Asked by Ayomide
                Evaluate 2 tan 240 degrees + 3 cos 12 degrees leaving your answer in surd form
            
            
        Answers
                    Answered by
            henry2,
            
    2*Tan240 + 3*Cos12 = 3.46 + 2.93 = 
    
                    Answered by
            Reiny
            
    Since it asked for "surd form", they would want exact values
2tan240 is easy
= 2tan60
= 2√3
3 cos 12
= 3cos(30-18)
= 3(cos30cos18 + sin30sin18)
we know cos30 = √3/2 and sin30 = 1/2, so that leaves the 18° angle
There are some interesting ways, find cos18 and sin18
Here is an algebraic way
let θ = 18°
5θ = 90°
2θ + 3θ = 90°
2θ = 90-3θ
sin(2θ) = sin(90-3θ)
sin 2θ = cos 3θ
2sinθcosθ = 4cos^3 θ - 3cosθ <---- basic identities
divide by cosθ and rearrange:
4cos^2 θ - 2sinθ - 3 = 0 , (since cosθ or cos18° ≠ 0, we can do that)
4(1 - sin^2 θ) - 2sinxθ - 3 = 0
4sin^2 θ + 2sinθ - 1 = 0
sin θ = sin18° = (-1 + √5)/4
and cos18° = √( 1 - (√5 - 1)^2 / 16 ) = √(10+2√5)/4
2 tan 240°+ 3 cos 12°
= 2√3 + 3(cos30cos18 + sin30sin18)
= 2√3 + 3( (√3/2) * √(10+2√5)/4 + (1/2) * (-1 + √5)/4 )
there!!! in surd form, whewww!!
    
2tan240 is easy
= 2tan60
= 2√3
3 cos 12
= 3cos(30-18)
= 3(cos30cos18 + sin30sin18)
we know cos30 = √3/2 and sin30 = 1/2, so that leaves the 18° angle
There are some interesting ways, find cos18 and sin18
Here is an algebraic way
let θ = 18°
5θ = 90°
2θ + 3θ = 90°
2θ = 90-3θ
sin(2θ) = sin(90-3θ)
sin 2θ = cos 3θ
2sinθcosθ = 4cos^3 θ - 3cosθ <---- basic identities
divide by cosθ and rearrange:
4cos^2 θ - 2sinθ - 3 = 0 , (since cosθ or cos18° ≠ 0, we can do that)
4(1 - sin^2 θ) - 2sinxθ - 3 = 0
4sin^2 θ + 2sinθ - 1 = 0
sin θ = sin18° = (-1 + √5)/4
and cos18° = √( 1 - (√5 - 1)^2 / 16 ) = √(10+2√5)/4
2 tan 240°+ 3 cos 12°
= 2√3 + 3(cos30cos18 + sin30sin18)
= 2√3 + 3( (√3/2) * √(10+2√5)/4 + (1/2) * (-1 + √5)/4 )
there!!! in surd form, whewww!!
                    Answered by
            dahiru
            
    2tan240+3cos120 in surd form
    
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