6 - Trigonometrical Equations Questions Answers

Asked By: UTTAM RAJ
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Joshi sir comment
2cos7x/cos(3)+sin(3)>2^cos2x
Asked By: AARSHI
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Joshi sir comment

Please submit what we have to do and also correct the question 

it is cos3 or cos3x 

(tanx)^2(tan3x)^2(tan4x)=(tanx)^2-(tan3x) ^2 + tan4x.
Asked By: AARSHI
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Joshi sir comment

I dont know what the question is but if the question is to solve this eq. then the method will be as given below

tan2x tan23x tan4x = tan2x  - tan23x + tan4x

so tan4x =  [tan2x  - tan23x] / [tan2x tan23x-1]

now split RHS terms by formula x2-y2= (x+y)(x-y)

so RHS = tan4xtan2x 

now solve

sinA + sinB + sinC = 4cosA/2.cosB/2.cosC/2

Asked By: AVI GARG
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Joshi sir comment

sinA + sinB + sinC = 2sin[(A+B)/2]cos[(A-B)/2] + 2sinC/2cosC/2

                                  = 2sin[(π-C)/2]cos[(A-B)/2] + 2sinC/2cosC/2

                                  = 2cosC/2cos[(A-B)/2] + 2sinC/2cosC/2

                                  = 2cosC/2* [cos[(A-B)/2]+[cos(A+B)/2]                     here sinC/2 = cos(A+B)/2

now use cosC + cosD = 2 cos[(C+D)/2]cos[(C-D)/2]

You should remember that the given formula is based on a triangle. 

A regular hexagon and a regular dodecagon are inscribed in the same circle. if the side of the dodecagon is (√3-1), then the side of hexagon is

 

Asked By: HEMA RANI
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Joshi sir comment

for dodecagon sin(2π/40) = (a/2)/r     so    r = (a/2)/ sin(π/20)

now for hexagon sin(2π/12) = (x/2)/r       so x/2 = r sin(π/6)  =  (a/2) sin(π/6)/ sin(π/20)  

so  x  = (√3-1) (1/2) / ((√5-1)/4)                      here sin(π/20) = (√5-1)/4

or x = 4(√3-1) / 2(√5-1) 

now solve it for further simplification

If     a sinx + b cos(x+θ) + b cos(x-θ) = d, then the minimum value of |cos θ| is equal to :

sir i have expanded cos(x+θ) & cos(x-θ) and after that i got the equation: (d sec x - a tanx)/2b = cos θ. now , can i write  minimum value of (d sec x -a tanx) = (d²-a²)^(1/2).. as all the options contain (d²-a²)^(1/2).......or do tell the other hint to solve it..

Asked By: AMIT DAS
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Joshi sir comment

Dear amit your starting steps are correct 

now let us consider that   d secx - a tanx  = y

on differentiating we will get d secx tanx - a sec2x = dy/dx

on comparing with 0 we get d tanx = a secx       or               sinx = a/d               so         secx = d/√(d2-a2)           and   tanx = a/√(d2-a2)

so minimum of given expression = [d2/ √(d2-a2)] - [a2/√(d2-a2)]  =        √(d2-a2)   

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