  # sine problem

<p>how do you find amplitude of y=sin2x+cos2x without using graphing caculator??
and also, amplitude of (sinx)(sinx)+(cosx)(cosx)=y???</p>

<p>theyre both 1 the amplitude is the number that comes before the function. so if you had y=2sinx the amplitude would be 2.</p>

<p>casablanca is wrong, but i don't know how you figure it out...the first amp is the square root of 2, and that's what it is when adding sin(yx) and cos(yx) every time. of course, if it was isolated, as in just sinx or just cosx, casablanca would be correct.</p>

<p>also, cosx squared plus sinx squared always equals one. that's one you never need a calculator for.</p>

<p>Ok. I'm not sure if this works for everything, but it's one way I came to a solution.
Set the problem equal to 0, so you have 0 = sin2x + cos2x . Now you can subtract cos2x to get -cos2x = sin2x . I believe you can divide both sides by 2 since both angles are doubled, so you have -cosx = sinx. Now you just have to figure out where they are equal. -cosx is equal to sinx at 3pi/4 and 7pi/4; they are both positive in the first quadrant and negative in the fourth quadrant, which is what you want. The sin(3pi/4) is the square root 0f 2/2 or about .707, as well as -cos(3pi/4). Added together they are the square root of 2, or about 1.414. So, the amplitude is +/- the square root of 2.</p>

<p>I agree that sin^2x + cos^2x is 1; It's a trig identity.</p>