2026/045) Prove that $(4\cos^2 9^{\circ}–3)(4\cos^2 27^{\circ}–3)=\tan 9^{\circ}$

We shall use formula for $\cos 3x$  

$\cos 3x = 4 \cos^3 x - 3 \cos x$

so we have $4 \cos ^2 x - 3 = \frac{\cos 3x}{ \cos x}$

Hence 

$4 \cos ^2 9^{\circ} - 3 = \frac{\cos 27^{\circ}}{\cos 9^{\circ}}$

And 

$4 \cos ^2 27^{\circ} - 3 = \frac{\cos 81^{\circ}}{\cos 27^{\circ}}$

so   $(4\cos^2 9^{\circ}–3)(4\cos^2 27^{\circ}–3)$ = $\frac{\cos 27^{\circ}}{\cos 9^{\circ}}$ *  $\frac{\cos 81^{\circ}}{\cos 27^{\circ}}$ 

= $\frac{\cos 81^{\circ}}{\cos 9^{\circ}}$

 = $\frac{\sin 9^{\circ}}{\cos 9^{\circ}}$ using $\cos \theta = \sin (90^{\circ}-\theta)$

 = $\tan 9^{\circ}$

Hence Proved  

 

2020/021) Evaluate $\tan^2 \frac{\pi}{7} + \tan^2 \frac{2\pi}{7} + \tan^2 \frac{3\pi}{7}$

Let $\theta = \frac{\pi}{7}$

We need to find $\tan^2 \theta + \tan^2 2\theta + \tan^2 3\theta$ 
Now $7\theta = \pi$

Or $4\theta = \pi - 3\theta$

Taking $\tan$ of both sides we get

Or $\tan 4\theta = \tan (\pi - 3\theta) = - \tan 3\theta$

Or $\frac{4\tan \theta - 4\tan^3 \theta}{1-6\tan ^2 \theta + \tan ^4 \theta} - \frac{3\tan \theta - \tan ^3\theta}{1-3\tan ^2 \theta}$

Or  $\frac{4 - 4\tan^2 \theta}{1-6\tan ^2 \theta + \tan ^4 \theta} + \frac{3 - \tan ^2\theta}{1-3\tan ^2 \theta}$

 Or  $(4 - 4\tan^2 \theta)(1-3\tan ^2 \theta) + (1-6\tan ^2 \theta + \tan ^4 \theta)(3 - \tan ^2\theta)$

Or $4- 16 \tan ^2 \theta + 12 \tan ^4 \theta + 3  -19 \tan ^2 \theta + 9\tan^4 \theta - \tan ^26\theta = 0$

Or $ \tan ^6\theta  - 21\tan ^4 \theta + 35 \tan ^2 \theta -7=0$

The above equation is a cubic equation in $\tan^2 \theta$ whose roots are $\tan \theta$, $\tan 2\theta$, and 

$\tan 3\theta$,

so using Vieta's formula $\tan^2 \theta + \tan^2 2\theta + \tan^2 3\theta= 21$ 

or $\tan^2 \frac{\pi}{7} + \tan^2 \frac{2\pi}{7} + \tan^2 \frac{3\pi}{7}= 21$ 

as a corollary adding 1 to each term of LHS and so adding 3 to RHS we get

 $\sec^2 \frac{\pi}{7}+ \sec^2 \frac{2\pi}{7} + \sec^2 \frac{3\pi}{7}= 24$

 

2015/023) show that $\tan\, 9^{\circ} - \tan\, 27^{\circ}- \tan\, 63^{\circ} + \tan\, 81^{\circ}= 4$

we have $\tan\, x + \cot\, x = \tan\, x + \dfrac{1}{\tan\, x}$

 = $\dfrac{tan ^2 x + 1}{\ tan\, x} = \dfrac{sec^2 x}{\tan,x}= \dfrac{1}{\cos\,x\sin\,x}=\dfrac{2}{2\cos\,x\sin\,x} = \dfrac{2}{\sin 2x} = 2 \csc 2x$

 so

 $\tan\, 9^{\circ} + \tan\, 81^{\circ} = 2 \csc\, 18^{\circ} $

$\tan\, 27^{\circ} + \tan\, 63^{\circ} = 2 \csc\,  54^{\circ} $

so given value
= $\tan\, 9^{\circ} - \tan\, 27^{\circ}- \tan\, 63^{\circ} + \tan\, 81^{\circ}=2( \csc\, 18^{\circ} - \csc\, 54^{\circ})$ 

