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Exercise 2.8 · Q3

Q.Find the value of (1+sin⁡π10+icos⁡π101+sin⁡π10−icos⁡π10)10\left(\dfrac{1+\sin\frac\pi{10}+i\cos\frac\pi{10}}{1+\sin\frac\pi{10}-i\cos\frac\pi{10}}\right)^{10}.

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With θ=π/10\theta=\pi/10, the numerator 1+sin⁡θ+icos⁡θ1+\sin\theta+i\cos\theta and denominator 1+sin⁡θ−icos⁡θ1+\sin\theta-i\cos\theta are conjugates; factoring each using 1+sin⁡θ=(cos⁡θ2+sin⁡θ2)21+\sin\theta=(\cos\frac\theta2+\sin\frac\theta2)^2 and cos⁡θ=cos⁡2θ2−sin⁡2θ2\cos\theta=\cos^2\frac\theta2-\sin^2\frac\theta2 turns the quotient into a single cis expression that de Moivre's theorem raises to the 10th power.

Step 1. Let θ=π10\theta=\dfrac\pi{10} and factor the numerator using half-angle identities. Since 1=cos⁡2θ2+sin⁡2θ21=\cos^2\frac\theta2+\sin^2\frac\theta2 and sin⁡θ=2sin⁡θ2cos⁡θ2\sin\theta=2\sin\frac\theta2\cos\frac\theta2,

1+sin⁡θ=cos⁡2θ2+sin⁡2θ2+2sin⁡θ2cos⁡θ2=(cos⁡θ2+sin⁡θ2)2.1+\sin\theta=\cos^2\frac\theta2+\sin^2\frac\theta2+2\sin\frac\theta2\cos\frac\theta2=\left(\cos\frac\theta2+\sin\frac\theta2\right)^2.

Also, cos⁡θ=cos⁡2θ2−sin⁡2θ2=(cos⁡θ2−sin⁡θ2)(cos⁡θ2+sin⁡θ2)\cos\theta=\cos^2\frac\theta2-\sin^2\frac\theta2=\left(\cos\frac\theta2-\sin\frac\theta2\right)\left(\cos\frac\theta2+\sin\frac\theta2\right).

Step 2. Factor out (cos⁡θ2+sin⁡θ2)\left(\cos\frac\theta2+\sin\frac\theta2\right) from the numerator.

1+sin⁡θ+icos⁡θ=(cos⁡θ2+sin⁡θ2)[(cos⁡θ2+sin⁡θ2)+i(cos⁡θ2−sin⁡θ2)].1+\sin\theta+i\cos\theta=\left(\cos\frac\theta2+\sin\frac\theta2\right)\left[\left(\cos\frac\theta2+\sin\frac\theta2\right)+i\left(\cos\frac\theta2-\sin\frac\theta2\right)\right].

Regroup the bracket by real/imaginary contributions of cos⁡θ2\cos\frac\theta2 and sin⁡θ2\sin\frac\theta2:

(cos⁡θ2+sin⁡θ2)+i(cos⁡θ2−sin⁡θ2)=cos⁡θ2(1+i)+sin⁡θ2(1−i).\left(\cos\frac\theta2+\sin\frac\theta2\right)+i\left(\cos\frac\theta2-\sin\frac\theta2\right)=\cos\frac\theta2(1+i)+\sin\frac\theta2(1-i).

Since 1−i=−i(1+i)1-i=-i(1+i), this is (1+i)[cos⁡θ2−isin⁡θ2](1+i)\left[\cos\frac\theta2-i\sin\frac\theta2\right]. So

1+sin⁡θ+icos⁡θ=(cos⁡θ2+sin⁡θ2)(1+i)(cos⁡θ2−isin⁡θ2).1+\sin\theta+i\cos\theta=\left(\cos\frac\theta2+\sin\frac\theta2\right)(1+i)\left(\cos\frac\theta2-i\sin\frac\theta2\right).

Step 3. Do the same for the denominator (conjugate throughout).

1+sin⁡θ−icos⁡θ=(cos⁡θ2+sin⁡θ2)(1−i)(cos⁡θ2+isin⁡θ2).1+\sin\theta-i\cos\theta=\left(\cos\frac\theta2+\sin\frac\theta2\right)(1-i)\left(\cos\frac\theta2+i\sin\frac\theta2\right).

Step 4. Divide and simplify. The common real factor (cos⁡θ2+sin⁡θ2)\left(\cos\frac\theta2+\sin\frac\theta2\right) cancels:

1+sin⁡θ+icos⁡θ1+sin⁡θ−icos⁡θ=1+i1−i⋅cos⁡θ2−isin⁡θ2cos⁡θ2+isin⁡θ2.\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}=\frac{1+i}{1-i}\cdot\frac{\cos\frac\theta2-i\sin\frac\theta2}{\cos\frac\theta2+i\sin\frac\theta2}.

Now 1+i1−i=(1+i)2(1−i)(1+i)=2i2=i=cis⁡π2\dfrac{1+i}{1-i}=\dfrac{(1+i)^2}{(1-i)(1+i)}=\dfrac{2i}2=i=\operatorname{cis}\dfrac\pi2, and cos⁡θ2−isin⁡θ2cos⁡θ2+isin⁡θ2=cis⁡(−θ)\dfrac{\cos\frac\theta2-i\sin\frac\theta2}{\cos\frac\theta2+i\sin\frac\theta2}=\operatorname{cis}\left(-\theta\right) (dividing cis⁡(−θ/2)\operatorname{cis}(-\theta/2) by cis⁡(θ/2)\operatorname{cis}(\theta/2)). So the whole quotient is

cis⁡π2⋅cis⁡(−θ)=cis⁡(π2−θ).\operatorname{cis}\frac\pi2\cdot\operatorname{cis}(-\theta)=\operatorname{cis}\left(\frac\pi2-\theta\right).

Step 5. Raise to the 10th power using de Moivre's theorem, with θ=π10\theta=\dfrac\pi{10}.

[cis⁡(π2−θ)]10=cis⁡(10(π2−π10))=cis⁡(5π−π)=cis⁡(4π)=cos⁡4π+isin⁡4π=1.\left[\operatorname{cis}\left(\frac\pi2-\theta\right)\right]^{10}=\operatorname{cis}\left(10\left(\frac\pi2-\frac\pi{10}\right)\right)=\operatorname{cis}\left(5\pi-\pi\right)=\operatorname{cis}(4\pi)=\cos4\pi+i\sin4\pi=1.

✓Final answer

1\boxed{1}.

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