# Find sin 2theta if theta is in the first quadrant and tan theta = 40/9 ??

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Given that ( \tan(\theta) = \frac{40}{9} ) and ( \theta ) is in the first quadrant, we can use the trigonometric identity relating tangent and sine:

[ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} ]

We know that ( \tan(\theta) = \frac{40}{9} ), so we can write:

[ \frac{40}{9} = \frac{\sin(\theta)}{\cos(\theta)} ]

Now, we also know that in the first quadrant, both sine and cosine are positive. So, we can find the cosine of ( \theta ) by using the Pythagorean identity:

[ \sin^2(\theta) + \cos^2(\theta) = 1 ]

We already have ( \sin(\theta) ) from the given information, so we can find ( \cos(\theta) ) and then find ( \sin(2\theta) ) using the double angle identity:

[ \sin(2\theta) = 2\sin(\theta)\cos(\theta) ]

Let's solve for ( \cos(\theta) ): [ \cos(\theta) = \frac{\sin(\theta)}{\tan(\theta)} ] [ \cos(\theta) = \frac{\frac{40}{9}}{\frac{40}{9}} ] [ \cos(\theta) = 1 ]

Now, we have ( \sin(\theta) = \frac{40}{41} ) and ( \cos(\theta) = 1 ), so we can find ( \sin(2\theta) ):

[ \sin(2\theta) = 2\sin(\theta)\cos(\theta) ] [ \sin(2\theta) = 2 \times \frac{40}{41} \times 1 ] [ \sin(2\theta) = \frac{80}{41} ]

So, ( \sin(2\theta) = \frac{80}{41} ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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