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Question Number 129360 by bramlexs22 last updated on 15/Jan/21

$$\{\begin{array}{l}\sin a+\sin b=\frac{1}{3}\\ \cos a+\cos b=\frac{5}{3}\end{array}$$$${\cos }^2(a+b)=?$$

Commented by bramlexs22 last updated on 15/Jan/21

$$\mathrm{nice}\mathrm{trigonometry}$$

Answered by liberty last updated on 15/Jan/21

$$\left(1\right)\cos (\mathrm{a}-\mathrm{b})=\frac{4}{9}$$$$\left(2\right)\sin (\mathrm{a}+\mathrm{b})+\sin (\mathrm{a}+\mathrm{b})\cos (\mathrm{a}-\mathrm{b})=\frac{5}{9}$$$$\Rightarrow \sin (\mathrm{a}+\mathrm{b})+\frac{4}{9}\sin (\mathrm{a}+\mathrm{b})=\frac{5}{9}$$$$\Rightarrow \frac{13}{9}\sin (\mathrm{a}+\mathrm{b})=\frac{5}{9}\Rightarrow \sin (\mathrm{a}+\mathrm{b})=\frac{5}{13}$$$$\Rightarrow \cos {}^2(\mathrm{a}+\mathrm{b})=1-\frac{25}{169}=\frac{144}{169}$$

Answered by bramlexs22 last updated on 15/Jan/21

$$\iff \frac{2\sin \left(\frac{\mathrm{a}+\mathrm{b}}{2}\right)\cos \left(\frac{\mathrm{a}-\mathrm{b}}{2}\right)}{2\cos \left(\frac{\mathrm{a}+\mathrm{b}}{2}\right)\cos \left(\frac{\mathrm{a}-\mathrm{b}}{2}\right)}=\frac{1}{5}$$$$\iff \tan \left(\frac{\mathrm{a}+\mathrm{b}}{2}\right)=\frac{1}{5}$$$$\Rightarrow \tan (\mathrm{a}+\mathrm{b})=\frac{2\tan \left(\frac{\mathrm{a}+\mathrm{b}}{2}\right)}{1-\tan {}^2\left(\frac{\mathrm{a}+\mathrm{b}}{2}\right)}$$$$\Rightarrow \tan (\mathrm{a}+\mathrm{b})=\frac{\frac{2}{5}}{1-\frac{1}{25}}=\frac{10}{24}=\frac{5}{12}$$$$\Rightarrow \sec {}^2(\mathrm{a}+\mathrm{b})=1+\frac{25}{144}=\frac{169}{144}$$$$\Rightarrow \cos {}^2(\mathrm{a}+\mathrm{b})=\frac{144}{169}$$

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