= $2(\dfrac{1}{\sin\,18^{\circ}} - \dfrac{1}{\sin\, 54^{\circ}})$

= $2 \dfrac{2 \cos\, 36^{\circ}  \sin\, 18^{\circ}}{ \sin \,18^{\circ}  \sin \,54^{\circ} }$
=  4

refer to https://in.answers.yahoo.com/question/index?qid=20111213202340AABYrgC

2011/081) prove that cos(3π/14)sin(π/7)cos(π/14) + sin(π/14)cos(π/7)cos(3π/14) + sin(3π/14)sin(π/14)sin(π/7) = cos(π/7)sin(3π/14)cos(π/14)

proof

cos(3π/14)sin(π/7)cos(π/14) + sin(π/14)cos(π/7)cos(3π/14) + sin(3π/14)sin(π/14)sin(π/7)

(knowing that sin (3π/14) is on the right so combine the terms not having sin(3π/14)

= cos(3π/14)[sin(π/7)cos(π/14) + sin(π/14)cos(π/7)] + sin(3π/14)sin(π/14)sin(π/7)

(using sin A cos B + cos A sin B = sn (A+B) in next line

= cos(3π/14)[sin(π/7+π/14] + sin(3π/14)sin(π/14)sin(π/7)

= cos(3π/14)sin(3π/14) + sin(3π/14)sin(π/14)sin(π/7)

= sin(3π/14)[ cos(3π/14) + sin(π/14)sin(π/7)]

= sin(3π/14)[ cos(π/7+π/14)+ sin(π/14)sin(π/7)] (as cos (A+B) shall cancel sin A sin B by contributing –ve of it)

= sin(3π/14)[ cos(π/7) cos (π/14) - sin(π/14)sin(π/7+ sin(π/14)sin(π/7)]

= cos(π/7)sin(3π/14)cos(π/14)

2011/080) prove that Tan 55 = tan35+ 2tan20

20 = 55 - 35

take tan of both sides

tan 20 = (tan 55 - tan 35)/(1+ tan 55 tan 35)

but as 55 + 35 = 90 so tan 35 = cot(90-35) = cot 55

so tan 20 = (tan 55 - tan 35)/(1+ tan 55 tan 35) = tan 20 = (tan 55 - tan 35)/(1+ tan 55 cot 55)
= (tan 55 - tan 35)/(1+ 1) = (tan 55 - tan 35)/2
or 2 tan 20 = tan 55 - tan 35

or tan 55 = tan 35 + 2 tan 20

2011/072) Prove that ( 1+ cos π/8)( 1+ cos 3π/8)( 1+ cos 5π/8)( 1+ cos 7π/8) =1/8?

(1 + cos π/8)(1 + cos 3π/8)(1 + cos 5π/8)(1 + cos 7π/8)
= (1 + cos π/8)(1 + cos 3π/8)(1 - cos 3π/8)(1 - cos π/8), since cos(π - t) = -cos t
= (1 - cos² π/8)(1 - cos² 3π/8)
= (1 - cos² π/8)(1 - sin² π/8), via cos(π/2 - t) = sin t
= (1 - cos² π/8)(1 - sin² π/8)
= sin ^2 π/8 cos ^2 π/8
= (1/4) (2 sin π/8 cos π/8)²
= (1/4) (sin 2π/8)², by double angle formula
= (1/4) (sin π/4)²
= (1/4) (1/√2)²
= (1/4)(1/2)
= 1/8.

2011/071) Show that 2 tan 20˚ + 4 tan 40˚ + 8 tan 80˚ = 9 (cot 10˚ - tan 10˚)

we first prove one relationship
cot x - tan x = cos x/ sin x - sin x/ cos x = (cos^2 x - sin ^2 x) /( sinx cos x) = 2 cot 2x

so

cot x - tan x = 2 cot 2x ...1

also
tan x = cot x - 2 cot 2x ...2

from 2 we generate following 3 relationships

2 tan 20= 2 cot 20 - 4 cot 40 ... 3
4 tan 40 = 4 cot 40 - 8 cot 40... 4
8 tan 80 = 8 cot 80 - 16 cot 160... 5
adding we get
2 tan 20˚ + 4 tan 40˚ + 8 tan 80 = 2 cot 20 - 16 cot 160
= 2 cot 20 + 16 cot(180-160)as cot x = - cot(180-x)
= 2 cot 20 + 16 cot 20
= 18 cot 20
= 9( 2 cot 20)
= 9( cot 10 - tan 10)from 1
proved

2011/044) show that $\sin \frac{\pi}{14} \sin \frac{3\pi}{14} \sin\frac{5\pi}{14} = \frac{1}{8}$

