use trigonometric and hyperbolic identities to solve and simplify the following:

4sin^2 theta +5cos theta =5
3tan^2 theta+5=7sec theta
prove that cosh^2x-sinh^2x=1
prove that sinh2x=2sinhxcoshx

any help with this would be much appreciated

thanks

Andy

1. Use the pythagorean identity, cos^2A + sin^2A = 1

You may use a substitution like cos A = x, then solve like a quadratic, plug back the trig function and solve for the angle.

2. Use another identity here. tan^2A + 1 =sec^2A and proceed with the same method as above.

Sorry, I have not yet done hyperbolic identities. :(

Hi
I don't understand trig identities at all please could you explain it step by step

Thanks

Andy

Well, it's just like a substitution.

Ok, I'll show you the first one:

4sin^2 \theta + 5 cos \theta = 5

sin^2\theta + cos^2\theta = 1

I'll use those two.

sin^2\theta + cos^2\theta = 1

sin^2\theta = 1 - cos^2\theta

Are you okay with that?
If yes, then watch this:

4(1 - cos^2\theta)+ 5 cos \theta = 5

Did you see? I replaced the sin^2 theta by the identity that I derived from the Pythagorean identity.

Now, expand;

4 - 4cos^2\theta+ 5 cos \theta = 5

Simplify;

4cos^2\theta- 5 cos \theta +1= 0

As I told you before, make use of a substitution, so as not to confuse yourself. Say, cos theta = x.

4(x^2)- 5(x) +1= 0

I'm keeping the terms I just substituted in brackets for you to see them.
Then solve like a quadratic. I guess you can do that. You'll have then:

4x - 1 = 0
x = 0.25

and

x - 1 = 0
x = 1

From those two, replace them by the initial trigonometric ratio:

cos\theta = 0.25 and cos\theta = 1

It's simple then. Grab your calculator, find the corresponding two values for each equation.

You should have: x = 0, 60, 300 or 360 (for x between 0 and 360 degrees inclusive)

prove that cosh^2x-sinh^2x=1
prove that sinh2x=2sinhxcoshx


These can be proved by simply substituting for the definitions of cosh(x) and sinh(x):


cosh(x) = \frac 1 2 (e^x + e ^{-x}) \\
sinh(x) = \frac 1 2 (e^x - e ^{-x})


So, for example:


cosh^2(x) - sinh^2 (x) = (\frac { e^x + e ^ {-x}} 2)^2 - ( \frac { e^x - e ^ {-x}} 2)^2 = \frac {e^ {2x} + 2 +e^ {-2x} - ( e^ {2x} - 2 + e^ {-2x})} 4


you can take it from here. The last proof can demonstrated using the same technique.

This was a double post, ebaines. Just to let you know.

https://www.askmehelpdesk.com/math-sciences/use-trigonometric-hyperbolic-identities-solve-simplify-following-432701.html