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Reduce 1-cosx + isins to the modulus amplitude form
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Answer 1

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Brainly.in

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express 1+cos(alpha)+isin(alpha) in modulus amplitude form

Answer · 40 votes

your answer for above question

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Toppr

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The amplitude of 1 + cos x - i sin x is

Answer · 6 votes

1 + cos x - i sin x = 2cos^2x/2 - 2i sinx/2cosx/2 = 2cosx/2 (cosx/2 - i sinx/2 ) = 2cosx/2 e^-ix/2 amplitude = -x/2

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Brainly.in

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Reduce 1 – cosx + isinx to the modulus amplitude form.

Answer · 2 votes

Given : 1 - cosx + isinx To Find : Reduce 1 - cosx + isinx to the modulus amplitude formSolution:modulus amplitude formz=∣z∣(cosθ+isinθ) 1 - cosx + isinx = 2sin²(x/2) + i (2sin(x/2)cos(x/2)= 2sin(x/2) ( sin(x/2) + icos(x/2))= 2sin(x/2) (cos(π/2 - x/2) + iSin(π/2 - x/2))2sin(x/2) (cos(π/2 - x/2) + iSin(π/2 - x/2)) is the required form ∣z∣ = 2sin(x/2) cosθ+isinθ = cos(π/2 - x/2) + iSin(π/2 - x/2)Learn More:If the area of the triangle formed by the vertices z, iz and z+iz is 50 ... brainly.in/question/9462497 Q. 60 z is a complex number such that z+1/z=2cos3°,then the value ... brainly.in/question/6666657

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Stack Exchange

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Express a complex number in modulus amplitude form 1+sinα+icosα My Attempt: rcosθ=1+sinα rsinθ=cosα Squaring and adding.. r2=(1+sinα)2+cos2α r2=2(1+sinα) tanθ=cosα1+sinα How to break up 1+sinα?

Answer · 2 votes

It is better to solve it this way: 1+sinα+icosα=1+cos(π2−α)+isin(π2−α)∗=2cos2(π4−α2)+i2sin(π4−α2)cos(π4−α2)=2cos(π4−α2)(cos(π4−α2)+isin(π4−α2))=2cos(π4−α2)ei(π/4−α/2) (∗), In this step I used the following formulas: cos(2x)=2cos2x−1 and sin(2x)=2sinxcosx In your method, I am not seeing how you get that expression for tanθ. Rather it should be tanθ=cosα1+sinα. Hence, tanθ=sin(π2−α)1+cos(π2−α)=2sin(π4−α2)cos(π4−α2)2cos2(π4−α2) ⇒tanθ=tan(π4−α2)

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Vedantu

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Express the complex number $1 + i\\sqrt 3 $ in modulus amplitude form.

Answer · 0 votes

Hint: Divide and multiply a number so that the complex number can be expressed in terms of sine & cosine of angles. Lets say, $x = 1 + i\sqrt 3$Multiply and divide the RHS of the above equation with $2$.$ x = 2\left( {\dfrac{{1 + i\sqrt 3 }}{2}} ight) = 2\left( {\dfrac{1}{2}} ight) + 2\left( {\dfrac{{i\sqrt 3 }}{2}} ight) \\ x = 2\cos \dfrac{\pi }{3} + i2\sin \left[ {\dfrac{\pi }{3}} ight] \\ $This above equation can be written in exponential formAs we know $\cos \theta + i\sin \theta = {e^{i\theta }}$Doing the same in the equation obtained we get,$x = 2{e^{i\dfrac{\pi }{3}}}$Hence, $2{e^{i\dfrac{\pi }{3}}}$ in modulus amplitude form.Note :- In these types of questions we have to obtain the given equation in the form of $\cos \theta + i\sin \theta = {e^{i\theta }}$ to convert it into modulus amplitude form. We should also be aware of trigonometric values needed to convert the equation in general form.

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