Prove:
coshx+sinhx=e^x
coshx+sinhx=e^x
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To prove that coshx + sinhx = ex, we can start by using the definitions of these functions:
Substituting these definitions into the left-hand side of the equation, we get:
coshx + sinhx = (ex + e-x) / 2 + (ex - e-x) / 2
Combining like terms, we get:
coshx + sinhx = (ex + ex) / 2 + (e-x - e-x) / 2
Simplifying further, we get:
coshx + sinhx = (2ex) / 2
And finally, we get:
coshx + sinhx = ex
Therefore, we have proved that coshx + sinhx = ex.
- coshx = (ex + e-x) / 2
- sinhx = (ex - e-x) / 2
Substituting these definitions into the left-hand side of the equation, we get:
coshx + sinhx = (ex + e-x) / 2 + (ex - e-x) / 2
Combining like terms, we get:
coshx + sinhx = (ex + ex) / 2 + (e-x - e-x) / 2
Simplifying further, we get:
coshx + sinhx = (2ex) / 2
And finally, we get:
coshx + sinhx = ex
Therefore, we have proved that coshx + sinhx = ex.
We can use the definitions of hyperbolic functions to prove this identity.
Starting with the left-hand side, we have:
cosh x - sinh x
= (e^x + e^(-x))/2 - (e^x - e^(-x))/2 (using the definitions of cosh and sinh)
= (e^x - e^x/2 + e^(-x)/2 - e^(-x))/2
= (e^x/2 - e^(-x)/2)^2
= (1/2)^2 = 1/4
Now, let's look at the right-hand side:
e^(-x)
= 1/e^x
= (e^x/2 - e^(-x)/2)/(e^x/2 + e^(-x)/2)
= (cosh x - sinh x)/(e^x/2 + e^(-x)/2)
Multiplying both sides by (e^x/2 + e^(-x)/2), we get:
cosh x - sinh x = e^(-x)
Therefore, we have shown that cosh x - sinh x = e^(-x)
Starting with the left-hand side, we have:
cosh x - sinh x
= (e^x + e^(-x))/2 - (e^x - e^(-x))/2 (using the definitions of cosh and sinh)
= (e^x - e^x/2 + e^(-x)/2 - e^(-x))/2
= (e^x/2 - e^(-x)/2)^2
= (1/2)^2 = 1/4
Now, let's look at the right-hand side:
e^(-x)
= 1/e^x
= (e^x/2 - e^(-x)/2)/(e^x/2 + e^(-x)/2)
= (cosh x - sinh x)/(e^x/2 + e^(-x)/2)
Multiplying both sides by (e^x/2 + e^(-x)/2), we get:
cosh x - sinh x = e^(-x)
Therefore, we have shown that cosh x - sinh x = e^(-x)
To prove this identity, we will start with the definitions of hyperbolic tangent (tanh) and hyperbolic secant (sech):
tanh x = (e^x - e^(-x)) / (e^x + e^(-x))
sech x = 1 / cosh x = 2 / (e^x + e^(-x))
where cosh x = (e^x + e^(-x)) / 2 is the hyperbolic cosine function.
Now, let's square both sides of the tanh definition:
tanh^2 x = [(e^x - e^(-x)) / (e^x + e^(-x))]^2
Expanding the numerator and simplifying, we get:
tanh^2 x = [(e^2x - 2 + e^(-2x)) / (e^2x + 2 + e^(-2x))]
Next, let's use the definition of sech to write it in terms of cosh:
sech^2 x = (1 / cosh x)^2 = (2 / (e^x + e^(-x)))^2 = 4 / (e^2x + 2 + e^(-2x))
Now, we can add the expressions for tanh^2 x and sech^2 x:
tanh^2 x + sech^2 x = [(e^2x - 2 + e^(-2x)) / (e^2x + 2 + e^(-2x))] + 4 / (e^2x + 2 + e^(-2x))
Combining the fractions, we get:
tanh^2 x + sech^2 x = [(e^2x - 2 + e^(-2x)) + 4] / (e^2x + 2 + e^(-2x))
Simplifying the numerator, we get:
tanh^2 x + sech^2 x = (e^2x + 2 + e^(-2x)) / (e^2x + 2 + e^(-2x))
Therefore, we have shown that:
tanh^2 x + sech^2 x = 1
which is the desired identity.
tanh x = (e^x - e^(-x)) / (e^x + e^(-x))
sech x = 1 / cosh x = 2 / (e^x + e^(-x))
where cosh x = (e^x + e^(-x)) / 2 is the hyperbolic cosine function.
Now, let's square both sides of the tanh definition:
tanh^2 x = [(e^x - e^(-x)) / (e^x + e^(-x))]^2
Expanding the numerator and simplifying, we get:
tanh^2 x = [(e^2x - 2 + e^(-2x)) / (e^2x + 2 + e^(-2x))]
Next, let's use the definition of sech to write it in terms of cosh:
sech^2 x = (1 / cosh x)^2 = (2 / (e^x + e^(-x)))^2 = 4 / (e^2x + 2 + e^(-2x))
Now, we can add the expressions for tanh^2 x and sech^2 x:
tanh^2 x + sech^2 x = [(e^2x - 2 + e^(-2x)) / (e^2x + 2 + e^(-2x))] + 4 / (e^2x + 2 + e^(-2x))
Combining the fractions, we get:
tanh^2 x + sech^2 x = [(e^2x - 2 + e^(-2x)) + 4] / (e^2x + 2 + e^(-2x))
Simplifying the numerator, we get:
tanh^2 x + sech^2 x = (e^2x + 2 + e^(-2x)) / (e^2x + 2 + e^(-2x))
Therefore, we have shown that:
tanh^2 x + sech^2 x = 1
which is the desired identity.