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serious [3.7K]
3 years ago
11

Find the exact value of sin2 theta given sin theta =5/13, 90°<theta <180°

Mathematics
1 answer:
Zepler [3.9K]3 years ago
3 0
Now, we know that 90°< θ <180°, that simply means the angle θ is in the II quadrant, where sine is positive and cosine is negative.

\bf sin(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{hypotenuse}{13}}\impliedby \textit{now let's find the \underline{adjacent side}}&#10;\\\\\\&#10;\textit{using the pythagorean theorem}\\\\&#10;c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a\qquad &#10;\begin{cases}&#10;c=hypotenuse\\&#10;a=adjacent\\&#10;b=opposite\\&#10;\end{cases}

\bf \pm\sqrt{13^2-5^2}=a\implies \pm\sqrt{144}=a\implies \pm 12 =a \implies \stackrel{II~quadrant}{-12=a}&#10;\\\\\\&#10;therefore \qquad cos(\theta )=\cfrac{\stackrel{adjacent}{-12}}{\stackrel{hypotenuse}{13}}&#10;\\\\&#10;-------------------------------\\\\&#10;sin(2\theta )\implies 2sin(\theta )cos(\theta )\implies 2\left(\frac{5}{13}  \right)\left( \frac{-12}{13} \right)\implies -\cfrac{120}{169}
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