Edexcel IAL Mathematics P2.6.1 core trigonometric identitiesQuestions reward identity work before solving: rewrite compound angles, use sin²θ + cos²θ = 1 and convert tan, sec or cosec forms carefully.SyllabusFirst assessment 2019CourseMathematics YMA01LevelAS
Exam pointsRewrite compound-angle expressions using sin2θ, cos2θ or sin4θ identities.Use sin²θ + cos²θ = 1 to find exact sin, cos or tan values from given data.Rearrange trig equations into stated quadratic forms in sin, cos or tan.
P2.6.1 - Core trigonometric identities question 1[Maximum number: 3] In this question you must show detailed reasoning.Solutions relying entirely on calculator technology are not acceptable.Show that the equation(3cosθ−tanθ)cosθ=2(3 \cos \theta-\tan \theta) \cos \theta=2(3cosθ−tanθ)cosθ=2can be written as3sin2θ+sinθ−1=03 \sin ^{2} \theta+\sin \theta-1=03sin2θ+sinθ−1=0Show Answercosθ(3cosθ−tanθ)=2⇒cosθ(3cosθ−sinθcosθ)=2\cos \theta(3 \cos \theta-\tan \theta)=2 \Rightarrow \cos \theta\left(3 \cos \theta-\frac{\sin \theta}{\cos \theta}\right)=2cosθ(3cosθ−tanθ)=2⇒cosθ(3cosθ−cosθsinθ)=2or e.g.cosθ(3cosθ−tanθ)=2⇒3cos2θ−sinθ=2\cos \theta(3 \cos \theta-\tan \theta)=2 \Rightarrow 3 \cos ^{2} \theta-\sin \theta=2cosθ(3cosθ−tanθ)=2⇒3cos2θ−sinθ=2or e.g.cosθ(3cosθ−tanθ)=2⇒3cos2θ−sinθcosθcosθ=2\cos \theta(3 \cos \theta-\tan \theta)=2 \Rightarrow 3 \cos ^{2} \theta-\frac{\sin \theta}{\cos \theta} \cos \theta=2cosθ(3cosθ−tanθ)=2⇒3cos2θ−cosθsinθcosθ=2M13(1−sin2θ)−sinθ=23\left(1-\sin ^{2} \theta\right)-\sin \theta=23(1−sin2θ)−sinθ=2M13sin2θ+sinθ−1=0∗3 \sin ^{2} \theta+\sin \theta-1=0 *3sin2θ+sinθ−1=0∗A1*(3)Add to Test