Trigonometry « Mathematics Hothouse

Category Archives: Trigonometry

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

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

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

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 Cauchy Schwarz inequality | 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 Loney problem, plane trigonometry problem | 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 complex, DeMoivre's theorem, nth roots of unity, roots | 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

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