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Molodets [167]
2 years ago
9

Write the trig form of the complex number 18i

Mathematics
1 answer:
ehidna [41]2 years ago
5 0

The trigonometric form of the complex number given in the task content is; 18(isin(π/2)).

<h3>What is the trigonometric form of the complex number?</h3>

If follows from the task content that the complex number whose trigonometric representation is to be determined is; 18i.

Hence, It follows that the trigonometric form is;

= 18(cos(π/2) + isin(π/2)). where; cos(π/2) = 0.

Hence, we have;

= 18(isin(π/2)).

Read more on complex numbers;

brainly.com/question/10662770

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In order to prove

\dfrac{\sin(x+y)\sin(x-y)}{\sin^2(x)\cos^2(x)}=1-\cot^2(x)\tan^2(y)

Let's write both sides in terms of \sin(x),\ \sin^2(x),\ \cos(x),\ \cos^2(x) only.

Let's start with the left hand side: we can use the formula for sum and subtraction of the sine to write

\sin(x+y)=\cos(y)\sin(x)+\cos(x)\sin(y)

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\sin(x-y)=\cos(y)\sin(x)-\cos(x)\sin(y)

So, their multiplication is

\sin(x+y)\sin(x-y)=(\cos(y)\sin(x))^2-(\cos(x)\sin(y))^2\\=\cos^2(y)\sin^2(x)-\cos^2(x)\sin^2(y)

So, the left hand side simplifies to

\dfrac{\cos^2(y)\sin^2(x)-\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}

Now, on with the right hand side. We have

1-\cot^2(x)\tan^2(y)=1-\dfrac{\cos^2(x)}{\sin^2(x)}\cdot\dfrac{\sin^2(y)}{\cos^2(y)} = 1-\dfrac{\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}

Now simply make this expression one fraction:

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