cho sinx cosx 1 2 t nh sin 3x cos 3x Use inverse trigonometric functions to find the solutions and check for extraneous solutions A basic trigonometric equation has the form sin x a cos x a tan x a cot x a The
3 8 2 1 9 32 23 32 1 cho sinx cosx 1 2 t nh a sinx cosx b sin 3x cos 3x c sin 4x cos 4x c u h i 1699662 hoidap247 Hoidap247 H i p online nhanh Using the identity i i 2cos you can express the sum of the cosine terms as cos x cos 3x cos 2n 1 x 1 2 nk 1 j 2k 1 x nk 1 j 2k 1 x The right hand
cho sinx cosx 1 2 t nh sin 3x cos 3x
cho sinx cosx 1 2 t nh sin 3x cos 3x
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Start on the left side Simplify Tap for more steps Apply Pythagorean identity Tap for more steps Rewrite 1 2cos x sin x 1 2 cos x sin x as 1 2sin x cos x 1 2 sin x cos x How do you factor sin3 X cos3 X Algebra Polynomials and Factoring Factor Polynomials Using Special Products 1 Answer Narad T Apr 21 2017 The answer is 1 2
Proof below Explanation Expansion of a cubic a3 b3 a b a2 ab b2 sin3x cos3x sinx cosx sinx cosx sin2x sinxcosx cos2x sinx cosx sin2x Solution LHS sinx cosx cos 3x sec 2xtanx sec 2x sec 2x tanx 1 1 RHS tan 3x tan 2x tanx 1 tanx tan 3x 1 tan 2x tanx 1 tan 2x 1 tan 2x
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cho sinx cosx 1 2 t nh sin 3x cos 3x - Proof below Explanation Expansion of a cubic a3 b3 a b a2 ab b2 sin3x cos3x sinx cosx sinx cosx sin2x sinxcosx cos2x sinx cosx sin2x