MHB Prove Statement: View Attachment

[anil86]

Please view attachment!View attachment 1675

 

[MarkFL]

We are given:

$$\sin\left(\theta+i\phi \right)=\tan(x)+i \sec(x)$$

Using the angle-sum identity for sine on the left, we have:

$$\sin(\theta)\cosh(\phi)+i \cos(\theta)\sinh(\phi)=\tan(x)+i \sec(x)$$

This implies:

$$\sin(\theta)\cosh(\phi)=\tan(x)$$

$$\cos(\theta)\sinh(\phi)=\sec(x)$$

Now, we have the following double-angle identities:

$$\cos(2\theta)=\cos^2(\theta)-\sin^2(\theta)$$

$$\cosh(2\phi)=\cosh^2(\phi)+\sinh^2(\phi)$$

Hence:

$$\cos(2\theta)\cosh(2\phi)=\cos^2(\theta)\cosh^2( \phi)+\cos^2(\theta)\sinh^2(\phi)- \sin^2(\theta)\cosh^2(\phi)- \sin^2(\theta)\sinh^2(\phi)$$

Using the implications we drew above, we may write:

$$\cos(2\theta)\cosh(2\phi)=\cos^2(\theta)\cosh^2( \phi)+\sec^2(x)-\tan^2(x)- \sin^2(\theta)\sinh^2(\phi)$$

Using a Pythagorean identity, there results:

$$\cos(2\theta)\cosh(2\phi)=\cos^2(\theta)\cosh^2( \phi)+1- \sin^2(\theta)\sinh^2(\phi)$$

Using Pythagorean identities, we have:

$$\cos(2\theta)\cosh(2\phi)=\left(1-\sin^2(\theta) \right)\cosh^2( \phi)+1-\left(1-\cos^2(\theta) \right)\sinh^2(\phi)$$

We may arrange this as follows:

$$\cos(2\theta)\cosh(2\phi)=1+\left(\cosh^2( \phi)-\sinh^2(\phi) \right)+\left(\cos^2(\theta)\sinh^2(\phi)- \sin^2(\theta)\cosh^2(\phi) \right)$$

Using the hyperbolic identity and our previous results, we then find:

$$\cos(2\theta)\cosh(2\phi)=1+1+1=3$$