## Prove that x^2 – x cos(A + B) + 1 is a factor of 2x^4 + 4x^3 sinA sinB – x^2(cos2A + cos2B) + 4xcosA cosB – 2

Q: Prove that x2 – x cos(A + B) + 1 is a factor of 2x4 + 4x3 sinA sinB – x2(cos2A + cos2B) + 4 x cosA cosB – 2 and also find the other factor.

Solution: Let 2x4 + 4x3 sinA sinB – x2 (cos2A + cos2B) + 4xcosA cosB – 2

= [x2 – xcos (A+B) + 1] [ax2 + bx + c]

= ax4 + (b – acos (A+B)) x3 + (a – bcos (A+B) + c) x2 + (b – c cos (A+B)) x + c

By comparing co-efficients,

a = 2, c = -2 and

b – a cos (A+B) = 4 sinA sinB …(1)

a – b cos (A+B) + c = -cos 2A – cos2B …(2)

b – c cos (A+B) = 4cosA cosB …(3)

solving, we get same value of b

b = 2cos (A – B),

hence the assumption is correct and by solving for the other factor ax2 + bx + c = 2x2 + 2cos (A – B) x – 2 = 2(x2 + cos(A- B)x – 1).

## Let A, B, C be three angles such that A = π/4  and tanB tanC = p. Find all possible values of p …

Q: Let A, B, C be three angles such that A = π/4  and tanB tanC = p. Find all possible values of p such that A, B, C are the angles of a triangle.

Solution : A + B + C = π

B + C = π – A

B + C = π – π/4 = 3π/4

0 < B , C < 3π/4

tanB tanC = p

$\large \frac{sinB sinC}{cosB cosC} = \frac{p}{1}$

$\large \frac{cosB cosC – sinB sinC}{cosB cosC + sinB sinC } = \frac{1-p}{1+p}$

$\large \frac{cos(B+C)}{cos(B-C)} = \frac{1-p}{1 + p}$

$\large \frac{cos(3\pi/4)}{cos(B-C)} = \frac{1-p}{1 + p}$

$\large cos(B-C) = \frac{1+p}{\sqrt{2}(p-1)}$ …(1)

Since B or C can vary from 0 to 3π/4

$\large -\frac{1}{\sqrt{2}} < cos(B-C) \le 1$

Equation (1) will now lead to

$\large -\frac{1}{\sqrt{2}} < \frac{1+p}{\sqrt{2}(p-1)} \le 1$ $1 + \frac{1+p}{(p-1)} > 0$

$\large \frac{2p}{p-1} > 0$

p < 0 , p > 1 …(2)

Also , $\large \frac{p+1 – \sqrt{2}(p-1)}{\sqrt{2}(p-1)} \le 0$

$\large \frac{(p-(\sqrt{2}+1)^2)}{p-1} \ge 0$

p < 1 or $p \ge (\sqrt{2}+1)^2$ …(3)

Combining (2) & (3) , we get

p < 0 or , $p \ge (\sqrt{2}+1)^2$

## If α + β + γ = π  and tan(β+γ-α)/4 . tan(γ+α -β)/4 .tan(α+β-γ)/4 = 1. Prove that 1 + cos α + cos β + cos γ = 0

Q: If α + β + γ = π  and $tan\frac{(β + γ-α)}{4} . tan\frac{(γ + α -β)}{4} .tan\frac{(α + β-γ)}{4} = 1$ . Prove that 1 + cos α + cos β + cos γ = 0 .

Solution : Let $A = tan\frac{(β + γ-α)}{4}$ , $B = tan\frac{(γ + α -β)}{4}$ and $C = tan\frac{(α + β-γ)}{4}$

Therefore , $tanA . tanB . tanC = 1$

$tanA . tanB = \frac{1}{tanC}$

$\large \frac{sinA}{cosA} . \frac{sinB}{cosB} = \frac{1}{tanC}$

$\large \frac{sinA sinB -cosA cosB}{sinA sinB + cosA cosB} = \frac{1-tanC}{1+tanC}$

$\large – \frac{cos(A+B)}{cos(A-B)} = \frac{sin(\frac{\pi}{4}-C)}{cos(\frac{\pi}{4}-C)}$

