Answers>Maths>IB>Article

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
Answered by Eno A. Maths tutor

12294 Views

See similar Maths IB tutors

Related Maths IB answers

All answers ▸

Find the constant term in the binomial expansion of (3x + 2/(x^2))^33


The sum of the first n terms of an arithmetic sequence is Sn=3n^2 - 2n. How can you find the formula for the nth term un in terms of n?


Solve the equation (2 cos x) = (sin 2 x) , for 0 ≤ x ≤ 3π .


Solve x^2 + 2x = 48 for all values of x


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences