# Video: Finding the Slope of a Tangent to a Trigonometric Curve

The graph of π¦ = β1 β (2 sin π₯) has a zero on the interval [π/2, 3π/2]. What is the slope of the tangent to the graph at that point?

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### Video Transcript

The graph of π¦ equals negative one minus two sin π₯ has a zero on the closed interval π by two to three π by two. What is the slope of the tangent to the graph at that point?

Well firstly, we know that, to find the slope of the tangent to a graph when given a function π of π₯, we need to evaluate the derivative at that point. The problem is, we donβt know where that point is yet. So we use the fact that the graph of π¦ equals negative one minus two sin of π₯ has a zero on the closed interval between π by two and three π by two radians. π has a zero. This tells us that the graph intersects the π₯-axis in this interval.

And so to find the value of π₯ at this point, we let π¦ be equal to zero. So we get negative one minus two sin of π₯ equals zero. We need to solve for π₯. To achieve this, we add two sin π₯ to both sides of our equation and then divide by two, so negative one-half equals sin of π₯. Then we take the inverse or arcsin of both sides of our equation. So we find π₯ is equal to the inverse sin of negative one-half.

Well, we know this is true for one value; specifically, thatβs π by six. The problem is, thatβs outside of our closed interval. However, we know that, for the sine curve, there is another solution. And we find that by subtracting π by six from π. And we can find that value either by sketching the graph of π¦ is equal to sin π₯ and looking for various points of symmetry or by using the CAST diagram.

Now we need to find the slope of the tangent to the graph at the point where π₯ equals five π by six. Well, the slope is found by differentiating π¦ with respect to π₯. We differentiate each term in turn. The derivative of negative one is zero. And when we differentiate negative two sin π₯, we get negative two cos π₯. Letβs evaluate this at the point where π₯ equals five π by six. So thatβs negative two times cos of five π by six, which is equal to the square root of three.

And so the slope of the tangent to the graph of π¦ equals negative one minus two sin π₯ at the point where π₯ equals five π by six is root three.