Re: Complementry Angle Theorem
intervade said:
I'm really confused on how to go about doing these.. First off can someone explain to me why sinA+cosA = 1? I'm using A for angle.
Secondly I have a problem:
sin80 * csc80...
I know that sin80 = cos10 and csc80 = sec10 but where do I go from there? The book gives me 1 for an answer, but I'm definitely not sure how they got it.
First of all, sin A + cos A is NOT equal to 1!
(sin A)^2 + (cos A)^2 = 1
Let's say you have a right triangle, with A as one of the acute angles, and C as the right angle. The side opposite angle A is "a", the hypotenuse is "c", and if B is the other acute angle, the side opposite angle B is "b".
sin A = opposite leg/hypotenuse, or a/c
cos A = adjacent leg/hypotenuse, or b/c
(sin A)^2 + (cos A)^2 = (a/c)^2 + (b/c)^2
(sin A)^2 + (cos A)^2 = (a^2 /c^2) + (b^2/c^2)
And (a^2 / c^2) + (b^2 / c^2) = (a^2 + b^2) / c^2............the denominators of the fractions are the same. Add the numerators, and put the result over the common denominator.
Now, if "a" and "b" are the legs of a right triangle, and "c" is the hypotenuse, the Pythagorean theorem tells us that a^2 + b^2 = c^2.
So, (a^2 + b^2) / c^2 can be written as c^2 / c^2, which is 1.
Therefore, (sin A)^2 + (cos A)^2 = 1
For your second question, remember that csc A = 1 / sin A
So, sin 80 * csc 80 is the same thing as sin 80 * (1 / sin 80)
Just as 3 * (1/3) = 1, sin 80 * (1/sin 80) = 1
