Category Archives: Trigonometry

References for IITJEE Foundation Mathematics and Pre-RMO (Homi Bhabha Foundation/TIFR)

January 14, 2019 – 6:44 am
  1. Algebra for Beginners (with Numerous Examples): Isaac Todhunter (classic text): Amazon India link: https://www.amazon.in/Algebra-Beginners-Isaac-Todhunter/dp/1357345259/ref=sr_1_2?s=books&ie=UTF8&qid=1547448200&sr=1-2&keywords=algebra+for+beginners+todhunter 
  2. Algebra for Beginners (including easy graphs): Metric Edition: Hall and Knight Amazon India link: https://www.amazon.in/s/ref=nb_sb_noss?url=search-alias%3Dstripbooks&field-keywords=algebra+for+beginners+hall+and+knight
  3. Elementary Algebra for School: Metric Edition: https://www.amazon.in/Elementary-Algebra-School-H-Hall/dp/8185386854/ref=sr_1_5?s=books&ie=UTF8&qid=1547448497&sr=1-5&keywords=elementary+algebra+for+schools
  4. Higher Algebra: Hall and Knight: Amazon India link: https://www.amazon.in/Higher-Algebra-Knight-ORIGINAL-MASPTERPIECE/dp/9385966677/ref=sr_1_6?s=books&ie=UTF8&qid=1547448392&sr=1-6&keywords=algebra+for+beginners+hall+and+knight
  5. Plane Trigonometry: Part I: S L Loney: https://www.amazon.in/Plane-Trigonometry-Part-1-S-L-Loney/dp/938592348X/ref=sr_1_16?s=books&ie=UTF8&qid=1547448802&sr=1-16&keywords=plane+trigonometry+part+1+by+s.l.+loney

The above references are a must. Best time to start is from standard VII or standard VIII.

-Nalin Pithwa.

By Nalin Pithwa | Also posted in algebra, IITJEE Foundation mathematics, Pre-RMO | Comments (0)

Some good trig practice problems for IITJEE Mains Mathematics

October 17, 2016 – 5:18 am

Problem:

Find the value of \cos^{4}\frac{\pi}{8} + \cos^{4}\frac{3\pi}{8}+\cos^{4}\frac{5\pi}{8}+\cos^{4}\frac{7\pi}{8}.

Problem:

Find the value of \cos{\theta}\cos{2\theta}\cos{4\theta}\ldots\cos{2^{n-1}}{\theta}

Problem:

Find the value of \cos{(2\pi/15)}\cos{(4\pi/15)}\cos{(8\pi/15)}\cos{(14\pi/15)}

Problem:

Prove that (\frac{\cos{A}+\cos{B}}{\sin{A}-\sin{B}})^{n}+(\frac{\sin{A}+\sin{B}}{\cos{A}-\cos{B}})^{n} = 2\cot^{n}(\frac{A-B}{2}), if n is even, and is zero, if n is odd.

More later,

Nalin Pithwa

By Nalin Pithwa | Also posted in IITJEE Advanced | Comments (0)

A problem on solutions of triangles (ambiguous case)

June 9, 2016 – 10:31 pm

Problem 1:

Given b=16, c=25, B= 33 \deg 13^{'}, prove that the triangle is ambiguous and find the other angles, using five figure tables (log/trig).

Kindly send your answers, comments, etc.

Nalin Pithwa

PS: Can you check/prove/verify the ambiguous case of solutions of triangles using plane geometry of your high school days?

By Nalin Pithwa | Also posted in IITJEE Mains, KVPY | Comments (0)

Solution of Triangles (Ambiguous Cases) : IIT JEE Maths

June 8, 2016 – 6:13 pm

The three sides a, b, c and the three angles A, B, C are called the elements of the triangle ABC. When any three of these six elements (except all the three angles) of a triangle are given, the triangle is known completely; that is, the other three elements can be expressed in terms of the given elements and can be evaluated. This process is called the solution of triangles.

