cos 2x sin 2x Since the right angled triangles and are similar we have now using simple geometry and elementary trig on right angled triangles we have HN cosx ON NP cosx NM cos x thus cos cos or cos2x 2cos x 1 but for all 1 cos x sin x giving cos2x 2cos x cos x sin x and the required result immediately follows Share
Maximum value of cos2 x 1 Minimum value of sin2 x 0 So this is the only case where you get cos2 x sin2 x 1 Hence cos2 x 1 and sin2 x 0 x n Or you could have used the formula cos2 x sin2 x cos 2x Hope the answer is clear Share One way is to use the complex definitions of sine and cosine sin theta frac e i theta e i theta 2i cos theta frac e i theta e i theta 2
cos 2x sin 2x
cos 2x sin 2x
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Remake Beweis Cos 2x cos 2 x sin 2 x Und Sin 2x 2cos x sin x
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Prove That Cos2x cos 2x Sin 2x Maths 13631643 Meritnation
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Alternatively note that if cos2 x sin2 x 1 cos2 x sin2 x we must have sin2 x 0 so that x n n Z MathematicsStudent1122 Jul 29 2017 at 1 44 1 Answers below and comment above are all correct But I d like to warn you no it s not possible to divide by 2 the way you did The reason is that cos 2x 2 cos x cos 6x 3 cos 2x sin 2x sin 6x which is also a noun without a verb to this line cos 6x sin 6x 3 cos 2x sin 2x Which has a verb is equal to and you moved the chunk 3 cos 2x sin 2 x to the other side of the equation But that assumes cos 6x 3 cos 2x sin 2x sin 6x had to have been equal to 0 all along
Beware of xyproblem info Wolfram Jan 29 2017 at 11 10 1 The only way to change sin 2x sin 2 x in terms of cos 2x cos 2 x is that sin 2x 1 cos2 2x sin 2 x 1 cos 2 2 x Take two cases one for the and one for and then solve the problem Joseph Quarcoo Click here point up 2 to get an answer to your question writing hand the range of fxcos2xsin2x contains the set
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Just subtract cos2x from both sides and you have your answer Now as to where sin2x cos2x 1 comes from Let s say you have a right triangle with legs a and b By the Pythagorean theorem the hypotenuse is a2 b2 Next sinx is defined as opposite hypotenuse and cosx is defined as adjacent hypotenuse 1 I am solving double integral 0 dx 2 cos 2x y dy 0 d x 2 c o s 2 x y d y I am not understanding how is the sin 2x pi 2 cos 2x and sin 2x pi I am aware that sin pi 2 1 rest is unclear to me sin a b sin a cos b sin b cos a sin a b sin a cos b sin b cos a
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cos 2x sin 2x - Beware of xyproblem info Wolfram Jan 29 2017 at 11 10 1 The only way to change sin 2x sin 2 x in terms of cos 2x cos 2 x is that sin 2x 1 cos2 2x sin 2 x 1 cos 2 2 x Take two cases one for the and one for and then solve the problem Joseph Quarcoo