$\large 2 sin(\frac{\pi}{4}-C) cos(A-B) + 2 cos(\frac{\pi}{4}-C) cos(A+B) = 0$

$sin(\frac{\pi}{4}-C + A-B) + sin(\frac{\pi}{4}-C – A+B) + cos(\frac{\pi}{4}-C + A+B) + cos(\frac{\pi}{4}-C – A-B) = 0$ …(1)

Since $A-B-C = \frac{\pi}{4}-\alpha$

Similarly , $B-C -A = \frac{\pi}{4}-\beta$

and , $C – A – B = \frac{\pi}{4}-\gamma$

and $C + A + B = \frac{\alpha + \beta + \gamma}{4} = \frac{\pi}{4}$

Equation (1) reduces to

$\large sin(\frac{\pi}{4} + \frac{\pi}{4} – \alpha) + sin(\frac{\pi}{4} + \frac{\pi}{4} – \beta) + cos(\frac{\pi}{4} – \frac{\pi}{4} + \gamma) + cos(\frac{\pi}{4} – \frac{\pi}{4} ) = 0$

$\large cos\alpha + cos\beta + cos\gamma + 1 = 0$

## Prove that in an acute angled triangle ABC, cos A cos B cos C ≤ 1/8

Q: Prove that in an acute angled triangle ABC, cos A cos B cos C ≤ 1/8

Solution : Let 8 (cos A cos B cos C) = y

4 (2 cos A cos B ) cos C = y

4 [cos(A + B) + cos(A-B)] cosC = y

(Since ,  A+B+C = π)

4 [cos(π-C) + cos(A-B)] cosC = y

– 4 cos2C + 4 cosC . cos(A-B) = y

4 cos2C – 4 cosC . cos(A-B) + y = 0

$\displaystyle cosC = \frac{-4cos(A-B) \pm \sqrt{16 cos^2 (A-B) – 16 y}}{8}$

Since cos C is real

y ≤ cos2 (A – B)

⇒ y ≤ 1

⇒ cos A cos B cos C ≤ 1/8

## Show that in an acute angled triangle ABC, ∑tanA tanB ≥  9

Q : Show that in an acute angled triangle ABC, ∑tanA tanB ≥  9

Sol. We have to prove that ∑tanA tanB ≥  9

tan A tan B + tan B tan C + tan C tan A ≥ 9

(tan A tan B – 1) + (tan B tan C – 1) + (tan C tan A – 1) ≥ 6

$\displaystyle \frac{tanA + tanB}{tanC} + \frac{tanB + tanC}{tanA} + \frac{tanC + tana}{tanB} \ge 6$

$\displaystyle (\frac{tanA}{tanC} + \frac{tanC}{tanA}) + (\frac{tanA}{tanB} + \frac{tanB}{tanA}) + (\frac{tanB}{tanC}+ \frac{tanC}{tanB}) \ge 6$

This is true.

## Solve for the equation cos(π/4(n^2 + 2n)) =sin(π/4(n^2 + n +1)) , where n is an integer

Q: Solve for n the equation $cos(\frac{\pi}{4}(n^2 + 2 n)) = sin(\frac{\pi}{4}(n^2 + n + 1))$ , where n is an integer .

Solution : The given equation is

$cos(\frac{\pi}{4}(n^2 + 2 n)) = sin(\frac{\pi}{4}(n^2 + n + 1))$

$cos(\frac{\pi}{4}(n^2 + 2 n)) = cos(\frac{\pi}{2} – \frac{\pi}{4}(n^2 + n + 1))$

$sin\frac{\pi}{8}(2n^2 + 3 n -1) sin\frac{\pi}{8}(n+1) = 0$

Either $sin\frac{\pi}{8}(n+1) = 0$ or , $sin\frac{\pi}{8}(2n^2 + 3 n -1) = 0$

CaseI : $sin\frac{\pi}{8}(n+1) = 0 = sink\pi$ ; k ∈ I

Case II : $sin\frac{\pi}{8}(2n^2 + 3 n -1) = 0 = sink\pi$; k ∈ I

2n2 + 3n – 1= 8 k , k ∈ I

Let n1 and n2 be two integral values of n such that

$2n_1^2 + 3 n_1 – 1 = 8 k_1$ ….(i)

and $2n_2^2 + 3 n_2 – 1 = 8 k_2$ …(ii)

Subtracting (ii) from (i), we get

(n1 – n2) [2(n1 + n2) + 3] = 8(k1 – k2).