  • If the three sides a, b,  and c are given, angle A is obtained from \tan{(A/2)}= \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} or \cos{A}=\frac{b^{2}+c^{2}-a^{2}}{2bc}. B and C can be obtained in a similar way.
  • If two sides b and c and the included angle A are given, then \tan{\frac{B-C}{2}}=\frac{b-c}{b+c}\cot{(A/2)} gives \frac{B-C}{2}. Also, \frac{B-C}{2}=90 \deg - \frac{A}{2}, so that B and C can be evaluated. The third side is given by a=b \frac{\sin{A}}{\sin{B}}, or, a^{2}=b^{2}+c^{2}-2bc \cos{A}.
  • If two sides b and c and the angle B (opposite to side b) are given, then \sin{C}=\frac{c}{b}\sin{B}. And, A=180\deg -(B+C), and a=\frac{b \sin{A}}{\sin{B}} give the remaining elements.

By applying the cosine rule, we have:

\cos{B}=\frac{a^{2}+c^{2} - b^{2}}{2ac}, or if we manipulate this, we get

a^{2}-(2c\cos{B})a+(c^{2}-b^{2})=0

or, a=c \cos{B} \pm \sqrt{b^{2}-(c\sin{B})^{2}}

This equation leads to the following cases:

Case I:

If b<c\sin{B}, no such triangle is possible.

Case II:

Let b=c\sin{B}. There are further following two cases:

Sub-case II a:

B is an obtuse angle, that is, \cos{B} is negative. There exists no such triangle.

Sub-case II b:

B is an acute angle, that is, \cos {B} is positive. There exists only one such triangle.

Case III:

Let b >c \sin{B}. There are following two cases further here also:

Sub-case IIIa:

B is an acute angle, that is, \cos {B} is positive. In this case, two values of a will exist if and only if c\cos{B} > \sqrt{b^{2}-(c \sin{B})^{2}} or, c>b, which means two such triangles are possible. If c<b, only one such triangle is possible.

Sub-case IIIb:

B is an obtuse angle, that is, \cos{B} is negative. In this case, triangle will exist if and only if \sqrt{b^{2}-(c \sin{B})^{2}} > c |\cos{B}| \Longrightarrow b > c. So, in this case, only one such triangle is possible. If b <c, there exists no such triangle.

Note:

If one side a and angles B and C are given, then A=180 \deg -(B+C), and b=a \frac{\sin{B}}{\sin{A}} and c=a\frac{\sin{C}}{\sin{A}}.

If the three angles A, B and C are given, we can only find the ratios of the three sides a, b, and c by using the sine rule(since there are infinite number of similar triangles possible).

More theory later,

Nalin Pithwa

 

By Nalin Pithwa | Also posted in IITJEE Mains, ISI Kolkatta Entrance Exam, KVPY | Comments (0)

A very tricky trig problem from China !! RMO or IITJEE Advanced mathematics training

May 1, 2016 – 7:47 pm

Problem: (China 2003)

Let n be a fixed positive integer. Determine the smallest positive real number \lambda such  that for any \theta_{1}, \theta_{2}, \theta_{3}, \ldots, \theta_{n} in the interval (0,\frac{\pi}{2}), if

\tan {\theta_{1}} \tan {\theta_{2}}\ldots \tan_{\theta_{n}}=2^{\frac{n}{2}}, then

\cos {\theta_{1}}+\cos{\theta_{2}}+\ldots+\cos{\theta_{n}} \leq \lambda.

Solution:

(by Yumin Huang)

The answer is:

\lambda = \frac{\sqrt{3}}{3}, if n=1

\lambda = \frac{\sqrt{2\sqrt{3}}}{3}, if n=2

\lambda = n-1, if n \geq 3

The case n=1 is trivial.

If n=2,we claim that \cos{\theta_{1}}+\cos{\theta_{2}} \leq \frac{2\sqrt{3}}{3} with  equality if and only if \theta_{1}=\theta_{2}=\arctan{\sqrt{2}}. It suffices to show that

\cos^{2}{\theta_{1}}+\cos^{2}{\theta_{2}}+2\cos{\theta_{1}}\cos{\theta_{2}} \leq \frac{4}{3}

or,

\frac{1}{1+\tan^{2}{\theta_{1}}}+\frac{1}{1+\tan^{2}{\theta_{2}}}+ 2 \sqrt{\frac{1}{(1+\tan^{2}{\theta_{1}})(1+\tan^{2}{\theta_{2}})}} \leq \frac{4}{3}

Because \tan{\theta_{1}}\tan{\theta_{2}}=2, we get

(1+\tan^{2}{\theta_{1}})(1+\tan^{2}{\theta_{2}})=5+\tan^{2}{\theta_{1}}+\tan^{2}{\theta_{2}}.

