## Category Archives: Trigonometry

### Tricky Trigonometry questions for IITJEE mains maths practice

Prove the following:

1. $\frac{\sin{A}}{1-\cos{A}} = \frac{1+\cos{A}-\sin{A}}{\sin{A}-1+\cos{A}}$
2. $\frac{1+\sin{A}}{\cos{A}}=\frac{1+\sin{A}+\cos{A}}{\cos{A}+1-\sin{A}}$
3. $\frac{\tan{A}}{\sec{A}-1}=\frac{\tan{A}+\sec{A}+1}{\sec{A}-1+\tan{A}}$
4. $\frac{1+\csc{A}+ \cot{A}}{1+\csc{A}-\cot{A}} = \frac{\csc{A}+\cot{A}-1}{\cot{A}-\csc{A}+1}$

Hint: Directly trying to prove LHS is RHS is difficult in all the above; or even trying to transform RHS to LHS is equally difficult; it is quite easier to prove the equivalent statement by taking cross-multiplication of the appropriate expressions. 🙂

Regards,

Nalin Pithwa.

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

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.

### Some good trig practice problems for IITJEE Mains Mathematics

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

### A problem on solutions of triangles (ambiguous case)

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).

Nalin Pithwa

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

### Solution of Triangles (Ambiguous Cases) : IIT JEE Maths

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, 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, 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 , 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

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

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)

$\lambda = \frac{\sqrt{3}}{3}$, 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 $latex$

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 CauchySchwarz 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

### Solution to another S L Loney problem for IITJEE Advanced Mathematics

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, 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.

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

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.

### Complex attitude!

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