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Hey all,

My "homework" for maths is to prove two things in the following triangle:

Triangle:
1)
(sin α)^2 + (cos α)^2 = 1 (always)

2)
tan α = 1/tan (90 - α)

I hope someone can help. I will get 2 A+'s if I could prove them both, which would make my average skyrocket. I will give Rep++ ofcourse :)
-Laurens.

well for the first 1 im not 100% how you would "prove" it but i belive the reson is because sine starts at 0 and cos folows the same curve at 90 so if you have an angle and you (sin a)^2 the angle you will get a figure lets say the angle is 50 then the result is 0.5868 and if you cos it you basicly get the same just added 90 deg's which will always be the other half to make it 1 so in this case (cos 50)^2 is 0.4132. i think this is A level? not sure and ive not covered this topic yet.. but from looking at it i know this is gonna be the anser im just not sure how the question whants me to "prove" it. correct me if im wrong any1, hope this helps in some way

SOH CAH TOA
Sin(x)=o/h
Or in this case a/b, so replace sin(x) with a/b. Do the same for cos.
Then solve the equation for b^2. In that form it should look very familiar.

For the second, do something similar. Replace tan(x) with o/a (a/c in this case). 90-x is the other angle, as all angles add up to 180, and the square is 90. So replace tan(90-x) with o/a for that angle, which is c/a. Then solve/reduce til you get 1=1.

The second one involves an identity, does it not? I don't remember what tan(90 - x) expands to, but it should expand to something easier to work with.

He can use either

the identities cot(x)=1/tan(x), and tan(90-x)=cot(x)
or
Sum of angles of triangle = 180, and basic sohcahtoa

However the first problem itself is an identity, so maybe that means they want him to avoid using higher level identities.

Okay thanks all, I'm going to school now and I wrote both bazzbarret's and Boreas' theory down :) Bye.

Put in some angles in for sin and cos. sin pie/4 and cos pie/4 = root 2 over 2 square those values and you get 2/4. 2/4 + 2/4 = 4/4 = 1. Use any angles they all work.