Question

If tan theta = 10/13 and cos theta > 0 then sin2theta is what ?

Answer 1

$sin2\alpha = \frac{260}{269}$

Step-by-step explanation:

We have , $Tan\alpha = \frac{Perpendicular}{Base} = \frac{10}{13}$,

We know that $sin\alpha = \frac{Perpendicular}{Hypotenuse} = \frac{Perpendicular}{\sqrt[2]{(Perpendicualr)^{2} + (Base)^{2})} }$

Substituting values of P & B , $sin\alpha = \frac{10}{\sqrt{10^{2} + 13^{2}} } = \frac{10}{\sqrt{269} }$

Now , $sin2\alpha = 2sin\alpha cos\alpha = 2sin\alpha \sqrt{1 - (sin\alpha)^{2} }$

⇒$sin2\alpha =$ $\frac{10}{\sqrt{269} }$ ×$\sqrt{1 - (\frac{10}{\sqrt{269} })^{2} }$×2

⇒ $sin2\alpha = \frac{20}{\sqrt{269} }( \sqrt{\frac{269 - 100}{269} } )$

⇒$sin2\alpha = \frac{20}{\sqrt{269} }( \sqrt{\frac{169 }{269} } )$

⇒$sin2\alpha = \frac{260}{\sqrt{269} }( \sqrt{\frac{1}{269} } )$

⇒$sin2\alpha = \frac{260}{269}$