LHS
$= \sin \frac{\pi}{14} \cos (\frac{\pi}{2} -\frac{3\pi}{14}) \cos(\frac{\pi}{2} -\frac{5\pi}{14})$

$= \sin \frac{\pi}{14} \cos (\frac{4\pi}{14})  \cos (\frac{2 \pi}{14})$

$= \sin \frac{\pi}{14} \cos (\frac{2\pi}{14})  \cos (\frac{4 \pi}{14})$
$= \dfrac{\cos \frac{\pi}{14} \sin \frac{\pi}{14} \cos (\frac{2\pi}{14})  \cos (\frac{4 \pi}{14})}{\cos \frac{\pi}{14}} $
$=\dfrac{1}{2}\frac{\sin \frac{2\pi}{14} \cos (\frac{2\pi}{14})  \cos (\frac{4 \pi}{14})}{\cos \frac{\pi}{14}} $
$=\dfrac{1}{4}\dfrac{\sin \frac{4\pi}{14}  \cos (\frac{4 \pi}{14})}{\cos \frac{\pi}{14}} $
$=\dfrac{1}{8}\dfrac{\sin \frac{8\pi}{14}}{\cos \frac{\pi}{14}} $
$=\dfrac{1}{8}\dfrac{\sin(\pi -  \frac{8\pi}{14})}{\cos \frac{\pi}{14}} $
$=\dfrac{1}{8}\dfrac{\sin(\frac{6\pi}{14}}{\cos \frac{\pi}{14}} $
$=\dfrac{1}{8}\dfrac{\cos(\frac{\pi}{2} -  \frac{6\pi}{14})}{\cos \frac{\pi}{14}} $
$=\dfrac{1}{8}\dfrac{\cos(\frac{\pi}{2} -  \frac{6\pi}{14})}{\cos \frac{\pi}{14}} $
 $=\dfrac{1}{8}\frac{\cos \frac{\pi}{14}}{\cos \frac{\pi}{14}} $
$=\frac{1}{8}$

edited the above based on the comment to keep the flow

2010/043) Given that 5cos x + 12 cos y = 13 ? maximum value of 5 sin x + 12sin y

What is the maximum value of 5 sin x + 12sin y , given that 5cos x + 12 cos y = 13 ?

we start with

(5 sin x + 12sin y)^2 + (5cos x + 12 cos y)^2 = 169 + 120 cos(x-y)

or (5 sin x + 12 sin y)^2 + 169 = 169 + 120 cos(x-y)

or 5 sin x + 12 sin y = 120 cos(x-y)

5 sin x + 12 sin y = sqrt( 120 cos(x-y))

this is maximum when cos(x-y) is maximum

theoritically cos(x-y) is maximum = 1 when x = y

but is is possible under given case that is x =y satisfies


then we get

5 cos x + 12 cos x = 13 so cos x = 13/17

it is possible

so maximum value of 5 sin x + 12sin y = sqrt(120)

2010/008) Show that there is exactly one value of x which satisfies the equation,?

2 cos^2 (x^3 + x) = 2^x + 2^(-x).

proof:
the maximum value of LHS = 2

as cos^2(x^3+x) <= 1

the minimum value of RHS

let 2^x = y

so y + 1/y = (sqrt(y) - 1/(sqrt(y))^2 + 2

so minum value of RHS = 2


so both sides are same when both are 2

RHS = 2 when y = 1 or x= 0

LHS = 2 when 2 cos^2(x^3+x) = 2

or cos^2(x^3+x) = 1

x = 0 satisies LHS

so x = 0 is the only value that satisfied equality

2010/007) prove 64 {Cos^8(x) + Sin^8(x)} = cos8 x + 28cos 4x + 35

We know

(a+b)^8 = a^8 + 8 a^7b + 28 a^6b^2 + 56 a^5b^3 + 70a^4b^4 + 56 a^3b^5 + 28 a^2b^6+8ab^7+b^8

And (a-b)^8 = = a^8 - 8 a^7b + 28 a^6b^2 - 56 a^5b^3 + 70a^4b^4 - 56 a^3b^5 + 28 a^2b^6 -8ab^7+b^8

So (a+b)^8 + (a-b)^8 = 2(a^8 + b^8) + 56 (a^6b^2+a^2b^6) + 140(a^4b^4)

Putting a = e^ix and b = e^-ix

LHS = (e^ix+e^-ix)^8 + (e^ix-e^-ix)^8 = 2(e^i8x + e^-i8x)+ 56(e^4ix + e^-4ix) + 140