By setting,

\tan^{2}{\theta_{1}}+\tan^{2}{\theta_{2}}=x, the last inequality becomes

\frac{2+x}{5+x}+2\sqrt{\frac{1}{5+x}} \leq \frac{4}{3}, or

2\sqrt{\frac{1}{5+x}} \leq \frac{14+x}{3(5+x)}

Squaring both sides and clearing denominators, we get 36(5+x) \leq 196+ 28x +x^{2}, that is,

0 \leq x^{2}-8x+16=(x-4)^{2}. This  establishes our claim.

Now, assume that n \geq 3. We claim that \lambda=n-1. Note that \lambda \geq n-1; by setting \theta_{2}=\theta_{3}= \ldots = \theta_{n}=\theta and letting \theta \rightarrow 0, we find that \theta_{1} \rightarrow \frac{\pi}{2}, and so the left hand side of the desired inequality approaches n-1. It suffices to show  that

\cos{\theta_{1}}+\cos{\theta_{2}}+\ldots+\cos{\theta_{n}} \leq n-1.

Without loss of generality, assume that \theta_{1} \geq \theta_{2} \geq \ldots \theta_{n}. Then,

\tan{\theta_{1}}\tan{\theta_{2}}\tan{\theta_{3}} \geq 2\sqrt{2}

It suffices to show that \cos{\theta_{1}}+\cos{\theta_{2}}+\cos{\theta_{3}} <2 relation *

But, because \sqrt{1-x^{2}} \leq 1 - \frac{1}{2}x^{2}, \cos{\theta_{i}}=\sqrt{1-\sin^{2}{\theta_{i}}} < 1-\frac{1}{2}\sin^{2}{\theta_{i}}

Consequently, by the arithmetic geometric mean inequality, 

\cos{\theta_{2}}+\cos{\theta_{3}} < 2-\frac{1}{2}(\sin^{2}{\theta_{2}}+\sin^{2}{\theta_{3}}) \leq 2 - \sin{\theta_{2}}\sin{\theta_{3}}

Because \tan^{2}{\theta_{1}} \geq \frac{8}{\tan^{2}{\theta_{2}}\tan^{2}{\theta_{3}}}

we have \sec^{2}{\theta_{1}} \geq \frac{8+\tan^{2}{\theta_{2}}\tan^{2}{\theta_{3}}}{\tan^{2}{\theta_{2}}\tan^{2}{\theta_{3}}}

or $latex $

By Nalin Pithwa | Comments (0)

A Trigonometry problem for IITJEE Advanced Mathematics or Mathematics Olympiad

April 9, 2016 – 10:22 am

Ref: Titu Andreescu and Zumin Feng.

Problem:

Let x, y, z be positive real numbers such that x+y+z=1. Determine the minimum value of

\frac{1}{x}+\frac{4}{y}+\frac{9}{z}

Solution:

An application of Cauchy–Schwarz inequality makes this as a one step problem. Nevertheless, we present a proof which involves only the easier inequality x^{2}+y^{2} \geq 2xy for real numbers x and y by setting first x=\tan {b} and y=2\tan{b} and second x=\tan{a} and y=\cot{a}.

Clearly, z is a real number in the interval [0,1]. Hence, there is an angle a such that z=\sin^{2} {a}. Then, x+y=1-\sin^{2}{a}=\cos^{2}{a}, or \frac{x}{\cos^{2}{a}}+\frac{y}{\cos^{2}{a}}=1. For an angle b, we have \cos^{2}{b}+\sin^{2}{b}=1. Hence, we can set x=\cos^{2}{a}\cos^{2}{b}, and y=\cos^{2}{a}\sin^{2}{b} for some b, it suffices to find the minimum value of

P = \sec^{2}{a} \sec^{2}{b}+4\sec^{2}{a}\csc^{2}{b}+9\csc^{2}{a}

or

P = (\tan^{2}{a}+1)(\tan^{2}{b}+1)+4(\tan^{2}{a}+1)(\cot^{2}{b}+1)+9(\cot^{2}{a}+1)

Expanding the right hand side gives

P = 14 + 5\tan^{2}{a}+9\cot^{2}{a}+(\tan^{2}{b}+4\cot^{b})(1+\tan^{2}{a})

