How is cos(90 - [degrees]) equivalent to sin([degrees])?

[Tigertigre2000]

I don't understand how..... cos(90-degree) is equivalent to sin degree ?
Can someone try explaining this?

Thanks, I'd appriciate it.

 

[stapel]

Look at the graphs. Take the sine graph, and move it back by ninety degrees (one-fourth of a period). What do you get?

Eliz.

 

[Mrspi]

I don't understand how..... cos(90-degree) is equivalent to sin degree ?
Can someone try explaining this?

Thanks, I'd appriciate it.

Draw a right triangle, with right angle at C. side c is the hypotenuse, and is opposite angle C. A and B are acute angles, with side a opposite angle A, and side b opposite angle B.

Ok...

sin A = a / c
cos B = a / c

sin A = cos B, right?

And how are angles A and B related?

m<A + m<B + m<C = 180
Since we know that <C is a right angle, m<C = 90 and

m<A + m<B + 90 = 180
m<A + m<B = 90
m<A = 90 - m<B

we just established that
sin A = cos B
sin (90 - m<B) = cos B

Does that help?

 

[Tigertigre2000]

Yes, it does. Thanks for explaining it.

 

[skeeter]

two angles are complementary if their sum is 90 degrees.

if \(\displaystyle \theta\) is an angle, then \(\displaystyle 90-\theta\) is its complement.

the word cosine is the shortened version of the term complementary sine, or sine of an angle's complement.