Or (2 cos x)^8 + (2i sin x)^8 = 2 ( 2 * cos 8x) + 56 * 2 * cos 4x + 140

Or 2^8(cos ^8x + sin ^8x) = 4 * ( cos 8x + 28 cos 4x + 35)

Or 2^6(cos ^8x + sin ^8x) = ( cos 8x + 28 cos 4x + 35)

Or 64(cos ^8x + sin ^8x) = ( cos 8x + 28 cos 4x + 35)

2010/006) equation sin(w) = (a^2 + b^2 + c^2)/(ab+bc+ca), where a,b,c are fixed nozero real numbers

has a solution for w
A) Whatever be a,b,c;
B) iff a^2 + b^2 + c^2 < 1;
C) iff a,b and c all lie in the interval (-1,1);
D) iff a = b = c;
Explain your answer.

ans

we know by GM AM enaquality a^2+b^2 >= 2ab
a^2+c^2 >= 2ac
and b^2+c^2 >= 2bc

adding all 3 we get 2(a^2+b^2+c^2) >= 2(ab+bc+ca)

or (a^2+b^2+c^2)/(ab+bc+ca)>= 1 and equal only when a= b= c

sin (w) cannot be > 1 so ans is when (a^2+b^2+c^2)/(ab+bc+ca) = 1 or a= b =c

that is d

2009/031) simplify sin10*sin50*sin60*sin70*sin90

we know sin 90 = 1 and sin 60 = sqrt(3)/2
hence
sin10*sin50*sin60*sin70*sin90 = sqrt(3/2) sin 10 sin 50 sin 70
= sqrt(3)/2 sin 10 sin (90-40) sin (90-20)
= sqrt(3)/2 sin 10 cos 40 cos 20 ( as sin (90-x) = cos x)
= sqrt(3)/2 sin 10 cos 20 cos 40 ..1

Knowing sin 2a =2 sin a cos a and 20 is double of 10 and 40 is double od 20 we proceed

now sin10 cos 20 cos 40 = (cos 10 sin 10 cos 20 cos 40)/cos 10
= (2 cos 10 sin 10 cos 20 cos 40)/(2 cos 10)
= ( sin 20 cos 20 cos 40)/( 2 cos 10
= (2 sin 20 cos 20 cos 40)/( 4 cos 10)
= ( sin 40 cos 40)/( 4 cos 10)
= ( 2 sin 40 cos 40)/( 8 cos 10)
= sin 80 /( 8 cos 10)
= cos 10/(8 cos 10) = 1/8 ...2

from 1 and 2 we get
sin10*sin50*sin60*sin70*sin90 = sqrt(3)/16

2009/027) If α = 2 arctan [(1 + x)/(1 - x)] and β = arcsin [(1 - x^2)/(1 + x^2)], then what is α + β

we know tan(pi/4+y) = tan (pi/4 + tan y)/(1 tan pi/4-tan y)

so tan (pi/4+y) = (1+ tan y)/(1-tan y))

so pi/4 + y = tan ^-1(1+tan y)/(1-tan y)

put x = tan y

so pi/4 + y = arctan ((1+x)/(1-x))

α = 2 arctan ((1+x)/(1-x)) = pi/2 + 2 tan ^- 1 x .1

again cos 2y = cos ^2 y - sin ^2 y

= cos^2 y(1- tan ^2 y)
= (1- tan ^2 y)/sec^2 y
= (1- tan ^2 y)/(1+ tan ^2y)

so put tan y = x

cos 2y = (1-x^2)/(1+x^2)
2y = arccos ((1-x^2)/(1+x^2))

so arcsin ((1-x^2)/(1+x^2)) = pi/2 - 2y = pi/2 - 2 tan ^- x .2

addimg 1 and 2 we get
α + β = π

2009/016) If a cos² x + b sin² x = c, express tan² x in terms of a, b and c

a cos² x + b sin² x = c
= c(sin ^2 x + cos ^2x) (knowing that 1 = sin ^2 x + cos ^ 2 x)
= c sin ^2 x + c cos ^2 x
or a cos^ 2 x - c cos^2 x = c sin ^2 x - b sin^ 2x
or (a-c) cos ^2x = (c-b) sin ^2 x

or (a-c)/(c-b) = sin ^2 x/ cos ^2 x = tan ^2 x

so tan ^2x = (a-c)/(c-b)

2009/011) Prove that (sin x)/x = cos(x/2)*cos(x/4)*cos(x/8)*cos(x/16)..

