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Question Number 206899 by MATHEMATICSAM last updated on 29/Apr/24

$$\mathrm{If}\tan \theta =\frac{2x(x+1)}{2x+1}\mathrm{then}\mathrm{find}\sin \theta \mathrm{and}$$$$\cos \theta .$$

Answered by mathzup last updated on 29/Apr/24

$$1+{tan}^2\theta =\frac{1}{{cos}^2\theta }\Rightarrow {cos}^2\theta =\frac{1}{1+{tan}^2\theta }$$$$=\frac{1}{1+\frac{4x^2{(x+1)}^2}{{(2x+1)}^2}}=\frac{{(2x+1)}^2}{{(2x+1)}^2+4x^2{(x+1)}^2}$$$$=\frac{4x^2+4x+1}{4x^2+4x+1+4x^2(x^2+2x+1)}$$$$=\frac{4x^2+4x+1}{4x^2+4x+1+4x^4+8x^3+4x^2}$$$$=\frac{4x^2+4x+1}{4x^4+8x^3+8x^2+4x+1}\Rightarrow$$$$\Rightarrow cos\theta =\stackrel{-}{+}\sqrt{\frac{4x^2+4x+1}{4x^4+8x^3+8x^2+4x+1}}$$$$sin\theta =\stackrel{-}{+}\sqrt{1-{cos}^2\theta }$$

Answered by Frix last updated on 29/Apr/24

$$c=\frac{1}{\sqrt{t^2+1}}\wedge s=\frac{t}{\sqrt{t^2+1}}$$$$\Rightarrow$$$$\cos \theta =\frac{2x+1}{2x^2+2x+1}$$$$\sin \theta =\frac{2x(x+1)}{2x^2+2x+1}$$

Answered by lepuissantcedricjunior last updated on 03/May/24

$$\mathit{t}\mathit{a}\mathit{n}\left(\mathit{\theta }\right)=\frac{2\mathit{x}(\mathit{x}+1)}{2\mathit{x}+1}$$$$\mathit{o}\mathit{r}\mathit{c}\mathit{a}\mathit{l}\mathit{c}\mathit{u}\mathit{l}\mathit{o}\mathit{n}\mathit{s}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathit{e}\mathit{t}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }$$$$\mathit{o}\mathit{n}\mathit{a}$$$$\mathit{t}\mathit{a}\mathit{n}\mathit{\theta }=\frac{\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }}{\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }}=\frac{2\mathit{x}(1+\mathit{x})}{2\mathit{x}+1}$$$$=>\frac{{\mathit{s}\mathit{i}\mathit{n}}^2\mathit{\theta }}{1-{\mathit{s}\mathit{i}\mathit{n}}^2\mathit{\theta }}={\left[\frac{2\mathit{x}(1+\mathit{x})}{2\mathit{x}+1}\right]}^2$$$$=>{\mathit{s}\mathit{i}\mathit{n}}^2\mathit{\theta }[4{\mathit{x}}^2+8{\mathit{x}}^2+4{\mathit{x}}^4+4{\mathit{x}}^2+4\mathit{x}+1]=\frac{{\left[2\mathit{x}(\mathit{x}+1)\right]}^2}{}$$$${\mathit{s}\mathit{i}\mathit{n}}^2\mathit{\theta }=\frac{{\left[2\mathit{x}(1+\mathit{x})\right]}^2}{{(2\mathit{x}+1)}^2+[2\mathit{x}{(1+\mathit{x})}^2}$$$$=>\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }=\mp \frac{2\mathit{x}(1+\mathit{x})}{\sqrt{{(2\mathit{x}+1)}^2+{\left[2\mathit{x}(\mathit{x}+1)\right]}^2}}$$$$=>\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }=\mp \frac{2\mathit{x}+1}{\sqrt{{(2\mathit{x}+1)}^2+{\left[2\mathit{x}(1+\mathit{x})\right]}^2}}$$$$........\mathit{l}\mathit{e}\mathit{p}\mathit{u}\mathit{i}\mathit{s}\mathit{s}\mathit{a}\mathit{n}\mathit{t}\mathit{D}\mathit{r}...................$$$$$$

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