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Question Number 38600 by ajfour last updated on 27/Jun/18

Commented by ajfour last updated on 27/Jun/18

$$Find\theta intermsof\mathit{c}.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 27/Jun/18

$$tan\theta =\frac{b}{1}sin\theta =\frac{b}{\sqrt{b^2+1}}$$$$sin\theta =\frac{c}{b+1}$$$$usingsimilartriangleproperty$$$$\frac{b}{c}=\frac{\sqrt{b^2+1}}{b+1}=\frac{1}{\sqrt{{(b+1)}^2-c^2}}$$$$b^2{(b+1)}^2=c^2(b^2+1)$$$$b^2(b^2+2b+1)=c^2{b}^2+c^2$$$$b^4+2b^3+b^2=c^2{b}^2+c^2$$$$b^4+2b^3+b^2(1-c^2)-c^2=0$$$$solvingwegetb=f\left(c\right)$$$$tan\theta =\frac{b}{1}=\frac{f\left(c\right)}{1}$$$$waitletmesolvebiquadraticeqn$$$$fromfigureitisclearthatb<1so{b}^2<<1$$$$letaprroximate$$$$\sqrt{1+b^2}\approx 1+\frac{1}{2}{b}^2$$$$\frac{b}{c}=\frac{\sqrt{b^2+1}}{b+1}$$$$\frac{b}{c}\approx \frac{1+\frac{1}{2}{b}^2}{b+1}$$$$b^2+b=c+\frac{{cb}^2}{2}$$$$2b^2+2b=2c+{cb}^2$$$$b^2(2-c)+2b-2c=0$$$$b=\frac{-2+\sqrt{4-4\times (2-c)(-2c)}}{2(2-c)}$$$$b=\frac{-2+\sqrt{4+8c(2-c)}}{2(2-c)}$$$$tan\theta =\frac{b}{1}=\frac{-2+\sqrt{4+8c(2-c)}}{2(2-c)}$$$$$$$$$$$$$$

Answered by behi83417@gmail.com last updated on 27/Jun/18

$${\mathit{c}}^2+{\left(\frac{\mathit{c}}{\mathit{t}\mathit{g}\mathit{\theta }}\right)}^2={(1+\mathit{t}\mathit{g}\mathit{\theta })}^2\Rightarrow$$$$\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }.(1+\mathit{t}\mathit{g}\mathit{\theta })=\mathit{c}$$

Answered by MJS last updated on 28/Jun/18

$$b^2+1=a^2$$$$c^2+d^2={(b+1)}^2\Rightarrow$$$$\Rightarrow d=\sqrt{{(b+1)}^2-c^2}$$$$b=\tan \theta$$$$c=d\tan \theta =bd$$$$c=\sqrt{{(1+b)}^2-c^2}b$$$$c^2=(1+2b+b^2-c^2)b^2$$$$b^4+2b^3+(1-c^2)b^2-c^2=0\left(\mathrm{I}\right)$$$$\mathrm{we}\mathrm{can}\mathrm{solve}\mathrm{this}\mathrm{for}b\mathrm{similar}\mathrm{to}\mathrm{a}\mathrm{former}$$$$\mathrm{problem}\mathrm{by}\mathrm{using}$$$$(b-\alpha -\beta \mathrm{i})(b-\alpha +\beta \mathrm{i})(b-\gamma -\delta )(b-\gamma +\delta )=0\left(\mathrm{II}\right)$$$$\mathrm{expand}\mathrm{it}\mathrm{for}\mathrm{be}\mathrm{and}\mathrm{solve}\mathrm{for}\alpha ,\beta ,\gamma ,\delta \mathrm{by}$$$$\mathrm{comparing}\mathrm{the}\mathrm{constants}\mathrm{of}\left(\mathrm{I}\right)\mathrm{and}\left(\mathrm{II}\right)$$$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{easy}\mathrm{to}\mathrm{get}\alpha \left(\gamma \right),\beta \left(\gamma \right)\mathrm{and}\delta \left(\gamma \right)\mathrm{but}\mathrm{for}\gamma$$$$\mathrm{we}\mathrm{get}\mathrm{a}\mathrm{polynome}\mathrm{of}{6}^{\mathrm{th}}\mathrm{degree},\mathrm{we}\mathrm{can}$$$$\mathrm{reduce}\mathrm{it}\mathrm{to}{3}^{\mathrm{rd}}\mathrm{degree}\mathrm{but}\mathrm{as}\mathrm{before}\mathrm{the}$$$$\mathrm{zeros}\mathrm{of}\mathrm{this}\mathrm{polynome}\mathrm{are}\mathrm{not}‘‘{\mathrm{nice}}^{″}\mathrm{and}$$$$\mathrm{we}{\mathrm{can}}^{\prime }\mathrm{t}\mathrm{give}\mathrm{a}\mathrm{closed}\mathrm{form}\mathrm{for}\theta =f\left(c\right).\mathrm{at}$$$$\mathrm{least}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{possible}\mathrm{to}\mathrm{calculate}\mathrm{it}...$$

