Question

These are known: A. Sin(-x)=-Sinx B. Cos(-x)=cosx C.Cos(x+y)= cosxcosy - sinxsiny D.Sin(x+y)= sinxcosy + cosxsiny

1. Sin^2x + cos^2x= 1 (use C and cos0=1)

Answer 1

Please read what you have typed in here. Have you posed a question? What is your goal?

Answer 2

I think you're supposed to use C. Cos(x+y)= cosxcosy - sinxsiny and the fact that cos(0) = 1 to prove Sin^2x + cos^2x= 1

I've never seen this way to prove that, so I'm genuinely curious

Answer 3

Let us assume that \[\large\rm cos²x+sin²x=1 \ \]is true
We check for the authority of our assumption by equating \[\large\rm x=0\ \]\[\large\rm (cos0)²+(sin0)²=1\ \]\[\large\rm 1²=1\ \]\[\large\rm 1=1\ \]
Now we will prove this generally
\[\large\rm cos²(x+1)+sin²(x+1)=1\ \]
let us take y=1
\[\large\rm cos²(x+y)+sin²(x+y)=1\ \]\[\large\rm [cos(x)cos(y)]²+[sin(x)sin(y)]²+[sin(x)cos(y)]²+[cos(x)sin(y)]²=1\ \]\[\large\rm cos²(x)cos²(y)+sin²(x)cos²(y)+cos²(x)sin²(y)+sin²(x)sin²(y)=1\ \]\[\large\rm cos²(y)(cos²x+sin²x)+sin²(y)(cos²x+sin²x)=1\ \]
From our elementary hypothesis \[\large\rm cos²x+sin²x=1\ \]\[\large\rm cos²y+sin²y=1\ \]
Recall our earlier substitution of\[\large\rm y=1\ \]\[\large\rm cos²1+sin²1=1\ \]\[\large\rm 0.29+0.71=1\ \]\[\large\rm 1=1\ \]

Answer 4

Thus we have proved that the hypothesis sin²x + cos²x= 1 is true

Answer 5

You can even just expand by using (x+y) as the angle and then use sin²x=1-cos²x to simplify the expression and attain 1 as the final value