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Question Number 112136 by weltr last updated on 06/Sep/20

$$find\frac{5}{6+7\sin 2\beta },if\tan \beta =0.2$$

Answered by bemath last updated on 06/Sep/20

$$\tan \beta =\frac{1}{5}\to \{\begin{array}{l}if\beta inI-quadrant\to \{\begin{array}{l}\sin \beta =\frac{1}{\sqrt{26}}\\ \cos \beta =\frac{5}{\sqrt{26}}\end{array}\\ if\beta inIII-quadrant\to \{\begin{array}{l}\sin \beta =-\frac{1}{\sqrt{26}}\\ \cos \beta =-\frac{5}{\sqrt{26}}\end{array}\end{array}$$$$\iff \frac{5}{6+14\sin \beta \cos \beta }=\frac{5}{6+14\left(\frac{5}{26}\right)}$$$$=\frac{5\times 26}{6\times 26+70}=\frac{65}{78+35}=\frac{65}{113}$$$$$$

Commented by weltr last updated on 06/Sep/20

$$thanks$$

Answered by $@y@m last updated on 06/Sep/20

$$\frac{5}{6+7\sin 2\beta }$$$$\frac{5}{6+7\sqrt{1-\cos {}^{2}2\beta }}$$$$\frac{5}{6+7\sqrt{1-{\left(\frac{1-\tan {}^2\beta }{1+\tan {}^2\beta }\right)}^2}}$$$$\frac{5}{6+7\sqrt{1-{\left(\frac{1-.04}{1+.04}\right)}^2}}$$$$\frac{5}{6+7\sqrt{1-{\left(\frac{.96}{1.04}\right)}^2}}$$$$\frac{5}{6+7\sqrt{1-{\left(\frac{12}{13}\right)}^2}}$$$$\frac{5}{6+7\sqrt{\frac{169-144}{{13}^2}}}$$$$\frac{5}{6+7\times \frac{5}{13}}$$$$\frac{5\times 13}{78+35}$$$$\frac{65}{113}$$$$$$

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