Intersection of the graphs are found when 2sin3x + 1 = -3/2cos6x - 1


Notice that 3cos2A + 2sinA = -4 is very similar. You have to rearrange it to match the above equation.

3cos2A + 4sinA = -4

3/2cos2A + 2sinA = -2

3/2cos2A + 2sinA + 1 = - 1

2sinA + 1 = -3/2cos2A - 1

2sin(3(⅓A)) + 1 = -3/2cos6(⅓A) - 1

Now compare this to the first equation,

Realise that x = ⅓A . So when you find the values of A from this equation, you can find the x coordinates for the first equation by simply dividing your values of A by 3.

tan(π/4 - θ)

= (tan π/4 - tanθ)/(1 + tan π/4 tanθ)

= (1 - tanθ)/(1 + tanθ)

= (1 - sinθ/cosθ)/(1 + sinθ/cosθ)

= (cosθ - sinθ)/(cosθ + sinθ)

(Multiply both numerator and denominator by cosθ)

= [(cosθ - sinθ)(cosθ + sinθ)]/[(cosθ + sinθ)(cosθ + sinθ)]

(Multiply both numerator and denominator by (cosθ + sinθ) )


= (cos²θ - sin²θ)/(cos²θ + 2sinθcosθ + sin²θ)

= cos2θ/(1 + sin2θ)

(Proved)





Since 1 + sin2θ = √3 cos2θ

1/√3 = cos2θ/(1 + sin2θ)


1/√3 = tan(π/4 - θ)

Solve this for the range given .

Or

tan(π/4 - θ)

= (tan π/4 - tanθ)/(1 + tan π/4 tanθ)

= (1 - tanθ)/(1 + tanθ)

= [(1 - tanθ)(1 + tanθ)]/[(1 + tanθ)(1 + tanθ)]

= (1 - tan²θ)/(1 + 2tanθ + tan²θ)

= (1 - sin²θ/cos²θ)/(1 + 2sinθ/cosθ + sin²θ/cos²θ)

= (cos²θ - sin²θ)/(cos²θ + 2sinθcosθ + sin²θ)
(Multiply both numerator and denominator by cos²θ)

= cos2θ/(1 + sin2θ)


(Proved)

Or

tan(π/4 - θ)

= (tan π/4 - tanθ)/(1 + tan π/4 tanθ)

= (1 - tanθ)/(1 + tanθ)

= (1 - sinθ/cosθ)/(1 + sinθ/cosθ)

= [(1 - sinθ/cosθ)(cos²θ + sinθcosθ)]/[(1 + sinθ/cosθ)(cos²θ + sinθcosθ)]


= (cos²θ + sinθcosθ - sinθcosθ - sin²θ)/(cos²θ + sinθcosθ + sinθcosθ + sin²θ)


= (cos²θ - sin²θ)/(1 + 2sinθcosθ)


= cos2θ/(1 + 2sinθ)


(Proved)