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# 2 Trig questions

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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
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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 )2 by 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
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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