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1.

Prove that:`tan6^0tan42^0tan66^0tan78^0=1.`

Answer» `L.H.S. = tan6^@tan42^@tan66^@tan78^@`
`=(sin6^@sin42^@sin66^@sin78^@)/(cos6^@cos42^@cos66^@cos78^@)`
`=((2sin6^@sin66^@)(2sin42^@sin78^@))/((2cos6^@cos66^@)(2cos42^@cos78^@))`
`=((cos60^@-cos72^@)(cos36^@-cos120^@))/((cos60^@+cos72^@)(cos36^@+cos120^@))`
`=((1/2-sin18^@)(cos36^@+1/2))/((1/2+sin18^@)(cos36^@-1/2))...[As costheta = sin(90-theta)]`
`=((1-2sin18^@)(2cos36^@+1))/((1+2sin18^@)(2cos36^@-1))`
Now, putting, `sin18^@ = (sqrt5-1)/4 and cos36^@ = (sqrt5+1)/4`
`=((1-2((sqrt5-1)/4))((2(sqrt5+1)/4)+1))/((1+2((sqrt5+1)/4))(2((sqrt5-1)/4)-1))`
`=((3-sqrt5)(sqrt5+3))/((1+sqrt5)(sqrt5-1))`
`=(9-5)/(5-1)`
`=4/4 =1 = R.H.S.`
Discussion
2.

Prove that:`cos^4pi/8+cos^4(3pi)/8+cos^4(5pi)/8+cos^4(7pi)/8=3/2`

Answer» `L.H.S. = cos^4((pi)/8)+cos^4((3pi)/8)+cos^4((5pi)/8)+cos^4((7pi)/8)`
`=cos^4((pi)/8)+cos^4((3pi)/8)+cos^4(pi-(3pi)/8)+cos^4(pi-(pi)/8)`
`=cos^4((pi)/8)+cos^4((3pi)/8)+cos^4((3pi)/8)+cos^4((pi)/8)...[As cos(pi-theta) = costheta]`
`=2(cos^4((pi)/8)+cos^4((3pi)/8))`
`=2(cos^4((pi)/8)+sin^4(pi/2-(3pi)/8))`
`=2(cos^4((pi)/8)+sin^4((pi)/8))`
`=2((cos^2(pi/8)+sin^2(pi/8))^2 - 2sin^2(pi/8)cos^2(pi/8))`
`=2(1-(2sin(pi/8)cos(pi/8))^2/2)`
`=2(1-(sin(2*pi/8))^2/2)`
`=2-sin^2(pi/4)`
`=2-(1/sqrt2)^2`
`=2-1/2`
`=3/2 = R.H.S.`
Discussion
3.

Prove that: `(1+cospi/8)(1+cos(3pi)/8)(1+cos(5pi)/8)(1+cos(7pi)/8)=1/8`

Answer» `L.H.S. =(1+cos(pi/8)) (1+cos((3pi)/8))(1+cos((5pi)/8)) (1+cos((7pi)/8)) `
`=(1+cos(pi/8)) (1+cos((3pi)/8))(1+cos(pi-(3pi)/8)) (1+cos(pi -(pi)/8)) `
`=(1+cos(pi/8)) (1+cos((3pi)/8))(1-cos((3pi)/8)) (1-cos((pi)/8)) `
`=(1-cos^2(pi/8)) (1-cos^2((3pi)/8)) `
`=sin^2(pi/8)sin^2((3pi)/8)`
`=sin^2(pi/8)cos^2(pi/2-(3pi)/8)`
`=sin^2(pi/8)cos^2(pi/8)`
`=1/4(2sin(pi/8)cos(pi/8))^2`
`=1/4(sin(2*pi/8))^2`
`=1/4(sin(pi/4))^2`
`=1/4(1/sqrt2)^2`
`=1/8 = R.H.S.`
Discussion
4.

