Question

Find sin 2theta if theta is in the first quadrant and tan theta=40/9. PLEASE HELP. No links

Answer 1

well, we know that angle θ is in the I Quadrant, namely that sine as well as cosine are both positive and most likely 2θ is in the II Quadrant, where sine is positive, so

`tan(\theta )=\cfrac{\stackrel{opposite}{40}}{\underset{adjacent}{9}}\qquad \textit{now let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=√(a^2+b^2) \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=√(9^2+40^2)\implies c=√(1681)\implies c=41 \\\\[-0.35em] ~\dotfill`

`sin(2\theta )\implies 2sin(\theta )cos(\theta )\implies 2\cdot \cfrac{\stackrel{opposite}{40}}{\underset{hypotenuse}{41}}\cdot \cfrac{\stackrel{adjacent}{9}}{\underset{hypotenuse}{41}}\implies \cfrac{720}{1681}`