Prove that (sinx)^2 + (cosx)^2 = 1
We start with the definitions of sine and cosine, which are, respectively: sinx = opposite/hypoteneuse and cosx = adjacent/hypoteneuse. We then square the analyzed expressions to get the following:
(opposite ^2)/(hypoteneuse ^2) + (adjacent ^2)/(hypoteneuse ^2)
And since the denominators are the same, we can add the fractions to get:
(opposite ^2) + (adjacent ^2) / (hypoteneuse ^2)
But recall the Pythagorean Theorem, according to which: (opposite ^2) + (adjacent ^2) = (hypoteneuse ^2). So we get:
[(hypoteneuse ^2)] / (hypoteneuse ^2) = 1. QED.
EA
Related Maths IB answers
All answers ▸Prove 2^(n+2) + 3^(2n+1) is a multiple of 7 for all positive integers of n by mathematical induction.
Given that sin(x) + cos(x) = 2/3, find cos(4x)
When finding single or multiple probabilities using the binomial distribution on the calculator, which function do I use respectively?
Find the area under the curve of f(x)=4x/(x^2+1) form x = 0 to x = 2.