## 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