**1. Euler's Formula** - \(cos x + isin x = {e}^{ix}\)

**2. Relation between Mean, Mode, Median** - \(Mode = 3*Median - 2*Mean\)

**3. Poisson's Distribution** - \(P(X=r) = {e}^{-\lambda}.\frac{{\lambda}^{r}}{r!}\) where \(\lambda(mean) = np(n = No. of trials, p = probability of success)\)

**4. RMS - Arithmetico-Geometric and Harmonic Mean inequality**

**5. Cauchy-Shwartz's inequality**

**6. Sum of sines with angles in AP**- \(sin\alpha + sin(\alpha + \beta) + ...... sin(\alpha + (n-1)\beta) = \frac{sin\frac{n\beta}{2}}{sin\frac{\beta}{2}}. sin(\alpha + (n-1)\frac{\beta}{2})\)

and similarly

**Sum of cosines with angles in AP** - which gives \(\frac{sin\frac{n\beta}{2}}{sin\frac{\beta}{2}}. cos(\alpha + (n-1)\frac{\beta}{2})\)

THANKS a lot in anticipation! And I believe that you have the answers and you must also be in need of some of the formulas. You are free to share them here to clear all your doubts. :)

## Comments

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TopNewestTo 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 :

\(cos(x)+isin(x)=e^{ix}\) – Ronak Agarwal · 2 years, 4 months ago

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– Deepanshu Gupta · 2 years, 4 months ago

I Love Your way To prove This equation ! Hats Off!! I 'am Amazed How Can any person down-votes it ??Log in to reply

Sorry if you didn't like the proof. – Ronak Agarwal · 2 years, 4 months ago

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– Sneha Sharma · 2 years, 3 months ago

I think when x=0 , then at the same time z cannot be equal to one.Log in to reply

– Kartik Sharma · 2 years, 4 months ago

Wow! Amazing! This made me become your fan! Thank You! Try other proofs too! And yes can you give me an explanation on why Taylor series makes sense?Log in to reply

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 \[\boxed{cosx+isinx=e^{ix}}\] – Sandeep Bhardwaj · 2 years, 4 months ago

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– Kartik Sharma · 2 years, 4 months ago

Can you tell me why does Taylor series make sense? How to find them? *if you don't mindLog in to reply

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 – Megh Choksi · 2 years, 4 months ago

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– Kartik Sharma · 2 years, 4 months ago

Nice Thanks! I solved that problem! Maybe I would give my solution too!Log in to reply

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– Kartik Sharma · 2 years, 4 months ago

Thanks! Nice! Try other ones too!Log in to reply

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 ! – Sandeep Bhardwaj · 2 years, 4 months ago

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– Kartik Sharma · 2 years, 4 months ago

Thanks for sharing! You are quite good at LatexLog in to reply

7. Proof for the integrating factor- \(I = {e}^{\int{p(x)} dx}\) where \(\frac{dy}{dx} + p(x)y = Q\) is the differential equation. – Kartik Sharma · 2 years, 4 months agoLog in to reply

If you have some more of which you need derivations of, you can share them here! – Kartik Sharma · 2 years, 4 months ago

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@Calvin Lin @Michael Mendrin @Sandeep Bhardwaj @Sanjeet Raria @Sharky Kesa @Krishna Ar @Satvik Golechha @Finn Hulse @Daniel Liu @megh choksi – Kartik Sharma · 2 years, 4 months ago

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– Kartik Sharma · 2 years, 4 months ago

and more.. (I can't write names of all of 'em).Log in to reply

You sir, can sometimes be so overt that it's covert. – Saurabh Chauhan · 2 years, 4 months ago

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– Kartik Sharma · 2 years, 4 months ago

ART OF IGNORING! (I also know it)Log in to reply

– Saurabh Chauhan · 2 years, 4 months ago

HUH! I bet you're gonna read this.Log in to reply

AoPS (Art of Problem Solving) has the answer to your question. That's where I learnt a lot of my formulae. – Sharky Kesa · 2 years, 4 months ago

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– Kartik Sharma · 2 years, 4 months ago

Yeah! But it would be better if I get them here.Log in to reply

– Sharky Kesa · 2 years, 4 months ago

Maybe, there should be wiki pages for these.Log in to reply

– Kartik Sharma · 2 years, 4 months ago

YEAH! Agree!Log in to reply