Proof:
Sin x = 2 cos (x/2) sin (x/2)
= 2 cos (x/2) (2 cos(x/4) sin (x/4))
= 2^n cos (x/2) cos(x/4) cos(x/8)cos(x/16) …. cos (x/2^n) sin (x/2^n)

So (sin x)/x = 2^n cos (x/2) cos(x/4) cos(x/8)cos(x/16) …. cos (x/2^n) sin (x/2^n)/ x
= cos (x/2) cos(x/4) cos(x/8)cos(x/16) …. cos (x/2^n) sin (x/2^n)/ (x/2^n)
as n goes to infinite sin (x/2^n)/ (x/2^n) goes to 1 and we get

(sin x)/x = cos (x/2) cos(x/4) cos(x/8)cos(x/16) ….


Proved

2009/010) Prove Cos pi/7 + cos 3pi/7 + cos 5pi/7=1/2

To prove
Cos pi/7 + cos 3pi/7 + cos 5pi/7=1/2

let z = cos pi/7 + i sin pi/7

z^7 = cos pi + i sin pi = - 1

so z^7+1 = 0
(z+1) (z^6-z^5+z^4 - z^3+z^2 -z + 1) = 0

as z is not - 1 so

z^6-z^5+z^4 - z^3+z^2 -z + 1 = 0

so z^6-z^5+z^4 - z^3+z^2 -z= -1

so z+z^3+ z^5 = 1 + (z^2+z^4+z6)
equating the real part

cos pi/7 + cos 3pi/7 + cos 5pi/ 7 = -1 - (cos 2pi/7+ cos 4pi/7 + cos 6pi/7)

but cos 6pi/7 = - cos(pi-6pi/7) = - cos pi/7
cos 4pi/7 = - cos 3pi/7
cos 2pi/7 = cos 5pi7

so
cos pi/7 + cos 3pi/7 + cos 5pi/ 7= 1 - (\cos pi/7 + cos 3pi/7 + cos 5pi/ 7)

or 2 (cos pi/7 + cos 3pi/7 + cos 5pi/ 7) = 1
or cos pi/7 + cos 3pi/7 + cos 5pi/ 7 = 1/2

2009/009) Prove that tan (A+B) = (tan A+ tan B)/(1-tan A tan B)

Proof:

We know that

the argument of a complex number z = x + yi is the angle to the real axis

then if we take the 1st number 1 + ai and argument is A and 2nd number 1+bi and argument is B

then a = tan A and b = tan B

1st number = 1 + ai = 1 + i tan A
2nd number = 1+ bi = 1 + i tan B


When we multiply the arguments add so angle is A + B

Now (1+ ai)(1+bi) = (1-ab + i(a+b))

As argument is A+B hence

tan (A+B) = (a+b)/(1-ab)= (tan A + tan B)/(1- tan A tan B) (note tan(arg) = y/x

2009/007) prove sin5A = 5sinA - 20 sin^3A + 16 sin ^5 A

We Know
Sin (A + B) = sin A cos B + cos A sin B
Putting B = 4A we get
sin 5A = sin ( A + 4 A)
= sin A cos 4A + cos A sin 4A
= sin A (1- 2 Sin^2 2A) + cos A 2. Sin 2A cos 2A
( putting cos 2x = 1 – 2 sin ^2 x and sin 2x = 2 sin x cos x and x = 2A)
= sin A(1- 8 Sin^2 A cos^2 A) + Cos A 4 Cos A sin A (1- 2 Sin^2 A)
(putting cos 2x = 1 – 2 sin ^2 x and sin 2x = 2 sin x cos x and x = A)
= sin A - 8 sin ^3A (1-sin^2 A) + 4 sin A Cos ^2 A(1- 2 sin ^2 A)
= sin A - 8 sin ^3 A + 8 Sin ^5 A + 4 sin A(1-sin^2A)(1-2 Sin ^2 A)
= sin A - 8 sin ^3 A + 8 sin ^5 A + 4 sin A ( 1- 3 sin ^2A + 2 Sin ^4 A)
= 5 SIn A - 20 sin ^3 A + 16 sin ^5 A

(note: edited the above based on 3rd comment below to keep the flow)

2009/004) Prove the identity sin^2Acos^2B-cos^2Asin^2B = sin^2A-sin^2B

sin^(2)Acos^(2)B-cos^(2)Asin^(2)B
= sin ^2 A(1- sin ^2 B) - cos^(2)Asin^(2)B
= sin ^2 A - sin ^2 A sin ^2 B - cos^(2)Asin^(2)B
= sin ^2 A - sin ^2 B(sin ^2 A + cos ^2 A)
= sin ^2 A - sin ^2 B