Additional Mathematics 0606 / Trigonometry / 3.1133.113: Horizontal tangents and a tan–cos identity0606/11/M/J/22 — Question 6 · 10 marksMark as done · Save for later · Show all solutions(a)Write down the values of kkk for which the line y=ky=ky=k is a tangent to the curve y=4 sin(x+π4)+10y=4\,\mathrm{sin}\left(x+\dfrac{\pi}{4}\right)+10y=4sin(x+4π)+10.[2]▸ Answer(b)(i)Show that(1+tan θ)(1+cos θ)+(1−tan θ)(1−cos θ)1−cos2θ=2(1+sin θ)sin2θ.\dfrac{(1+\mathrm{tan}\,\theta)(1+\mathrm{cos}\,\theta)+(1-\mathrm{tan}\,\theta)(1-\mathrm{cos}\,\theta)}{1-\mathrm{cos}^{2}\theta}=\dfrac{2(1+\mathrm{sin}\,\theta)}{\mathrm{sin}^{2}\theta}.1−cos2θ(1+tanθ)(1+cosθ)+(1−tanθ)(1−cosθ)=sin2θ2(1+sinθ).[4]▸ Answer(b)(ii)Hence solve the equation(1+tan θ)(1+cos θ)+(1−tan θ)(1−cos θ)1−cos2θ=3\dfrac{(1+\mathrm{tan}\,\theta)(1+\mathrm{cos}\,\theta)+(1-\mathrm{tan}\,\theta)(1-\mathrm{cos}\,\theta)}{1-\mathrm{cos}^{2}\theta}=31−cos2θ(1+tanθ)(1+cosθ)+(1−tanθ)(1−cosθ)=3for 0∘⩽θ⩽360∘0^{\circ}\leqslant\theta\leqslant 360^{\circ}0∘⩽θ⩽360∘.[4]▸ Answer▸ Official mark scheme← 3.23: Normal to tan curve: midpoint of AB3.24: Sine graph: find a, b, c →
