Tag Archives: plane trigonometry problem

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.

A Cute Complex Problem

Question:

If w=\cos{\frac{\pi}{n}}+i\sin{\frac{\pi}{n}}, then find the value of 1+w+w^{2}+w^{3}+\ldots+w^{n-1}.

Solution:

We have S=1+w+w^{2}+w^{3}+\ldots+w^{n-1}=\frac{1-w^{n}}{1-w}.

But, w^{n}=\cos{\frac{n\pi}{n}}+i\sin{\frac{n\pi}{n}}=-1

Thus, S=\frac{2}{1-w}

but, 1-w=1-\cos{\frac{\pi}{n}}-i\sin{\frac{\pi}{n}} which equals

2\sin^{2}{\frac{\pi}{2n}}-2i\sin{\frac{\pi}{2n}}\cos{\frac{\pi}{2n}}

that is, -2i\sin{\frac{\pi}{2n}}[\cos{\frac{\pi}{2n}}+i\sin{\frac{\pi}{2n}}].

Thus, S=\frac{-2}{2i\sin{\frac{\pi}{2n}}}[\cos{\frac{\pi}{2n}}+i\sin{\frac{\pi}{2n}} ]^{-1}=1+i\cot{\frac{\pi}{2n}}.

Hope you are finding it useful,

More later,

Nalin Pithwa

Solutions of Triangles — a tricky IITJEE problem

Question (IITJEE 1978). Suppose p_{1},p_{2},p_{3} are the altitudes through vertices A, B, C of a triangle ABC with area \Delta.

Prove that: 

\frac {1}{p_{1}} + \frac {1}{p_{2}} - \frac {1}{p_{3}}=\frac {2ab}{(a+b+c) \Delta} \cos^{2} (C/2)

Proof:

The RHS looks daunting. But, if we bring the factor \Delta in its denominator to the LHS, then the problem unfolds itself. Since, \Delta =(1/2)ap_{1}=(1/2)bp_{2}=(1/2)cp_{3}, the problem is equivalent to the following:

prove: a+b-c=\frac {4ab}{a+b+c}\cos^{2}(C/2)

If we write 2\cos^{2}(C/2) as 1+\cos C, and multiply both the sides of by a+b+c, then the problem is reduced to  proving that

(a+b)^{2}-c^{2}=2ab+2ab \cos C.

which is the same as the cosine formula 🙂 🙂 🙂

More later…

Nalin Pithwa