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HELPPPPP PLEASEEEEEEEE. 25pts!!!!

HELPPPPP PLEASEEEEEEEE. 25pts!!!!-example-1

2 Answers

4 votes

Answer:

sin A = 12/13, so cos A = 5/13 and

tan A = 12/5

tan 2A = 2(12/5)/(1 - (12/5)²)

= (24/5)/(1 - 144/25)

= (24/5)/(-119/25)

= (24/5)(-25/119)

= -120/119

User Bogdana Zadic
by
7.5k points
3 votes

Answer:


\sf tan(2\theta) = - (120)/(119)

Explanation:

Since θ is in the first quadrant, sinθ and cosθ are both positive.


\sf sin \theta = (12)/(13)

We know the following identity to find tan 2θ:


\sf tan(2\theta) = (2tan(\theta))/(1 - tan^2(\theta))

First, let's find tan 2θ using the information given:


\sf tan(\theta) = (sin(\theta))/(cos(\theta))

We already know that :


\sf sin(\theta) = (12)/(13)

To find, cosθ we can use the Pythagorean identity:


\sf \cos^2(\theta) = 1 - \sin^2(\theta)


\sf \cos^2(\theta)= 1 - \left((12)/(13)\right)^2


\sf \cos^2(\theta) =1 - (144)/(169)


\sf \cos^2(\theta)= (169 - 144)/(169)


\sf \cos(\theta)= \sqrt{(25)/(169)}


\sf cos(\theta)= (5)/(13)

Now, we can find tan θ.


\begin{aligned}\sf tan(\theta) &\sf= (sin(\theta))/(cos(\theta)) \\\\&\sf = ((12)/(13))/((5)/(13)) \\\\&\sf = (12)/(5) \end{aligned}

Now, we can use tan θ to find tan 2θ using the double-angle formula:


\begin{aligned}\sf tan(2\theta) &\sf = (\tan(\theta))/(1 - tan^2(\theta)) \\\\&\sf= (2\left((12)/(5)\right))/(1 - \left((12)/(5)\right)^2) \\\\&\sf = ((24)/(5))/(1 - (144)/(25)) \\\\&\sf = ((24)/(5))/(1 - (144)/(25)) \\\\&\sf = ((24)/(5))/((25)/(25) - (144)/(25)) \\\\ &\sf = ((24)/(5))/((-119)/(25)) \\\\&\sf = (24)/(5) \cdot (-25)/(119)\\\\&\sf = -(120)/(119) \end{aligned}

So,


\sf tan(2\theta) = - (120)/(119)

User Morteza Kavakebi
by
7.8k points
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