30 - Trigonometry Questions Answers

if x is any real number and cosA=x^2+1/2x then cos A value is 

 

Asked By: RAMACHANDRAM
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Joshi sir comment

x2+1 > 2x so cosA>1 which is not possible

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

if x=2Cosθ-Cos2θ and y=2Sinθ-Sin2θ prove that dy/dx=tan(3θ/2)

Asked By: SARABJEET
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Joshi sir comment

calculate dx/dθ and dy/dθ

then dy/dx = [2cosθ-2cos2θ]/[-2sinθ+2sin2θ]

now solve

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

and sinC+sinD = 2sin[(C+D)/2]cos[(C-D)/2]

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. 

If cos-1x - cos-1(y/2) = α , then 4x2-4xycosα + y2 is equal to ?

Asked By: AMIT DAS
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Joshi sir comment

according to the given condition 

cos-1x  =  cos-1(y/2)  +  a

so x = cos ( cos-1y/2   +  a)

so x = y/2 cosa  -  sin cos-1y/2 + sina

so x = y/2 cosa  - [1-(y2/4)]1/2 + sina

so x - sina - y/2 cosa = -[1-(y2/4)]1/2

now square both side and solve

 

If  sin-1a + sin-1b+ sin-1c = π , then the value of { a√(1-a2) + b√(1-b2) + c√(1-c2)}

Asked By: AMIT DAS
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Joshi sir comment

let sin-1a = A ,    sin-1b = B    and    sin-1c= C 

so  { a√(1-a2) + b√(1-b2) + c√(1-c2)}   = sinAcosA + sinBcosB + sinCcosC   = 1/2 (sin2A+sin2B+sin2C)

                                                                                                          = 1/2 (4sinAsinBsinC)

                                                                                                          = 2abc                                                    

If    [cot‾1x] + [cos‾1x] = 0 , then complete set of value of x is ( [ * ] is GIF)?

Asked By: AMIT DAS
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Joshi sir comment

These are the graphs for cot-1x and cos-1x   

violet for cos and green for cot

from graph it is clear that cos part with integer function will be 0 for cos1<x≤1

and cot part with integer function will be 0 for cot1<x<∞

so answer will be cot1<x≤1

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

What are trigonometrical identities?

Asked By: SIDDHANT KAPOOR
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Joshi sir comment

sin2A + cos2A = 1 and two similar formulae in terms of tanA, cotA, secA and cosecA 

these are identities because these are true for all values of A

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