\geq 14 + 5\tan^{2}{a}+9\cot^{2}{a}+2\tan{b}.2\cot{b}(1+\tan^{2}{a})

= 18 + 9(\tan^{2}{a}+\cot^{2}{a}) \geq 18 + 9.2\tan{a}\cot{a}=36

Equality holds when \tan{a}=\cot{a} and \tan{b}=2\cot{b}, which implies that \cos^{2}{a}=\sin^{2}{a} and 2\cos^{2}{b}=\sin^{b}. Because \sin^{2}{\theta}+\cos^{2}{\theta}=1, equality holds when \cos^{2}{a}=\frac{1}{2} and \cos^{2}{b}=\frac{1}{3}, that is, x=\frac{1}{6}, y=\frac{1}{3}, z=\frac{1}{2}.

More later.

Nalin Pithwa

 

By Nalin Pithwa | Also posted in IITJEE Advanced Mathematics | Tagged | Comments (0)

Solution to another S L Loney problem for IITJEE Advanced Mathematics

January 2, 2016 – 6:48 pm

Exercise XXVII. Problem #32. The sides of a triangle are in AP and the greatest and least angles are \theta and \phi. Prove that

4(1-\cos{\theta})(1-\cos{\phi})=\cos{\theta}+\cos{\phi}

Proof:

Let a, b, c be in AP. Hence, a<b<c, which in turn implies c is greatest and a is the least.

Hence, \theta=\angle{C} and \angle{A}.

Want:

4(1-\cos{A})(1-\cos{C})=\cos{A}+\cos{C}

Given:

b-a=c-b

(b-a)^{2}=(b-c)^{2}. Hence,

b^{2}-2ab+a^{2}=b^{2}+c^{2}-2bc

b^{2}+a^{2}-c^{2}=b^{2}-2bc+2ab

2b=a+c

4b^{2}=a^{2}+c^{2}+2ac

a^{2}+c^{2}-b^{2}=3b^{2}-2ac

a^{2}+b^{2}-c^{2}=b^{2}-2bc+2ab

b^{2}+c^{2}-a^{2}=b^{2}-2ab+2bc

Now, we have 1-\cos{A}=1-\frac{b^{2}+c^{2}-a^{2}}{2bc} and this is equal to the following:

\frac{2bc-(b^{2}+c^{2}-a^{2})}{2bc}=\frac{2bc-(b^{2}-2ab+2bc)}{2bc}

= \frac{2bc-b^{2}+2ab-2bc}{2bc}=\frac{2ab-b^{2}}{2bc}.

Also, similarly, we have the following:

1-\cos{C}=1-\frac{a^{2}+b^{2}-c^{2}}{2ab}=1-\frac{b^{2}-2bc+2ab}{2ab}=\frac{2bc-b^{2}}{2ab}

\cos{A}+\cos{C}=\frac{b^{2}+c^{2}-a^{2}}{}+\frac{a^{2}+b^{2}-c^{2}}{2ab}

= \frac{b^{2}-2ab+2bc}{2bc}+\frac{b^{2}-2bc+2ab}{2ab}

=\frac{a(b^{2}-2ab+2bc)+c(b^{2}-2bc+2ab)}{2abc}

=\frac{ab^{2}-2a^{2}b+2abc+b^{2}c-2bc^{2}+2abc}{2abc}

=\frac{ab^{2}+b^{2}c-2a^{2}b-2c^{2}b+4abc}{2abc}

= \frac{ab+bc-2a^{2}-2c^{2}+4ac}{2ac}

=\frac{b(a+c)-2(a-c)^{2}}{2ac}

=\frac{2b^{2}-2(a-c)^{2}}{2ac}

=\frac{b^{2}-(a-c)^{2}}{2ac}

\frac{(b+a-c)}{(b-a+c)}{ac}

but it is given that b-a=c-b, hence, a+c=2b, a+c-b=b. So the above expression changes to

= \frac{(b+a-c)(c-b+c)}{ac}

= \frac{(2c-b)(a+b-c)}{ac}

= \frac{(2c-b)(a+b-\overline{2b-a})}{ac}

= \frac{(2c-b)(a-b+c)}{ac}

= \frac{(2c-b)(2a-b)}{ac}

= \frac{4ac-2bc-2ab+b^{2}}{ac}

= 4(1-\cos{A})(1-\cos{C}).

QED.

