Prove:cosh^2 (x) - sinh^2 (x) = 1 ?
The definitions of cosh(x) and sinh(x) are: cosh(x) = (e^(x) + e^(-x)) / 2 sinh(x) = (e^(x) - e^(-x)) / 2 cosh²(x) = (e^(2x) + 2 + e^(-2x)) / 4 sinh²(x) = (e^(2x) - 2 + e^(-2x)) / 4 cosh²(x) - sinh²(x) = (e^(2x) + 2 + e^(-2x))/4 - (e^(2x) - 2 + e^(-2x)) / 4 = (e^(2x) + 2 + e^(-2x) - e^(2x) + 2 - e^(-2x)) / 4 = 4/4 = 1
Prove cosh(2x) = cosh²x + sinh²x?
Use the definition. cosh(x) = [e^x+e^(-x)]/2 and sinh(x) = [e^x-e^(-x)]/2. So cosh(2x) = [e^(2x)+e^(-2x)]/2. Also cosh^2(x) + sinh^2(x) = [e^(2x)+2+e^(-2x)]/4 + [e^(2x)-2+e^(2x)]/4 = [2e^(2x) + 2e^(-2x)]/4 = [e^(2x)+e^(-2x)]/2. So they're the same.
How can i prove that : 1 - (tanh^2 x) = sech^2 x ??
Since (tanh^2 x) = (sinh^2 x)/(cosh^2 x), 1-(sinh^2 x)/(cosh^2 x)= [(cosh^2 x) - (sinh^2 x)]/(cosh^2 x) since (cosh^2 x) - (sinh^2 x) = 1 1/(cosh^2 x) = (sech^2 x) (sech^2 x) = (sech^2 x) Hope you understand.
How can 2cosh^2x-1 = cosh2x?
There are multiple approaches to verifying this. Probably the simplest thing is to use the ID for non-hyperbolic trig functions cos(2Θ) = 2cos²(Θ) - 1 and then just appeal to the fact that cosh(x) = cos(ix) where i = √(-1). Take the above identity and substitute Θ = ix; your identity falls out immediately. If you want to work a little harder, you can use exponentials. 2cosh²(x) - 1 = 2[(e^x + e^(-x))/2]² - 1 = ½[e^(2x) - 2 + e^(-2x)] - 1 = ½[e^(2x) + e^(-2x)] - 1 + 1 = ½[e^(2x) + e^(-2x)] = cosh(2x).
Prove the following hyperbolic identity sinh^2 x =1/2(cosh 2x-1)?
(i) sinh^2(2x) = [(1/2)(e^x - e^(-x))]^2 = (1/4)[e^(2x) - 2 + e^(-2x)]. (ii) (1/2)[cosh(2x) - 1] = (1/2)[(1/2)(e^(2x) + e^(-2x)) - 1] = (1/2)(1/2)[(e^(2x) + e^(-2x)) - 2], factoring 1/2 from each term = (1/4)[e^(2x) - 2 + e^(-2x)]. Since both (i) and (ii) reduce to the same result, the identity now follows. I hope this helps!