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Exercise 2.8 · Q4

Q.If 2cos⁡α=x+1x2\cos\alpha=x+\dfrac1x and 2cos⁡β=y+1y2\cos\beta=y+\dfrac1y, show that

(i) xy+yx=2cos⁡(α−β)\dfrac xy+\dfrac yx=2\cos(\alpha-\beta)
(ii) xy−1xy=2isin⁡(α+β)xy-\dfrac1{xy}=2i\sin(\alpha+\beta)
(iii) xmyn−ynxm=2isin⁡(mα−nβ)\dfrac{x^m}{y^n}-\dfrac{y^n}{x^m}=2i\sin(m\alpha-n\beta)
(iv) xmyn+1xmyn=2cos⁡(mα+nβ)x^my^n+\dfrac1{x^my^n}=2\cos(m\alpha+n\beta).
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The substitution x=cis⁡α, y=cis⁡βx=\operatorname{cis}\alpha,\ y=\operatorname{cis}\beta automatically satisfies x+1x=2cos⁡αx+\frac1x=2\cos\alpha and y+1y=2cos⁡βy+\frac1y=2\cos\beta (since cis⁡α+cis⁡(−α)=2cos⁡α\operatorname{cis}\alpha+\operatorname{cis}(-\alpha)=2\cos\alpha), turning every part of this question into a direct application of the product/quotient/power rules for cis expressions.

Step 1. Justify the substitution. Take x=cis⁡αx=\operatorname{cis}\alpha. Then 1x=cis⁡(−α)\dfrac1x=\operatorname{cis}(-\alpha), so x+1x=cis⁡α+cis⁡(−α)=2cos⁡αx+\dfrac1x=\operatorname{cis}\alpha+\operatorname{cis}(-\alpha)=2\cos\alpha, matching the given 2cos⁡α=x+1x2\cos\alpha=x+\frac1x. Similarly take y=cis⁡βy=\operatorname{cis}\beta, so y+1y=2cos⁡βy+\frac1y=2\cos\beta.

Step 2. Part (i): compute xy+yx\dfrac xy+\dfrac yx.

xy=cis⁡αcis⁡β=cis⁡(α−β),yx=cis⁡(β−α)=cis⁡(−(α−β)).\frac xy=\frac{\operatorname{cis}\alpha}{\operatorname{cis}\beta}=\operatorname{cis}(\alpha-\beta),\qquad \frac yx=\operatorname{cis}(\beta-\alpha)=\operatorname{cis}(-(\alpha-\beta)).

xy+yx=cis⁡(α−β)+cis⁡(−(α−β))=2cos⁡(α−β).\frac xy+\frac yx=\operatorname{cis}(\alpha-\beta)+\operatorname{cis}(-(\alpha-\beta))=2\cos(\alpha-\beta).

Step 3. Part (ii): compute xy−1xyxy-\dfrac1{xy}.

xy=cis⁡α⋅cis⁡β=cis⁡(α+β),1xy=cis⁡(−(α+β)).xy=\operatorname{cis}\alpha\cdot\operatorname{cis}\beta=\operatorname{cis}(\alpha+\beta),\qquad \frac1{xy}=\operatorname{cis}(-(\alpha+\beta)).

xy−1xy=cis⁡(α+β)−cis⁡(−(α+β))=2isin⁡(α+β),xy-\frac1{xy}=\operatorname{cis}(\alpha+\beta)-\operatorname{cis}(-(\alpha+\beta))=2i\sin(\alpha+\beta),

using cis⁡ϕ−cis⁡(−ϕ)=2isin⁡ϕ\operatorname{cis}\phi-\operatorname{cis}(-\phi)=2i\sin\phi.

Step 4. Part (iii): compute xmyn−ynxm\dfrac{x^m}{y^n}-\dfrac{y^n}{x^m}. By de Moivre's theorem, xm=cis⁡(mα)x^m=\operatorname{cis}(m\alpha) and yn=cis⁡(nβ)y^n=\operatorname{cis}(n\beta), so …

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