Commented by MJS last updated on 28/Jun/18

$$b^4+2b^3+(1-c^2)b^2-c^2=0$$$$b^4-2(\alpha +\gamma )b^3+({\alpha }^2+4\alpha \gamma +{\beta }^2+{\gamma }^2-{\delta }^2)b^2-2({\alpha }^2\gamma +\alpha {\gamma }^2-\alpha {\delta }^2+{\beta }^2\gamma )b+({\alpha }^2+{\beta }^2)({\gamma }^2-{\delta }^2)=0$$$$$$$$\left(1\right)\alpha +\gamma +1=0$$$$\left(2\right){\alpha }^2+4\alpha \gamma +{\beta }^2+{\gamma }^2-{\delta }^2-1+c^2=0$$$$\left(3\right){\alpha }^2\gamma +\alpha {\gamma }^2-\alpha {\delta }^2+{\beta }^2\gamma =0$$$$\left(4\right){\alpha }^2{\gamma }^2-{\alpha }^2{\delta }^2+{\beta }^2{\gamma }^2-{\beta }^2{\delta }^2+c^2=0$$$$$$$$\left(1\right)\alpha =-\gamma -1$$$$\left(2\right){\beta }^2-2{\gamma }^2-2\gamma -{\delta }^2+c^2=0\Rightarrow \beta =\sqrt{2{\gamma }^2+2\gamma +{\delta }^2-c^2}$$$$\left(3\right)2{\gamma }^3+3{\gamma }^2+2\gamma {\delta }^2+(1-c^2)\gamma +{\delta }^2=0\Rightarrow$$$$\Rightarrow \delta =\sqrt{\frac{\gamma (-2{\gamma }^2-3\gamma -1+c^2)}{2\gamma +1}}$$$$\left(4\right)\frac{16{\gamma }^6+48{\gamma }^5-8(c^2-7){\gamma }^4-16(c^2-2){\gamma }^3+{(c^2-3)}^2{\gamma }^2+{(c^2+1)}^2\gamma +c^2}{{(2\gamma +1)}^2}=0e$$$$$$$${\gamma }^6+3{\gamma }^5-\frac{c^2-7}{2}{\gamma }^4-(c^2-2){\gamma }^3+\frac{{(c^2-3)}^2}{16}{\gamma }^2+\frac{{(c^2+1)}^2}{16}\gamma +\frac{c^2}{16}=0$$$$\gamma =u-\frac{1}{2}$$$$u^6-\frac{2c^2-1}{4}{u}^4+\frac{c^2(c^2+6)}{16}{u}^2-\frac{c^4}{64}=0$$$$u=\sqrt{v}$$$$v^3-\frac{2c^2-1}{4}{v}^2+\frac{c^2(c^2+6)}{16}v-\frac{c^4}{64}=0$$$$v=w+\frac{2c^2-1}{12}$$$$w^3-\frac{c^4-14c^2+1}{48}+\frac{c^6+33c^4+21c^2-1}{864}=0$$$$p=-\frac{c^4-14c^2+1}{48};q=\frac{c^6+33c^4+21c^2-1}{864}$$$$m=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}};n=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}=$$$$w_1=m+n$$$$w_2=(-\frac{1}{2}+\frac{\sqrt{3}}{2}\mathrm{i})m+(-\frac{1}{2}-\frac{\sqrt{3}}{2}\mathrm{i})n$$$$w_3=(-\frac{1}{2}-\frac{\sqrt{3}}{2}\mathrm{i})m+(-\frac{1}{2}+\frac{\sqrt{3}}{2}\mathrm{i})n$$$$v_i=w_i+\frac{2c^2-1}{12}$$$$u_i=\sqrt{v_i}$$$${\gamma }_i=u_i-\frac{1}{2}$$$${\alpha }_i=-{\gamma }_i-1$$$${\beta }_i=\sqrt{\frac{2{\gamma }_i^3+3{\gamma }_i^2+(1-c^2){\gamma }_i-c^2}{2{\gamma }_i+1}}$$$${\delta }_i=\sqrt{\frac{{\gamma }_i(-2{\gamma }_i^2-3{\gamma }_i-1+c^2)}{2{\gamma }_i+1}}$$$$b={\alpha }_i\pm {\beta }_i\mathrm{i}\vee {\gamma }_i\pm {\delta }_i$$$$\theta =\arctan b$$

Commented by ajfour last updated on 29/Jun/18

$$IundestandSir,plentifulthanks!$$

Commented by MJS last updated on 29/Jun/18

$$\mathrm{I}\mathrm{tried}\mathrm{several}\mathrm{ways}\mathrm{but}\mathrm{they}\mathrm{all}\mathrm{lead}\mathrm{to}\mathrm{the}$$$$\mathrm{same}\mathrm{equation}...$$

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