Locked on 2 'prove identity' questions for days now

[Mathlocked]

I been trying to work these 2 problems for what seems forever

1) Prove: Sec A + Tan A = Tan(pi / 4 + A / 2)

I leave the Sec A + Tan A alone and work the Tan( pi / 4 + A / 2) part.

I keep getting to (1 + Tan A / 2) / (1 - Tan A / 2) and getting stuck there.

The next one been kick my butt for days

2) Prove: (1 + Cos theta + iSin theta)^n + (1 + Cos theta - i Sin theta)^n

= 2^(n+1) Cos^n theta/2 Cos n(theta)/2

I am completely lost in the woods. Even a gentle push in the right direction would be appriecated. Thank you!

 

[soroban]

Hello, Mathlocked!

1) Prove: \(\displaystyle \,\sec A \,+ \,\tan A \;= \;\tan\left(\frac{\pi}{4}\,+\,\frac{A}{2}\right)\)

Your first step is correct . . .

\(\displaystyle \L\tan\left(\frac{\pi}{4}\,+\,\frac{A}{2}\right) \;= \;\frac{\tan\left(\frac{\pi}{4}\right)\,+\,\tan\left(\frac{A}{2}\right)}{1 - \tan\left(\frac{\pi}{4}\right)\tan\left(\frac{A}{2}\right)} \;=\;\frac{1\,+\,\tan\left(\frac{A}{2}\right)}{1\,-\,\tan\left(\frac{A}{2}\right)} \;= \;\frac{1\,+\,\frac{\sin\left(\frac{A}{2}\right)}{\cos\left(\frac{A}{2}\right)} }{1\,-\,\frac{\sin\left(\frac{A}{2}\right)}{\cos\left(\frac{A}{2}\right)} }\)


Multiply top and bottom by \(\displaystyle \cos\left(\frac{A}{2}\right):\)

. . \(\displaystyle \L\frac{\cos\left(\frac{A}{2}\right)\,\cdot\,\left[1\,+\,\frac{\sin\left(\frac{A}{2}\right)}{\cos\left(\frac{A}{2}\right)}\right] }{\cos\left(\frac{A}{2}\right)\,\cdot\,\left[1\,-\,\frac{\sin\left(\frac{A}{2}\right)}{\cos\left(\frac{A}{2}\right)}\right] } \;= \;\frac{\cos\left(\frac{A}{2}\right) \,+ \,\sin\left(\frac{A}{2}\right)}{\cos\left(\frac{A}{2}\right) \,- \,\sin\left(\frac{A}{2}\right)}\)


Multiply top and bottom by \(\displaystyle \cos\left(\frac{A}{2}\right)\,+\,\sin\left(\frac{A}{2}\right)\)

. . \(\displaystyle \L\frac{\cos\left(\frac{A}{2}\right) \,+ \,\sin\left(\frac{A}{2}\right)}{\cos\left(\frac{A}{2}\right) \,+\,\sin\left(\frac{A}{2}\right)} \,\cdot\,\frac{\cos\left(\frac{A}{2}\right)\,+\,\sin\left(\frac{A}{2}\right)}{\cos\left(\frac{A}{2}\right)\,-\,\sin\left(\frac{A}{2}\right)} \;= \;\frac{\cos^2\left(\frac{A}{2}\right)\,+\,2\cos\left(\frac{A}{2}\right)\sin\left(\frac{A}{2}\right)\,+\,\sin^2\left(\frac{A}{2}\right)}{\cos^2\left(\frac{A}{2}\right)\,-\,\sin^2\left(\frac{A}{2}\right)}\)


. . \(\displaystyle \L=\;\frac{1\,+\,\sin A}{\cos A} \;=\;\frac{1}{\cos A}\,+\,\frac{\sin A}{\cos A} \;= \;\sec A\,+\,\tan A\)

 

[soroban]

Hello again, Mathlocked!

No good news . . . I'm stuck on #2, too.
I tried to apply DeMoivre's Theorem, but . . .

2) Prove: \(\displaystyle \1\,+\,\cos\theta\,+\,i\sin\theta)^n\:+\1\,+\,\cos\theta\,-\,i\sin\theta)^n \;= \;2^{n+1}\,\cos^n\left(\frac{\theta}{2}\right)\,\cos\left(n\frac{\theta}{2}\right)\)

Let: \(\displaystyle \begin{array}{cc}z\,=\,\cos\theta\,+\,i\sin\theta \\ \overline z\,=\,\cos\theta\,-\,\imath\sin\theta\end{array}\)

We have: \(\displaystyle \:\left(1\,+\,z)^n\,+\,(1\,+\,\overline z)^n\)


Use the Binomial Theorem:

. . \(\displaystyle (1\,+\,z)^n\;=\;1\,+\,nz\,+\,{n\choose 2}z^2\,+\,{n\choose3}z^3\,+\,\cdots\,+\,z^n\)
. . \(\displaystyle (1\,+\,\overline z)^n\;=\;1\,+\,n\overline z\,+\,{n\choose 2}\overline{z}^2\,+\,{n\choose3}\overline{z}^3\,+\,\cdots\,+\,\overline{z}^n\)

Add: \(\displaystyle \:2\,+\,n(z\,+\,\overline z)\,+\,{n\choose2}(z^2\,+\,\overline z^2)\,+\,{n\choose3}(z^3\,+\,\overline z^3)\,+\,\cdots\,+\,(z^n\,+\,\overline z^n)\;\) [1]

And I thought I was on my way . . .


Since: \(\displaystyle \begin{array}{cc}z\:=\:\cos z\,+\,i\sin z\\ \overline z \:=\:\cos z\,-\,i\sin z\end{array}\)

. . by DeMoivre's Theorem: \(\displaystyle \:\begin{array}{cc}z^n\:=\:\cos(n\theta)\,+\,i\sin(n\theta) \\ \overline z^n\:=\:\cos(n\theta)\,-\,i\sin(n\theta)\end{array}\)

Then: \(\displaystyle \:z^n\,+\,\overline z^n\:=\:2\cdot\cos(n\theta)\)


Hence [1] becomes: \(\displaystyle \:2\,+\,2n\cos\theta\,+\,2{n\choose2}\cos(2\theta)\,+\,2{n\choose3}\cos(3\theta)\,+\,\cdots\,+\,2\cos(n\theta)\)

. . . Now what?