How do I simplify sin Xtanx cos Xcotx sin2x

Reichel Mathematics 8, textbook

209 II.1 "Classical" Algebra - Equation Solving II 813 Solving in the Standard Intervaøø 1 [0; 2 π [, 2 [0 °; 360 ° [! a cos x + cot x = 1 + sinx b sin x + tanx = 1 + cos xc sin x + cos x = 1 / cos xd sin x + cos x = 1 / sinx e sin x + cot x = 1 / sinx f cos x + tanx = 1 / cos xg sin x - cot x = 1 / (4 sin x) h sin x + 2 · cot x = ‒1 / (4 sin x) 814 Solve in the standard interval [0; 2 π [! a tan x + 3 cot x ª 4 b tan x + cot x º 4/9 _ 3 c 3 sin 2 x + cos 2 x º 2 d 5 sin 2 x + cos 2 x ª 4 815 Solve in the standard intervaøø 1 [0; 2 π [, 2 [0 °; 360 ° [! a sin2 x + cos 2 x = ‒1 b sin2 x - cos 2 x = ‒1 c sin2 x + cos 2 x = 1/2 d sin2 x - cos 2 x = 1/2 e (cos x + sin x) 2 = cos 2 xf (cos x + sin x) 2 = ‒cos 2 xg (cos x - sin x) 2 = sin2 xh (cosx + sinx) 2 = ‒sin2x 816 solve in the standard intervaøø [0; 2 π [! a tan 4 x - 4 tan 2 x + 3 = 0 b cot 4 x - 2 cot 2 x - 3 = 0 817 Solve 1 graphically, 2 mathematically to two decimal places in the interval [0; 2 π [! a cos x _ 2 - sin2 x = 0 b sin x _ 2 = sin2 x 818 Solve for G = R! a sin 3 x + cos 2 x - 2 sin x + 1 = 0 b sin 3 x - 3 cos 2 x + 3 sin x + 4 = 0 819 Simplify as much as possible and verify your result for x = 5 π / 6! a (1 + cot 2 x) · cos “π _ 2 - x § b (1 + tan 2 x) · sin“ π _ 2 - x § 820 Prove that for aøøe x * R giøt: a sin x + cos x ≠ 1.5 b † sin x + cos x † ª 9 _ 2 c † tan x † + † cot x † º 2 821 proofs! a 2 tanx _____ 1 + ta n 2 x = sin2 xb 1 - ta n 2 x _____ 1 + ta n 2 x = cos 2 xc 2 cos 2 x - tanx tan (π / 2 - x) = cos2 xd 2 cos 2 x - cot x cot (π / 2 - x) = cos 2 x 822 Derive from the (aøs correctly assumed) formø for cos (α + β ) find a form for cos (α - β)! Verify them (without calculator!) For α = 30 ° and β = 90 °! (Vgø. Buch 5. Kø. S. 217!) 823 Derive a form for sin (α - β) from the (aøs correctly assumed) formula for sin (α + β)! Verify them (without calculator!) For α = 60 ° and β = 90 °! (Vgø. Buch 5. Kø. S. 217!) 824 Calculate mitteø's addition theorems in two ways exactly a sin75 °, b cos75 °! 825 Solve arithmetically as a function of the integration constant c in the Intervaøø [- π; π]! For which values ​​of c do you get double solutions? Interpret the task geometrically! a: s in x · dx = ‒cos 2 x b: - sin x · dx = sin 2 x 826 For some value of the integration constant c the calibration has a double solution in [0; 2 π]? Solve the problem 1 arithmetically, 2 graphically, 3 aøøein by graphical considerations! Enter this! a: s in x · dx = sinx b: c os x · dx = cos x 827 1 Interpret the following calibration for x * [0; π / 2] geometrically and then mathematically! 2 Does the function on the left or the right of the calibration grow more snowy? Prove your claim arithmetically and interpret the result geometrically! a: 0 x s in t · dt = sin x b: 0 x c os t · dt = 1 - cos x 828 Interpret the calibration geometrically and solve it arithmetically in the interval [0; 2 π [! a d sinx ____ dx = sin 2 x b d cosx ____ dx = cos 2 x For testing purposes only - property of the publisher öbv

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