Since 2(n1 + n3) + 3 is odd, n1 – n2 must be a multiple of 8

i.e. n1 – n2 = 8p or n1 = 8p + n2.

For k2 = 1, (ii) gives n2 = -3. Hence n = 8p – 3.

Hence from case I and II, the solution is n = 8p – 1 or n = 8p – 3 for p ∈ I.

## If 2cosA = x + 1/x , 2cosB = y + 1/y . Show that 2cos(A-B) = x/y + y/x

Q: If $\displaystyle 2 cosA = x + \frac{1}{x}$ , $\displaystyle 2 cosB = y + \frac{1}{y}$ . Show that $\displaystyle 2 cos(A-B) = \frac{x}{y} + \frac{y}{x}$

Solution : $\displaystyle 2 cosA = x + \frac{1}{x}$

Since $\displaystyle 4 sin^2 A = 4 – 4 cos^2 A$

$\displaystyle4 sin^2 A = 4 – (x + \frac{1}{x})^2$

$\displaystyle 4 sin^2 A = – [(x + \frac{1}{x})^2 – 4]$

$\displaystyle 4 sin^2 A = i^2 (x-\frac{1}{x})^2$

$\displaystyle 2sinA = i (x-\frac{1}{x})$

$\displaystyle 2 cosB = y + \frac{1}{y}$

$\displaystyle 2sinB = i (y-\frac{1}{y})$

Now $\displaystyle 2 cos(A – B)= 2(cosA cosB + sinA sinB)$

$\displaystyle = 2 \times \frac{1}{4}[(x + \frac{1}{x})(y + \frac{1}{y}) + i^2 (x-\frac{1}{x})(y-\frac{1}{y})]$

$\displaystyle = \frac{1}{2}[xy + \frac{1}{xy} + \frac{y}{x} + \frac{x}{y}] – \frac{1}{2} [xy + \frac{1}{xy} – \frac{y}{x} – \frac{x}{y}]$

$\displaystyle = \frac{1}{2}. 2 (\frac{y}{x}+ \frac{x}{y})$

$\displaystyle = (\frac{y}{x}+ \frac{x}{y})$

## Value of the expression 2sinx – cos2x is

Q: Value of the expression 2sinx – cos2x is

(A) greater than -3/2

(B) equal to – 3/2

(C) less than -3/2

(D) none of these

Sol. 2 sinx – cos2x = 2 (sin2x + sinx) – 1

$\large = 2 [(sinx + \frac{1}{2})^2 – \frac{1}{4}] – 1$

$\large = 2(sinx + \frac{1}{2})^2 – \frac{3}{2} \ge -\frac{3}{2}$

⇒ 2 sinx – cos2x ≥ – 3/2

Hence (A), (B) are the correct answers.

## If 4 sinA + secA = 0 then tanA equals to

Q: If 4 sinA + secA = 0 then tanA equals to

(A) 4 + √2

(B) – 2 + √3

(C) 2 + 4√3

(D) –2 –√3

Sol.  4sinA + secA = 0

$\large 4 sinA + \frac{1}{cosA} = 0$

$\large \frac{4sinA cosA + 1}{cosA} = 0$

4 sinA cosA + 1 = 0

2(2 sinA cosA ) = -1

2( sin2A ) = -1

⇒ sin2A = -1/2

$\large \frac{2 tanA}{1 + tan^2A} = – \frac{1}{2}$

tanA = – 2 ± √3

Hence (B), (D) are the correct answers.

## If Tn = sin^nθ + cos^nθ , then $\frac{T_6 – T_4}{T_6} = m$ holds for values of m satisfying

Q: If Tn = sinnθ + cosnθ , then $\frac{T_6 – T_4}{T_6} = m$ holds for values of m satisfying

(A) $\large m \in [-1 , \frac{1}{3}]$

(B) $\large m \in [0 , \frac{1}{3}]$

(C) $\large m \in [-1 , 0]$

(D) $\large m \in [-1 , -\frac{1}{2}]$

Sol: $\large \frac{T_6 – T_4}{T_6} = \frac{-sin^2 2\theta}{4-3sin^2 2\theta} = m$

$\large sin^2 2\theta = \frac{4 m}{-1 + 3 m}$

$\large 0 \le \frac{4 m}{-1 + 3 m} \le 1$

$\large m \in [-1 , 0]$

Hence (C), (D) are the correct answers