Need derviation of some formulas! Help!

To prove $(i)$ we have $z=cos(x)+isin(x)$ where $z$ is a function of $x$

Differentiating $z$ with respect to $x$ we get :

$\frac{dz}{dx}=-sin(x)+icos(x)$

Since ${i}^{2}=-1$ hence we have :

$\frac{dz}{dx}={i}^{2}sin(x)+icos(x)=i(cos(x)+isin(x))$

Since $cos(x)+isin(x)=z$ hence we have :

$\frac{dz}{dx}=iz$

Seperating variables and integrating both sides we have :

$\int { \frac { dz }{ z } } =\int { idx }$

$ln(z)=ix+C$

When $x=0,z=1$ hence we get $C=0$

$\Rightarrow ln(z)=ix$

$\Rightarrow z={e}^{ix}$

Finally :

$cos(x)+isin(x)=e^{ix}$

To prove $1$ Euler's Formula, simple use the expansion of $e^x$ , i.e. $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+.....$

$\implies e^{ix}=1+x.i+\frac{x^2.i^2}{2!}+\frac{x^3.i^3}{3!}+\frac{x^4.i^4}{4!}+\frac{x^5.i^5}{5!}+.....$

$\quad \quad \quad \quad =1+x.i-\frac{x^2}{2!}-i.\frac{x^3}{3!}+\frac{x^4}{4!}+i.\frac{x^5}{5!}+........$

$\quad \quad \quad \quad =(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+.....)+i.(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+......)$

$\quad \quad \quad \quad =cosx +i.sinx$

Because we know that $cosx=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+.....$ $sinx=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+......$

Hence proved that $\boxed{cosx+isinx=e^{ix}}$

5) and 6) are simple, i don't know about rest

$[\sum_{i = 1}^{n} a_{i}b_{i}]^{2} \leq [\sum_{i = 1}^{n} a_{i}^{2}]^{2}[\sum_{i = 1}^{n} b_{i}^{2}]^{2}$

and the equality holds if $\frac{a_{1}}{b_{1}} = \frac{a_{2}}{b_{2}} = .....=\frac{a_{n}}{b_{n}}$

Let$A = [\sum_{i = 1}^{n} a_{i}^{2}]^{2} , B =[\sum_{i = 1}^{n} b_{i}^{2}]^{2} , C = [\sum_{i = 1}^{n} a_{i}b_{i}]^{2}$

then $C^{2} \leq AB$

If B = 0 then $b_{i} = 0$ for i = 1 , 2, . , n .Hence C = 0 and is true .Therefore it is sufficient to consider the case $B\neq 0$ This implies B > 0 .

$0 \leq \sum_{i = 1}^{n} (Ba_{i} - Cb_{i})^{2} = \sum_{i = 1}^{n} ( B^{2}a_{i}^{2} - 2BCa_{i}b_{i} + C^{2}b_{i}^{2}$

distributing the sum,

$= B(AB - C^{2})$

Sinc eB> 0 we get $AB - C^{2} > 0$ which is the inequality $C^{2} \leq AB$. Moreover , equality holds if $\sum_{i = 1}^{n} (Ba_{i} - Cb_{i})^{2} = 0$

This is equivalent to

$\frac{a_{i}}{b_{i}} = \frac{C}{B}$ for i = 1 , 2 , .... , n

Simple application and beautiful application

If a, b , c and d are positive real numbers such that $c^{2} + d^{2} = (a^{2} + b^{2})^{3}$,

prove that $\frac{a^{3}}{c} + \frac{b^{3}}{d}$ if ad = bc .

Using above inequality we get ,

$(a^{2} + b^{2})^{2} = (\sqrt{\frac{a^{3}}{c}}\sqrt{ac} + \sqrt{\frac{b^{3}}{d}}\sqrt{bd} )^{2}$

$\leq ( \frac{a^{3}}{c} + \frac{b^{3}}{d})(ac + bd)$ , where the equality holds

Thus $\leq ( \frac{a^{3}}{c} + \frac{b^{3}}{d})(ac + bd) \geq ( a^{2} + b^{2})^{3}$

$= ( a^{2} + b^{2})^{1/2} ( a^{2} + b^{2})^{3/2}$

$= ( a^{2} + b^{2})^{1/2}( c^{2} + d^{2})^{1/2} \geq ac + bd$

again using the inequality , the equality holds in the last step ( above) if $\frac{a}{c} = \frac{b}{d}$

Combining both we get $\frac{a^{3}}{c} + \frac{b^{3}}{d} \geq 1$ and the equality holds if ad = bc.

