Question

real part of (sinx+icosx)™5 is equal to​

Answer 1

Correct option is

D

1Ω

Effective resistance of 2Ω and 2Ω in parallel R

AB

=

2+2

2×2

=1Ω

∴ Equivalent resistance between P and Q, R

PQ

=2∥(1+R

AB

)=2∥2

⟹ R

PQ

=1Ω

Answer 2

Correct question: The real part of $(sinx+icosx)^{5}$ is equal to​ ____, where $i=\sqrt{-1}$.

Answer:

The real part of the expression $(sinx+icosx)^{5}$ is $sin5x$.

Given,

The expression: $(sinx+icosx)^{5}$.

To Find,

The real part of the given expression.

Solution,

The method of finding the real part of the expression is as follows -

We know by the Euler formula that $e^{i\theta}=cos\theta+i \ sin\theta$.

Now we will simplify the given expression.

$(sinx+icosx)^{5} = (cos(\frac{\pi}{2}-x )+ i \ sin(\frac{\pi}{2}-x ))^{5}$

$=(e^{i(\frac{\pi}{2} -x)} )^{5}$ [Here $\theta = (\frac{\pi}{2} -x)$]

$=(e^{i(\frac{5\pi}{2} -5x)} )$ [By the laws of exponents $(a^{m})^{n}=a^{mn}$]

$= cos(\frac{5\pi}{2}-5x )+ i \ sin(\frac{5\pi}{2}-5x )$

$=sin5x+i \ cos5x$

So, the real part is $sin5x$.

Hence, the real part of the expression $(sinx+icosx)^{5}$ is $sin5x$.

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