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If \cos A= \frac{\sqrt{10+2\sqrt5}}{4} then solve   \frac{\sin A}{\tan \frac{A}{2}}

Option: 1

1- \frac{\sqrt5-1}{4}


Option: 2

1+ \frac{\sqrt5-1}{4}


Option: 3

1- \frac{\sqrt{10+2\sqrt5}}{4}


Option: 4

1+ \frac{\sqrt{10+2\sqrt5}}{4}


Answers (1)

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Value Trigonometric Ratio of some Particular Angle (Applications) (Part 1) -

Value Trigonometric Ratio of some Particular Angle

 

(Applications) (Part 1)

1. sin 180

\\\begin{array}{l}{\text { Let } \theta=18^{\circ}, \text { then } 5 \theta=90^{\circ} \therefore 2 \theta+3 \theta=90^{\circ}} \\ {\text { or } 2 \theta=90^{\circ}-3 \theta \therefore \sin 2 \theta=\sin \left(90^{\circ}-3 \theta\right)} \\ {\text { or } \sin 2 \theta=\cos 3 \theta \text { or } 2 \sin \theta \cos \theta=4 \cos ^{3} \theta-3 \cos \theta} \\ {\text { or } 2 \sin \theta=4 \cos ^{2} \theta-3 \quad \quad[\text { dividing by } \cos \theta]}\end{array}\\\begin{array}{l}{\text { or } 2 \sin \theta=4\left(1-\sin ^{2} \theta\right)-3=1-4 \sin ^{2} \theta} \\ {\text { or } 4 \sin ^{2} \theta+2 \sin \theta-1=0} \\ {\therefore \sin \theta=\frac{-2 \pm \sqrt{4+16}}{8}=\frac{-2 \pm 2 \sqrt{5}}{8}=\frac{-1 \pm \sqrt{5}}{4}} \\ {\text { Thus } \sin \theta=\frac{-1+\sqrt{5}}{4}, \frac{-1-\sqrt{5}}{4}}\end{array}

 

\begin{array}{l}{\because \theta=18^{\circ}} \\ {\therefore \sin \theta=\sin 18^{\circ}>0, \text { for } 18^{\circ} \text { lies in the } 1 \text { st quadrant }} \\ {\therefore \sin \theta \text { i.e., } \sin 18^{\circ}=\frac{\sqrt{5}-1}{4}}\end{array}

 

2. cos 180

\begin{aligned} \cos ^{2} 18^{\circ} &=1-\sin ^{2} 18^{\circ}=1-\left(\frac{\sqrt{5}-1}{4}\right)^{2} \\ &=1-\frac{5+1-2 \sqrt{5}}{16}=1-\frac{6-2 \sqrt{5}}{16} \\ &=\frac{16-6+2 \sqrt{5}}{16}=\frac{10+2 \sqrt{5}}{16} \\ \therefore \quad \cos 18^{\circ} &=\frac{1}{4} \sqrt{10+2 \sqrt{5}} &\left[\because \cos 18^{\circ}>0\right] \end{aligned}

 

3. sin 720 and cos 720

\begin{array}{l}{ \sin 72^{\circ}=\sin \left(90^{\circ}-18^{\circ}\right)=\cos 18^{\circ}=\frac{1}{4} \sqrt{10+2 \sqrt{5}}} \\ {\cos 72^{\circ}=\cos \left(90^{\circ}-18^{\circ}\right)=\sin 18^{\circ}=\frac{\sqrt{5}-1}{4}}\end{array}

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\frac{\sin A}{\tan \frac{A}{2}}=\frac{\sin A}{\frac{\sin A}{1+ \cos A}}\ \ \ \because We\ know\ \tan \frac{A}{2}=\frac{\sin A}{1+ \cos A}\\ =1+cos A\\ =1+ \frac{\sqrt{10+2\sqrt5}}{4}\frac{\sin A}{\tan \frac{A}{2}}=\frac{\sin A}{\frac{\sin A}{1+ \cos A}}\ \ \ \because We\ know\ \tan \frac{A}{2}=\frac{\sin A}{1+ \cos A}\\ =1+cos A\\ =1+ \frac{\sqrt{10+2\sqrt5}}{4}

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