6) One of my favorite

let s be equal to the given series , $\alpha =a , \beta = b$

$2Xsin(b/2) = 2sinasin(b/2) + 2sin(a + b)sin(b/2) ............$

$= cos( a - b/2) - cos( a + b/2) + cos(a + b/2) - cos( a + 3b/2) .... + cos[(a + (n - 3/2)b)] - cos[a + (n - 1/2)b]$

$= cos(a - b/2) - cos(a + (n - 1/2)b]$

$= 2sin[a +((n-1)/2)b]sin(nb/2)$

$X = \huge{\boxed{\frac{sin[a + \frac{n - 1}{2}b]sin\frac{nb}{2}}{sin\frac{b}{2}}}}$

Similarly for the cosine series it can be done , try it ,its interesting and enjoyable.

Lovely Application of this can be found in THIS QUESTION

Enjoy@Kartik Sharma

I am giving proof of $6th$ formula : $S(let)=sin\alpha + sin(\alpha + \beta) + ...... sin(\alpha + (n-1)\beta) = \frac{sin\frac{n\beta}{2}}{sin\frac{\beta}{2}}. sin(\alpha + (n-1)\frac{\beta}{2})$

We know that $\boxed{2.sinA.sinB=cos(A-B)-cos(A+B)}$ $\implies 2.sin\alpha.sin\frac{\beta}{2}=cos(\alpha-\frac{\beta}{2})-cos(\alpha+\frac{\beta}{2})$

$\implies 2.sin(\alpha+\beta).sin\frac{\beta}{2}=cos(\alpha+\frac{\beta}{2})-cos(\alpha+\frac{3\beta}{2})$

$\implies 2.sin(\alpha+2\beta).sin\frac{\beta}{2}=cos(\alpha+2\beta-\frac{\beta}{2})-cos(\alpha+2.\beta+\frac{\beta}{2})$

$\bullet$

$\bullet$

$\bullet$

$\bullet$

$\bullet$

$\implies 2.sin(\alpha+(n-1)\beta).sin\frac{\beta}{2}=cos(\alpha+(n-1)\beta-\frac{\beta}{2})-cos(\alpha+(n-1).\beta+\frac{\beta}{2})$

By adding, we get

$2.sin\frac{\beta}{2}[sin\alpha+sin(\alpha+\beta)+.....+sin(\alpha+(n-1)\beta)]=cos(\frac{\alpha-\beta}{2})-cos(\alpha+(\frac{n-1}{2}).\beta)$

$\implies 2.sin\frac{\beta}{2} \times S=2sin(\alpha+(\frac{n-1}{2}).\beta).sin\frac{n.\beta}{2}$ $\implies S=sin\alpha + sin(\alpha + \beta) + ...... sin(\alpha + (n-1)\beta) = \frac{sin\frac{n\beta}{2}}{sin\frac{\beta}{2}}. sin(\alpha + (n-1)\frac{\beta}{2})$

In the similar way,we can prove $cos\alpha + cos(\alpha + \beta) + ......+cos(\alpha + (n-1)\beta) = \frac{sin\frac{n\beta}{2}}{sin\frac{\beta}{2}}. cos(\alpha + (n-1)\frac{\beta}{2})$

To prove this, we will use the formula $\boxed{2.cosA.cosB=sin(A+B)-sin(A-B)}$

enjoy !

@Calvin Lin @Michael Mendrin @Sandeep Bhardwaj @Sanjeet Raria @Sharky Kesa @Krishna Ar @Satvik Golechha @Finn Hulse @Daniel Liu @megh choksi

If you have some more of which you need derivations of, you can share them here!

7. Proof for the integrating factor - $I = {e}^{\int{p(x)} dx}$ where $\frac{dy}{dx} + p(x)y = Q$ is the differential equation.

You sir, can sometimes be so overt that it's covert.

AoPS (Art of Problem Solving) has the answer to your question. That's where I learnt a lot of my formulae.