By Nalin Pithwa | Also posted in Cnennai Math Institute Entrance Exam, IITJEE Advanced Mathematics | Comments (0)

A solution to a S L Loney Part I trig problem for IITJEE Advanced Math

December 30, 2015 – 4:26 pm

Exercise XXVII. Problem 30.

If a, b, c are in AP, prove that \cos{A}\cot{A/2}, \cos{B}\cot{B/2}, \cos{C}\cot{C/2} are in AP.

Proof:

Given that b-a=c-b

TPT: \cos{B}\cot{B/2}-\cos{A}\cot{A/2}=\cot{C/2}\cos{C}-\cos{B}\cot{B/2}. —— Equation 1

Let us try to utilize the following formulae:

\cos{2\theta}=2\cos^{2}{\theta}-1 which implies the following:

\cos{B}=2\cos^{2}(B/2)-1 and \cos{A}=2\cos^{2}(A/2)-1

Our strategy will be reduce LHS and RHS of Equation I to a common expression/value.

LHS=(\frac{2s(s-b)}{ac}-1)(\frac{\sqrt{\frac{(s)(s-b)}{ac}}}{\sqrt{\frac{(s-a)(s-c)}{ac}}})-\cos{A}\cot{(A/2)}

which is equal to

(\frac{2s(s-b)}{ac}-1))(\frac{\sqrt{\frac{(s)(s-b)}{ac}}}{\sqrt{\frac{(s-a)(s-c)}{ac}}})-(2\cos^{2}(A/2)-1)\frac{\cos{A/2}}{\sin{A/2}}

which is equal to

(\frac{2s(s-b)}{ac}-1))(\frac{\sqrt{\frac{(s)(s-b)}{ac}}}{\sqrt{\frac{(s-a)(s-c)}{ac}}})-(\frac{2s(s-a)}{bc}-1)\frac{\sqrt{\frac{s(s-a)}{bc}}}{\sqrt{\frac{(s-b)(s-c)}{bc}}}

which is equal to

(\frac{2s(s-b)}{ac}-1))\sqrt{s(s-b)}{(s-a)(s-c)}-(\frac{2s(s-a)}{bc}-1)\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}

which in turn equals

\sqrt{\frac{s}{(s-a)(s-b)(s-c)}}((\frac{2s(s-b)}{ac}-1)(s-b)-(\frac{2s(s-a)}{bc}-1)(s-a))

From the above, consider only the expression, given below. We will see what it simplifies to:

\frac{2s(s-b)^{2}}{ac}-(s-b)-\frac{2s(s-a)^{2}}{bc}+(s-a)

=\frac{2s(s-b)^{2}}{ac}-\frac{2s(s-a)^{2}}{bc}+b-a

=(\frac{2s}{c})(\frac{(s-b)^{2}}{a}-\frac{(s-a)^{2}}{b})+b-a

=\frac{2s(s-b)^{2}}{ca}-\frac{2s(s-a)^{2}}{bc}+c-b —- Equation II.

Now, consider RHS of Equation I. Let us see if it also boils down to the above expression after simplification.

RHS=\cot{(C/2)}\cos{C}-\cos{B}\cot{(B/2)}

=(2\cos^{2}{(C/2)}-1)\cot{(C/2)}-(2\cos^{2}({B/2})-1)\cot{(B/2)}

=(\frac{2s(s-c)}{ab}-1)\frac{\sqrt{\frac{s(s-c)}{ab}}}{\sqrt{\frac{(s-b)(s-a)}{ab}}}-(\frac{2s(s-b)}{ac})\frac{\sqrt{\frac{s(s-b)}{ac}}}{\sqrt{\frac{(s-a)(s-c)}{ac}}}

= \sqrt{\frac{s}{(s-a)(s-b)(s-c)}}((\frac{2s(s-c)}{ab}-1)(s-c)-(\frac{2s(s-b)}{ac}-1)(s-b))

From equation II and above, what we want is given below:

