In Exercises 7–14, use the given information to find the exact va...

Question

In Exercises 7–14, use the given information to find the exact value of each of the following:

c. tan 2θ

2
sin θ = ﹣ -------- , θ lies in quadrant III.
3

Accepted Answer

Hey, everyone in this problem, we're asked to determine the exact value of the expression using the provided data. The expression we're given is tan of two beta. And we're told that sign of beta is equal to negative three quarters and that beta lies in quadri three. We're given four answer choices. Option A is negative three multiplied by the square root of seven divided by seven. Option B is the same as A but it is positive. Option C is negative three multiplied by the square root of seven. And option D is the same as C but it is positive. So let's start with what we're trying to find and what we're trying to find is up two beta. Now let's think about our double angle formula here because we have this double angle inside, we have two beta inside of the brackets. But the information we're given is only about sine beta. They have single angle. So let's write this in terms of our double angle formula and recall that 10 of two beta can be written as two multiplied by 10 beta divided by one minus squared. So in order to calculate 10 of two beta, we need to calculate tan beta so that we can use it in this formula that we've written on the right hand side. Now, let's recall, but of beta, it's going to be equal to Y divided by X. If we think about plotting this on our coordinate plane, the tangent value is going to be the Y coordinate divided by the X coordinate. All right. So we need to figure out what Y is. We need to figure out what X is. Now we're told about sign, we're told that sign of beta is equal to negative three quarters. And recall that sign is going to be equal to Y divided by R. All right. So we know that Q what's beta sorry lies in quadrant three. And we can imagine, let me quickly draw a little Cartesian plane. We know that the top right is quadrant one, top left quadrant two, bottom right is quadrant three and bottom, right or sorry. Bottom left is quadrant three, bottom right is quadrant four. The quad three is what we're looking at. And you can see that both the X and Y values are going to be negative here. So we have that be light in quad three. So why is Nick? And what that tells us is that from our ratio sine beta is equal to negative three quarters, which we know is equal to Y divided by R. We can write that Y is equal to negative three and R is equal to four. OK? And there are other combinations of Y and R that could give you this ratio of negative three quarters, but they're all going to have that same ratio. They're all gonna simplify down to the same value. OK? So we're gonna take Y is equal to negative three and R is equal to four. Now we have our Y value. That's great. And we wanted to find Y to substitute into our tangent equation. But we still need to find X. And let's recall how X Y and R are related. You can imagine drawing a triangle in quadrant three here with our draw this in black. So it's easier to see. And we can use the Pythagorean theorem to find R K X is the one side Y is the other side and R is the hypo. So we can use the Pythagorean theorem and write that R squared is equal to X squared plus Y squared substituting in our values. We have that four squared is equal to X squared plus negative three squared. We're just gonna keep simplifying until we find this value of X. We have 16 is equal to X word 49, which tells us that X squared is equal to seven and subtracting nine from both sides to isolate X. When we take the square root, we get that X is equal to plus or minus the square root of seven. Now again, beta is in Q three, OK. We've said that, that means that Y is negative but that also means that X is negative as well. And quadrant three is to the left of the Y axis, the X values are negative. And so we're gonna take the negative squared of set, correct. So now we have or X value and we have our Y value and we can substitute them in for 10 beta. So now we have an expression for 10 beta that is gonna be equal to negative three divided by negative squared of seven and those negatives will divide out. So we can write this as three divided by the square of seven. So we have this expression, but we have to remember what we're actually looking for. We weren't looking for tan of beta, we were looking for tan of two beta. So let's get back to that equation. We have that of two beta is equal to two can be divided by one minus 10 squared beta and we can substitute in the value for 10 beta that we found evaluate this expression. We have two multiplied by three divided by this word of seven, divided by one minus three, divided by the squared of seven squared. OK. Simplifying this is gonna give us six divided by the square of seven in the numerator. In the denominator, we have one minus nine divided by seven. All right. In the denominator, we want to simplify, we wanna create a common denominator so that we can do this subtraction. We have one minus nine divided by seven and we can write one as seven divided by seven. And the denominator will therefore be negative two divided by seven. OK. So we can write this as six divided by the square root of seven divided by negative to des. And we're done the hard part here. We've already figured out our trick functions. We've figured out tan beta, we've used our double angle identity. This is just a matter of simplifying this fraction down at this point. OK. All right. So when we're multiplying a fraction, sorry, no, we're multiplying when we're dividing a fraction, we can multiply by the reciprocal. Instead we have six divided by the square root of seven multiplied by seven divided by negative two. Now six and two, those are both the visible by two. So we can divide both by two. I get that this is going to be equal to three, multiplied by seven, divided by the square root of seven multiplied by negative one. We can write this as negative three, multiplied by seven. You're moving the ne negative up to the nuer divided by the square root of seven. Now we have a square root in the denominator. We don't like to leave those. We wanna rationalize this denominator. So we're gonna multiply the numerator and denominator by the square root of seven. OK. It's just like multiplying by one So it doesn't change the answer. This is gonna be negative three multiplied by seven, multiplied by the square root of seven, divided by the square root of seven multiplied by the squared of seven, which gives us seven. And now these sevens will divide out and we're left with just negative three multiplied by the square root of seven. OK. So we have lots of simplifying to do there, but we've got it done. We've written this in an exact value and we found that the expression is equivalent to negative three multiplied by the square root of seven. If we compare this with our answer choices, we can see that this corresponds with answer choice. C Thanks everyone for watching. I hope this video helped see you in the next one.

In Exercises 7–14, use the given information to find the exact value of each of the following: c. tan 2θ 2 sin θ = ﹣ -------- , θ lies in quadrant III. 3

Key Concepts

Trigonometric Functions

Double Angle Formulas

Quadrants and Angle Signs

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