Prove that:`"cot"pi/(24)=sqrt(2)+sqrt(3)+sqrt(4)+sqrt(6)`

Answer» `cot (pi/24) = cos (pi/24)/sin(pi/24)`
`=cos (pi/24)/sin(pi/24)**(2cos(pi/24))/(2cos(pi/24))`
`=(2cos^2 (pi/24))/(2sin(pi/24)cos(pi/24))`
`=(1+cos(pi/12))/(sin(pi/12))`
`=(1+cos(pi/4-pi/6))/(sin(pi/4-pi/6))`
`=(1+cos(pi/4)cos(pi/6)+sin(pi/4)sin(pi/6))/(sin(pi/4)cos(pi/6)-cos(pi/4)sin(pi/6))`
`=(1+(1/sqrt2)(sqrt3/2)+(1/sqrt2)(1/2))/((1/sqrt2)(sqrt3/2)-(1/sqrt2)(1/2))`
`=(2sqrt2+sqrt3+1)/(sqrt3-1)`
`=(2sqrt2+sqrt3+1)/(sqrt3-1)**(sqrt3+1)/(sqrt3+1)`
`=(2sqrt6+3+sqrt3+2sqrt2+sqrt3+1)/(3-1)`
`=sqrt6+sqrt2+sqrt3+2`
`=sqrt2+sqrt3+sqrt4+sqrt6 = R.H.S.`
Discussion
5.

Prove that `(tan3x)/(tanx)`never lies between `1/3a n d3`.

Answer» `(tan3x)/tanx = (3tan - tan^3x)/(tanx(1-3tan^2x))`
`=(3-tan^2x)/(1-3tan^2x)`
Now, let `(3-tan^2x)/(1-3tan^2x) = a`
`=>3-tan^2x = a-3atan^2x`
`=>tan^2x(3a-1) = a-3`
`=>tan^2x = (a-3)/(3a-1)`
It means, `(a-3)/(3a-1) gt 0.`So, `a-3 gt 0 and 3a-1 lt 0`
`=> a gt 3 and a lt 1/3`
So, `(tan3x)/tanx` never lies between `1/3` and `3.`
Discussion
6.

Prove that:`sin(pi/10) sin((13pi)/10)=-1/4`

Answer» Here, we will use,
`sin 18^@ = (sqrt5-1)/4 and cos 36^@ = (sqrt5+1)/4`
Now, `L.H.S. = sin(pi/10)sin((13pi)/10)`
`=sin18^@sin(pi+(3pi)/10)`
`=sin18^@(-sin((3pi)/10))`
`=-sin18^@sin54^@`
`=-sin18^@sin(90^@-36^@)`
`=-sin18^@cos36^@`
`=-((sqrt5-1)/4)( (sqrt5+1)/4) `
`=-((5-1)/16)`
`=-4/16`
`=-1/4 = R.H.S.`
Discussion
7.

Prove that `cos^3A+cos^3(120^0+A)+cos^3(240^0+A)=3/4cos3A`

Answer» `cosA + cos(120^@+A) +cos(240^@+A)`
`= cosA + (2cos((120+A+240+A)/2)cos((120+A-240-A)/2))`
`= cosA + (2cos(180+A)cos(-60^@))`
`= cosA + (2cos(180+A)cos60^@)`
`= cosA + (-2cosA(1/2))`
`=cosA - cosA = 0`
We know, when, `a+b+c = 0`,
`a^3+b^3+c^3 = 3abc`
`:. L.H.S. = cos^3A + cos^3(120^@+A) +cos^3(240^@+A) = 3cosAcos(120^@+A)cos(240^@+A)`
`=3cosAcos(180-(60-A))cos(180+(60+A))`
`=3cosAcos(-(60-A))cos(-(60+A))`
`=3cosAcos(60-A)cos(60+A)`
Using `cos(C-D)cos(C+D) = cos^2C - sin^2D`
`=3cosA[cos^2 60^@-sin^2A]`
`=3cosA[1/4-(1-cos^2A)]`
`=3cosA[-3/4+cos^2A]`
`=3/4(4cos^3A - 3cosA)`
`=3/4cos3A = R.H.S.`
Discussion
8.

