Video: Finding the Equation of the Tangent to the Curve of a Trigonometric Function at a Point

Find the equation of the tangent to the curve 𝑦 = 7 cos π‘₯ βˆ’ 3 sec π‘₯ at π‘₯ = πœ‹/6.

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Video Transcript

Find the equation of the tangent to the curve 𝑦 is equal to seven cos of π‘₯ minus three sec of π‘₯ at π‘₯ is equal to πœ‹ by six.

We’re given a curve 𝑦 is equal to some function of π‘₯. And we need to find the equation of the tangent to this curve at the point where π‘₯ is equal to πœ‹ by six. Let’s start by recalling how we find the equation of a tangent line to the curve 𝑦 is equal to 𝑓 of π‘₯. First, we recall the equation of a line passing through the point 𝑦 one, π‘₯ one with a slope of π‘š is given by 𝑦 minus 𝑦 one is equal to π‘š times π‘₯ minus π‘₯ one. Next, we need to remember that a tangent line to a curve at a point must have the same slope as the curve at that point. In other words, if we call our curve 𝑦 is equal to 𝑓 of π‘₯, then the slope of our tangent line π‘š must be the same as the slope of 𝑓 of π‘₯ when π‘₯ is equal to πœ‹ by six.

So to find the slope of our tangent line, we’re going to need to differentiate 𝑦 with respect to π‘₯. That’s the derivative of seven cos of π‘₯ minus three sec of π‘₯ with respect to π‘₯. And we can evaluate this derivative term by term. First, we recall the derivative of the cos of π‘₯ with respect to π‘₯ is equal to negative the sin of π‘₯. So if we multiply this by seven, we get the derivative of seven cos of π‘₯ with respect to π‘₯ is equal to negative seven sin of π‘₯.

We can do something similar to differentiate our second term. We need to recall the following reciprocal trigonometric derivative result. The derivative of the sec of π‘₯ with respect to π‘₯ is equal to the sec of π‘₯ times the tan of π‘₯. So by multiplying this by negative three, we get the derivative of negative three sec of π‘₯ is equal to negative three sec of π‘₯ times the tan of π‘₯. And now we’re ready to find the slope of our tangent line when π‘₯ is equal to πœ‹ by six. We just need to substitute π‘₯ is equal to πœ‹ by six into our expression for d𝑦 by dπ‘₯. Substituting π‘₯ is equal to πœ‹ by six into our expression for d𝑦 by dπ‘₯, we get the slope of our tangent line π‘š is equal to negative seven times the sin of πœ‹ by six minus three times the sec of πœ‹ by six multiplied by the tan of πœ‹ by six.

And to evaluate this expression, it might help to recall the following trigonometric identity. The sec of πœƒ is equivalent to one divided by the cos of πœƒ for any value of πœƒ. So multiplying by the sec of πœ‹ by six is the same as dividing by the cos of πœ‹ by six. We can now evaluate this expression. First, the sin of πœ‹ by six is equal to one-half. So our first term is negative seven over two. Next, in our second term, multiplying by the sec of πœ‹ by six is the same as dividing by the cos of πœ‹ by six. And the cos of πœ‹ by six is root three over two. Also, the tan of πœ‹ by six is root three over three. This gives us that π‘š is equal to negative seven over two minus three times one over root three over two multiplied by root three over three. And we can simplify this expression.

First, we have three divided by three in our second term. We can simplify this to give us one. Next, instead of dividing by the function root three over two, we can instead multiply by the reciprocal two divided by root three. So this gives us that π‘š is equal to negative seven over two minus two over root three multiplied by root three. And of course, we can simplify this. Root three divided by root three is equal to one. So we get that π‘š is equal to negative seven over two minus two, which we can evaluate is negative 11 divided by two. Now that we found the value of π‘š, we just need to find the coordinates of a point which our tangent line passes through.

To do this, we just need to remember the tangent lines to a curve at a point must pass through this point. So in our case, our tangent line must pass through the point on the curve where π‘₯ is equal to πœ‹ by six. And we directly know the coordinates at this point as π‘₯-coordinate will be πœ‹ by six and the 𝑦-coordinate on our curve when π‘₯ is πœ‹ by six is given by 𝑓 of πœ‹ by six. So all that’s left to do is substitute π‘₯ is equal to πœ‹ by six into our curve. Substituting π‘₯ is equal to πœ‹ by six into our curve, we see that 𝑦 one will be equal to seven cos of πœ‹ by six minus three sec of πœ‹ by six. And we can just evaluate this expression.

First, the cos of πœ‹ by six is root three over two. Next, remember the sec of πœ‹ by six is one divided by the cos of πœ‹ by six. And we already explained, instead of dividing by the cos of πœ‹ by six, which is root three over two, we can multiply by the reciprocal which is two over root three. So we’ve shown 𝑦 one is seven times root three over two minus three times two over root three. And we can simplify this. First, in our second term, we’re going to rationalize the denominator. To do this, we need to multiply both the numerator and the denominator by the square root of three. We can then multiply out denominator. The square root of three multiplied by the square root of three gives us three. But then we can see we have three divided by three, which is of course just equal to one.

So we simplified our expression for 𝑦 one. It’s equal to seven root three over two minus two root three. Then, by taking out the shared factor of root three and simplifying, we can see that 𝑦 one is equal to three root three divided by two. Now all that’s left to do is substitute our values of 𝑦 one, π‘₯ one, and π‘š into our equation for a line. To do this, we’ll first clear some space and then substitute π‘₯ one is equal to πœ‹ by six, 𝑦 one is equal to three root three over two, and π‘š is equal to negative 11 over two into our equation for a line. This gives us 𝑦 minus three root three over two is equal to negative 11 over two multiplied by π‘₯ minus πœ‹ by six. And we could just leave our answer like this. However, we’re going to distribute negative 11 over two over our parentheses.

By doing this, we can see the first term in our expansion will be negative 11π‘₯ over two and the second term in our expansion will be 11πœ‹ over 12. This gives us 𝑦 minus three root three over two is equal to negative 11π‘₯ over two plus 11πœ‹ by 12. And the last piece of simplification we’ll do is rearrange our equation so all of our terms are on the same side of the equation. And by doing this and rearranging, we’re able to get our final answer.

Therefore, we were able to show the equation of the tangent line to the curve 𝑦 is equal to seven cos of π‘₯ minus three sec of π‘₯ at π‘₯ is equal to πœ‹ by six is given by 𝑦 plus 11π‘₯ over two minus 11πœ‹ by 12 minus three root three over two is equal to zero.

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