Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 59246 by hovea cw last updated on 06/May/19

Commented by MJS last updated on 06/May/19

$$\left(1\right){\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{seem}\mathrm{to}\mathrm{have}\mathrm{a}‘‘{\mathrm{nice}}^{″}\mathrm{solution}$$$$\mathrm{I}\mathrm{get}\sqrt{3}\tan 10°\cos 40°\approx .233956$$

Commented by maxmathsup by imad last updated on 07/May/19

$$letA=sin\left({10}^o\right)+cos\left({70}^o\right).tan\left({190}^o\right)lettransformtoradians$$$$\pi \to 180$$$$x\to 10°\Rightarrow 180x=10\pi \Rightarrow x=\frac{10\pi }{180}=\frac{\pi }{18}also70=\frac{70\pi }{180}=\frac{7\pi }{18}$$$$190=\frac{190\pi }{180}=\frac{19\pi }{18}\Rightarrow A=sin\left(\frac{5\pi }{90}\right)+cos\left(\frac{7\pi }{18}\right).tan(\pi +\frac{\pi }{18})$$$$=sin\left(\frac{\pi }{18}\right)+cos\left(\frac{7\pi }{18}\right)tan\left(\frac{\pi }{18}\right)butcos\left(\frac{7\pi }{18}\right)=sin(\frac{\pi }{2}-\frac{7\pi }{18})=sin\left(\frac{2\pi }{18}\right)=sin\left(\frac{\pi }{9}\right)$$$$\Rightarrow cos\left(\frac{7\pi }{18}\right)tan\left(\frac{\pi }{18}\right)=2sin\left(\frac{\pi }{18}\right)cos\left(\frac{\pi }{18}\right)\frac{sin\left(\frac{\pi }{18}\right)}{cos\left(\frac{\pi }{18}\right)}=2{sin}^2\left(\frac{\pi }{18}\right)=1-cos\left(\frac{\pi }{9}\right)\Rightarrow$$$$A=sin\left(\frac{\pi }{18}\right)+1-cos\left(\frac{\pi }{9}\right)=2{sin}^2\left(\frac{\pi }{18}\right)+sin\left(\frac{\pi }{18}\right)resttocalculatesin\left(\frac{\pi }{18}\right)...$$

Commented by maxmathsup by imad last updated on 07/May/19

$$2)itseazy\frac{{cos}^2\theta -{sin}^2\theta }{cos\theta +sin\theta }=\frac{(cos\theta -sin\theta )(cos\theta +sin\theta )}{cos\theta +sin\theta }=cos\theta -sin\theta$$$$and\theta \ne -\frac{\pi }{4}+2k\pi$$

Commented by maxmathsup by imad last updated on 07/May/19

$$cos\left(3a\right)=cos(2a+a)=cos\left(2a\right)cos\left(a\right)-sin\left(2a\right)sin\left(a\right)$$$$=(2{cos}^{2}a-1)cos\left(a\right)-2{cosasin}^{2}a=2{cos}^{3}a-cosa-2cosa(1-{cos}^{2}a)$$$$=2{cos}^{3}a-cosa-2cosa+2{cos}^{3}a=4{cos}^{3}a-3cosa\Rightarrow$$$$cos\left(3a\right)=4{cos}^{3}a-3cosa.$$

Answered by MJS last updated on 06/May/19

$$\left(2\right)\mathrm{is}\mathrm{easy}$$$$\frac{{\mathrm{c}}^2-{\mathrm{s}}^2}{\mathrm{c}+\mathrm{s}}=\frac{(\mathrm{c}-\mathrm{s})(\mathrm{c}+\mathrm{s})}{\mathrm{c}+\mathrm{s}}=\mathrm{c}-\mathrm{s}$$

Answered by MJS last updated on 06/May/19

$$\left(3\right)$$$$\sin (\alpha +\beta )=\cos \alpha \sin \beta +\sin \alpha \cos \beta$$$$\cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta$$$$$$$$\cos 3\alpha =\cos (\alpha +2\alpha )=\cos \alpha \cos 2\alpha -\sin \alpha \sin 2\alpha =$$$$=\cos \alpha \cos (\alpha +\alpha )-\sin \alpha \sin (\alpha +\alpha )=$$$$=\cos \alpha ({\cos }^2\alpha -{\sin }^2\alpha )-\sin \alpha \left(2\cos \alpha \sin \alpha \right)=$$$$=\cos \alpha ({\cos }^2\alpha -(1-{\cos }^2\alpha ))-2\cos \alpha (1-{\cos }^2\alpha )=$$$$={2\cos }^3\alpha -\cos \alpha -2\cos \alpha +{2\cos }^3\alpha =$$$$={4\cos }^3\alpha -3\cos \alpha$$

Answered by tanmay last updated on 07/May/19

$$1)sin{10}^o+cos{70}^o.tan\left({190}^o\right)$$$$=sin{10}^o+sin{20}^o\times tan({180}^o+{10}^o)$$$$=sin{10}^o+sin{20}^o\times tan{10}^o$$$$=sin{10}^o(1+\frac{sin{20}^o}{cos{10}^o})$$$$=sin{10}^o\left(\frac{cos{10}^o+sin{20}^o}{cos{10}^o}\right)$$$$=sin{10}^0\times \left(\frac{cos{10}^o+cos{70}^o}{cos{10}^o}\right)$$$$=sin{10}^o\times \frac{2cos{40}^o\times cos{30}^o}{cos{10}^o}$$$$=\sqrt{3}\times tan{10}^{o}cos{40}^o$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com