S SpiderX1X Active Member Joined Sep 3, 2019 Messages 137 Gender Undisclosed HSC 2021 Oct 27, 2019 #1
StudyOnly Active Member Joined Oct 16, 2018 Messages 171 Gender Male HSC 2019 Oct 27, 2019 #2 Is the answer cosθ ? If so, just sub tan(θ/2) for t and you get 1 + (tan(θ/2))^2 at bottom which can be equated using the trig identity to (sec(θ/2))^2 and take it from there Last edited: Oct 27, 2019
Is the answer cosθ ? If so, just sub tan(θ/2) for t and you get 1 + (tan(θ/2))^2 at bottom which can be equated using the trig identity to (sec(θ/2))^2 and take it from there
K klmtutor New Member Joined Aug 19, 2019 Messages 4 Gender Female HSC N/A Oct 27, 2019 #3 cos θ not cos 2θ
StudyOnly Active Member Joined Oct 16, 2018 Messages 171 Gender Male HSC 2019 Oct 27, 2019 #4 klmtutor said: cos θ not cos 2θ Click to expand... Yes you're right, just rushed the answer and forgot about θ/2.
klmtutor said: cos θ not cos 2θ Click to expand... Yes you're right, just rushed the answer and forgot about θ/2.
S SpiderX1X Active Member Joined Sep 3, 2019 Messages 137 Gender Undisclosed HSC 2021 Oct 29, 2019 #5
S SpiderX1X Active Member Joined Sep 3, 2019 Messages 137 Gender Undisclosed HSC 2021 Oct 29, 2019 #6 what about that one , just wondering if i got that right
K klmtutor New Member Joined Aug 19, 2019 Messages 4 Gender Female HSC N/A Oct 30, 2019 #7 Solutions when cos 2θ = 0 or tan (3θ) - 1 = 0 for 0 ≤ θ ≤ 360. 8 solutions (cos: 45, 135, 225, 315 tan:15, 75, 135, 195, 255, 315) Last edited: Oct 30, 2019
Solutions when cos 2θ = 0 or tan (3θ) - 1 = 0 for 0 ≤ θ ≤ 360. 8 solutions (cos: 45, 135, 225, 315 tan:15, 75, 135, 195, 255, 315)