If `acos2theta+bsin2theta=ch a salphaa n dbeta`as its roots , then prove that`tanalpha+t a nbeta=(2a b)/(a+c)``tanalpha+t a nbeta=(c-a)/(c+a)``tan(alpha+beta)=b/a`

Answer» Here, `acos2theta+bsin2theta = c`
`=>a((1-tan^2theta)/(1+tan^2theta)) +b((2tantheta)/(1+tan^2theta)) = c`
`=>a(1-tan^2theta)+2btantheta = c(1+tan^2theta)`
`=>(c+a)tan^2theta-2btantheta+(c-a) = 0->(1)`
Now, given equation has roots `alpha` and `beta`.
So, equation (1), will have roots `tan alpha` and `tan beta`.
`:.` Sum of the roots `= (-(-2b))/(c+a)`
`:. tanalpha+tanbeta = (2b)/(a+c)`
Discussion
9.

Prove that: `|sintheta"sin"(60-theta)sin(60+theta)|lt=1/4`for all values of `theta`.

Answer» `|sinthetasin(60-theta)sin(60+theta)|`
`=1/2|sintheta(2sin(60-theta)sin(60+theta))|`
Using, `cosC-cosD = -2sin((C+D)/2)((C-D)/2),`
`=1/2|sintheta(cos2theta - cos120^@)|`
`=1/2|sinthetacos2theta - sintheta(-sin30^@)|`
`=1/2|sinthetacos2theta +1/2sintheta|`
`=1/4|2sinthetacos2theta +sintheta|`
Using `sinC-sinD = 2cos((C+D)/2)sin((C-D)/2)`
`= 1/4|sin3theta - sintheta+sintheta|`
`=1/4|sin3theta|`
Now, we know, `-1 le sin3theta le 1`
`=>|sin3theta| le 1`
`=>1/4|sin3theta| le 1/4`
`:. |sinthetasin(60-theta)sin(60+theta)| le 1/4.`
Discussion
10.

If `theta=pi/(2^n+1)`, prove that: `2^ncosthetacos2thetacos2^2 cos2^(n-1)theta=1.`

Answer» `L.H.S. = 2^ncosthetacos2theta...cos2^(n-1)theta`
`=2costheta*2cos2theta*2cos2^2theta...2cos2^(n-1)theta`
`=1/sintheta[2sinthetacostheta*2cos2theta*2cos2^2theta...2cos2^(n-1)theta]`
`=1/sintheta[sin2theta*2cos2theta*2cos2^2theta...2cos2^(n-1)theta]`
`=1/sintheta[sin2^2theta*2cos2^2theta...2cos2^(n-1)theta]`
If we solve it in this way, we finally come to the below solution,
`=1/sintheta[2sin2^(n-1)thetacos2^(n-1)theta]`
`=sin2^ntheta/sintheta`
`=sin(pi-2^ntheta)/sintheta`...As `sin(pi-theta) = sintheta`
`=sin(pi-2^n(pi/(2^n+1)))/sintheta`
`=sin((2^npi+pi-2^npi)/(2^n+1))/sintheta`
`=sin((pi)/(2^n+1))/sintheta`
`=sintheta/sintheta`
`=1 = R.H.S.`
Discussion
11.

Prove that: `cos5A=16cos^5A-20cos^3A+5cosA`

Answer» `cos5A = cos(3A+2A) = cos3Acos2A - sin3Asin2A`
`=(4cos^3A-3cosA)(2cos^2A -1) - (3sinA - 4sin^3A)(2sinAcosA)`
`=(8cos^5A-10cos^3A+3cosA)- (3-4sin^2A)(2sin^2AcosA)`
`=(8cos^5A-10cos^3A+3cosA)- (3-4(1-cos^2A))(2(1-cos^2A)cosA)`
`=(8cos^5A-10cos^3A+3cosA)- (3-4+4cos^2A)(2cosA - 2cos^3A)`
`=(8cos^5A-10cos^3A+3cosA)- (-8cos^5A+10cos^3A- 2cosA)`
`=16cos^5A-20cos^3A+5cosA= R.H.S.`
Discussion
12.

