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Given that π one is equal to five cos five π plus π sin five π, π two is equal to cos four π plus π sin four π, tan π is equal to four-thirds, and π is greater than zero and less than or equal to π by two, find π one over π two.
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Weβve been given two complex numbers written in trigonometric or polar form.
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And we have a few little extra bits of information.
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But ultimately, weβre looking to find the quotient of these two numbers.
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We need to recall what we know about finding the quotient of two complex numbers in polar form.
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To divide them, we divide their moduli and we subtract their arguments.
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The general polar form of a complex number is π cos π plus π sin π, where π is the modulus of this number and π is its argument.
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Comparing our complex numbers with the general form, we can see that the modulus of our first complex number, π one, is five.
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And whilst itβs not explicitly written, we can see that the modulus of our second complex number, π two, is one.
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Then, we can see that the argument of π one is five π.
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And the argument of π two is four π.
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To find the modulus of π one over π two, weβre going to divide π one by π two.
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Thatβs five divided by one, which is simply five.
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And to find the argument of π one divided by π two, weβre going to subtract their arguments.
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Thatβs five π minus four π, which is of course π.
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And we can now substitute this back into the general polar form of our complex number.
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And we see that π one over π two is five cos π plus π sin π.
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But where does this extra piece of information coming?
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Weβre not going to try to solve the equation tan π is equal to four-thirds by finding the arctan of both sides.
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Instead, we recall that tan π is equal to opposite over adjacent.
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And since π is greater than zero and less than or equal to π by two, we can form a right-angle triangle with an included angle π.
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The side opposite to this angle is four units.
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And the side adjacent to the angle is three units.
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We also know that the hypotenuse of this right-angle triangle must be five units.
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This is the most well-known Pythagorean triple.
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Three squared plus four squared equals five squared.
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And since sin π is equal to opposite over hypotenuse, we can say that, for this triangle, sin π is equal to four over five.
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Itβs four-fifths.
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We can also say that since cos π is adjacent over hypotenuse, itβs in this triangle three-fifths.
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And we can now substitute these values for sin π and cos π into our complex number.
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Itβs five multiplied by three-fifths plus four-fifths π.
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And if we distribute this bracket, we see that the fives cancel.
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And weβre simply left with three as the real part and four as the imaginary component of this complex number.
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π one over π two is therefore equal to three plus four π.