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1. Value of cos2theta when tantheta = 3/4 and pi < theta < 3pi/2

2. Solve 5cos^2x + 4 cos x = 1 if 0 <= x <= 2pi

 May 17, 2019
edited by Guest  May 17, 2019
 #1
avatar+8778 
+3

1.  Find the value of  cos( 2θ )     when     tan( θ )  =  3/4     and     π  <  θ  <  3π/2

 

To use the double-angle formula for cos, let's find sin( θ ) and cos( θ ) with a sketch and the Pythagorean theorem.

 


 

sin( θ )  =  opposite / hypotenuse  =  -3/5

cos( θ )  =  adjacent / hypotenuse  =  -4/5

 

cos( 2θ )  =  cos2( θ ) - sin2( θ )____by the double-angle formula for cos
cos( 2θ )  =  ( -4/5 )2 - ( -3/5 )2by subsituting  ( -4/5 )  in for  cos( θ )  and  ( -3/5 )  in for  sin( θ )
cos( 2θ )  =  16/25 - 9/25 
cos( 2θ )  =  7/25 
 May 17, 2019
 #2
avatar+8778 
+3

2.  Solve   5cos2(x) + 4cos(x)  =  1   if   0  ≤  x  ≤  2π

 

5cos2(x) + 4cos(x)  =  1____  
  Subtract  1  from both sides of the equation. 
5cos2(x) + 4cos(x) - 1  =  0 Notice that this is a quadratic equation. 
  Split  4cos(x) into two terms so that the product of their coefficients is  -5 
5cos2(x) + 5cos(x) - 1cos(x) - 1  =  0   
  Factor  5cos(x)  out of first two terms, factor  -1  out of last two terms. 
5cos(x)(cos(x) + 1) - 1(cos(x) + 1)  =  0   
  Factor  (cos(x) + 1)  out of both remaining terms. 
( cos(x) + 1 )( 5cos(x) - 1 )  =  0   
  Set each factor equal to zero. 
cos(x) + 1 = 0__or__5cos(x) - 1 = 0  

 

cos(x) = -1 cos(x) = 1/5  

 

 

x = π

 

x ≈ 1.369

x ≈ 4.914

 Only use values of  x  in the interval given

 

 May 17, 2019
edited by hectictar  May 17, 2019

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