If `tanalpha=1/7,sinbeta=1/(sqrt(10)),`prove that`alpha+2beta=pi/4`, where `0

Answer» `sin beta = 1/sqrt10`
`:. cos beta = sqrt(1-(1/sqrt10)^2) = 3/sqrt10`
`:. tanbeta = sinbeta/cosbeta = 1/3`
Now, `tan2beta = (2tanbeta)/(1-tan^2beta) = (2/3)/(1-1/9) = 6/8 = 3/4`
Now, `tan(alpha+2beta) = (tanalpha+tan2beta)/(1-tanalphatan2beta)`
`=(1/7+3/4)/(1-1/7(3/4)) = (25/28)/(25/28) = 1`
`:. tan(alpha+2beta) = tan (pi/4)`
`:. alpha+2beta = pi/4.`
Discussion
13.

`sin5A=5cos^4AsinA-10cos^2Asin^3A+sin^5A`

Answer» `L.H.S. = sin5A = sin(3A+2A) = sin3Acos2A+sin2Acos3A`
`=(3sinA-4sin^3A)(1-2sin^2A)+(4cos^3A-3cosA)(2sinAcosA)`
`=3sinA-6sin^3A-4sin^3A+8sin^5A+8cos^4AsinA - 6sinAcos^2A`
`=sin^5A+5cos^4AsinA-10cos^2Asin^3A = R.H.S.`
Discussion
14.

Prove that:`(2cos2^ntheta+1)/(2costheta+1)=(2costheta-1)(2cos2theta-1)(2cos2^2theta-1)(2cos2^(n-1)theta-1)`

Answer» `R.H.S. = (2costheta-1)(2cos2theta-1)...(2cos2^(n-1)theta - 1)`
`=1/(2costheta+1)[(2costheta+1)(2costheta-1)(2cos2theta-1)...(2cos2^(n-1)theta - 1)]`
`=1/(2costheta+1)[(4cos^2theta-1)(2cos2theta-1)...(2cos2^(n-1)theta - 1)]`
`=1/(2costheta+1)[(2(2cos^2theta)-1)(2cos2theta-1)...(2cos2^(n-1)theta - 1)]`
`=1/(2costheta+1)[(2(1+cos2theta)-1)(2cos2theta-1)...(2cos2^(n-1)theta - 1)]`
`=1/(2costheta+1)[(2cos2theta+1)(2cos2theta-1)...(2cos2^(n-1)theta - 1)]`
`=1/(2costheta+1)[(4cos^ 2theta -1)...(2cos2^(n-1)theta - 1)]`
`=1/(2costheta+1)[(2cos4theta+1)(2cos4theta-1)...(2cos2^(n-1)theta - 1)]`
Similarly, if we solve this, it will become,
`=1/(2costheta+1)[(2cos2^(n-1)theta + 1)(2cos2^(n-1)theta - 1)]`
`=1/(2costheta+1)[4cos^2 2^(n-1)theta-1]`
`=1/(2costheta+1)[2(cos2^n theta+1)-1]`
`=(2cos2^n theta+1)/(2costheta+1) = R.H.S.`
Discussion
15.

The value of `(sin5alpha-sin3alpha)/(cos5alpha+2cos4alpha+cos3alpha)` is

Answer» Here, we will use,
`sinC - sinD = 2sin((C-D)/2) sin((C+D)/2)`
`cosC + cosD = 2cos((C+D)/2)cos((C-D)/2)`
`L.H.S. = (sin5alpha-sin3alpha)/(cos5alpha+2cos4alpha+cos3alpha)`
`=(2sinalphacos4alpha)/(2cos4alphacosalpha+2cos4alpha)`
`=sinalpha/(1+cos alpha)`
`=(2sin(alpha/2)cos(alpha/2))/(2cos^2(alpha/2))`
`=tan(alpha/2)`
`:. (sin5alpha-sin3alpha)/(cos5alpha+2cos4alpha+cos3alpha) = tan(alpha/2)`
Discussion