\frac{2s(s-c)^{2}}{ab}-(s-c)-\frac{2s(s-b)^{2}}{ac}+(s-b)=\frac{2s(s-b)^{2}}{ac}-\frac{2s(s-a)^{2}}{bc}+c-b

that is, want to prove that c(s-c)^{2}+a(s-a)^{2}=2b(s-b)^{2}

but, it is given that a+c=2b and hence, c=2b-a, which means a+c-b=b and b-a=c-b

that is, want to prove that

c(a+b-c)^{2}+a(b+c-a)^{2}=2b(a+c-b)^{2}=2b^{3}

i.e., want: c(a+b-c)^{2}+a(b+c-a)^{2}=2b^{3}

i.e., want: (2b-a)(a+b-2b+a)^{2}+a(b-a+2b-a)^{2}=2b^{3}

i.e., want: (2b-a)(2a-b)^{2}+a(3b-2a)^{2}=2b^{3}

Now, in the above, LHS=(2b-a)(4a^{2}+b^{2}-4ab)+a(9b^{2}+4a^{2}-12ab)

= 8a^{2}b+2b^{3}-8ab^{2}-4a^{3}-ab^{2}+4ab^{2}+9ab^{2}+4a^{3}-12a^{2}b

= 2b^{3}.

Hence, LHS+RHS.

QED.

By Nalin Pithwa | Also posted in Cnennai Math Institute Entrance Exam, IITJEE Advanced Mathematics, ISI Kolkatta Entrance Exam, KVPY, RMO | Tagged , | Comments (0)

Complex attitude!

October 22, 2015 – 8:58 pm

Let us a discuss yet one more complex number based IITJEE mains problem.

Problem:

Let z_{1} and z_{2} be nth roots of unity, which subtend a right angle at the origin. Then, integer n must be of the form __________________. (fill in the blank).

Solution:

nth roots of unity are given by

\cos{\frac{2m\pi}{n}}+ i \sin{\frac{2m\pi}{n}}= e^{\frac{2m\pi}{n}}

where m=0,1,2, \ldots, n-1.

Let z_{1}=\cos{\frac{2m_{1}\pi}{n}}+ i\sin{\frac{2m_{1}\pi}{n}}=e^{2m_{1}\pi i/n}

Let z_{2}=e^{2m_{2}\pi i/n} where 0 \leq m_{1}, m_{2}< n, m_{1} \neq m_{2}

As the join of z_{1} and z_{2} subtends a right angle at the origin, we deduce that \frac{z_{1}}{z_{2}} is purely imaginary.

\Longrightarrow \frac{e^{2m_{1}\pi i/n}}{e^{2m_{2}\pi i/n}}=ik, for some real k

\Longrightarrow e^{2(m_{1}-m_{2})\pi i/n}=ik

\Longrightarrow n=4(m_{1}-m_{2}). Thus, n must be of the form 4k.

More later,

Nalin Pithwa

 

By Nalin Pithwa | Also posted in Complex Numbers, IITJEE Mains | Tagged , , , | Comments (0)

De Moivre’s Theorem application

July 28, 2015 – 11:23 pm

Question:

If f_{r}(\alpha)=(\cos{\frac{\alpha}{r^{2}}}+i\sin{\frac{\alpha}{r^{2}}}) \times (\cos{\frac{2\alpha}{r^{2}}}+i\sin{\frac{2\alpha}{r^{2}}}) \ldots (\cos{\frac{\alpha}{r}}+i\sin{\frac{\alpha}{r}}), then

\lim_{n \rightarrow \infty}f_{n}{\pi} equals

(a) -1

(b) 1

(c) -i

(d) i

Solution.

Using De Moivre’s theorem,

f_{r}{\alpha}=e^{i\frac{\alpha}{r^{2}}}e^{i\frac{2\alpha}{r^{2}}}\ldots e^{i\frac{\alpha}{r}}

which in turn equals e^{(i \frac{\alpha}{r^{2}})(1+2+\ldots+r)}=e^{(i\frac{\alpha}{r^{2}})(\frac{r(r+1)}{2})}=e^{i(\frac{\alpha}{2})(1+\frac{1}{r})}

Hence, \lim_{n \rightarrow \infty}f_{n}(\pi)=\lim_{n \rightarrow \infty}e^{(i)(\frac{\pi}{2})(1+\frac{1}{n})}=e^{i(\frac{\pi}{2})}=\cos{\frac{\pi}{2}}+i\sin{\frac{\pi}{2}}=i.

More complex stuff to be continued in next blog (pun intended) 🙂

Nalin Pithwa

By Nalin Pithwa | Also posted in calculus, Cnennai Math Institute Entrance Exam, Complex Numbers, IITJEE Mains, ISI Kolkatta Entrance Exam | Tagged , , | Comments (0)