta=90^\circ-\alpha$

The red line corresponds to $\displaystyle \sin(\alpha)$ and the blue line corrsponds to $\displaystyle \sin(\beta)=sin(90^\circ-\alpha)$

According to the diagram is: $\displaystyle (\mbox{red})^2+(\mbox{blue})^2=1$

Plug in the sine values and you'll get the property you have detected.

The blue line corresponds to $\displaystyle \cos(\alpha)$. So now you can easily complete the Pythagoran rule of the Sine and Cosine function.

Greetings

EB
• Aug 6th 2006, 10:58 AM
spiritualfields
spiritualfields
earboth, thanks. What program did you use to draw that?
• Aug 7th 2006, 01:38 AM
earboth
Quote:

Originally Posted by spiritualfields
earboth, thanks. What program did you use to draw that?

Hello,

have a look here: http://www.dynageo.de

Greetings

EB