# Question Video: Differentiating Trigonometric Functions Using Double-Angle Identities Mathematics • Higher Education

Find the first derivative of the function π¦ = 9 cos (π₯/3) cos (π₯/6) sin (π₯/6).

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### Video Transcript

Find the first derivative of the function π¦ is equal to nine cos of π₯ over three times the cos of π₯ over six multiplied by the sin of π₯ over six.

In this question, weβre asked to find the first derivative of a given function, and we can see that our function is given in terms of the variable π₯. So we need to differentiate this with respect to π₯. And there are a few different ways of doing this. For example, we can notice that our function is the product of three differentiable functions, so we could do this by using the product rule. However, differentiating the product of three functions is quite difficult. So instead, letβs see if we can simplify our function. And to simplify this function, we can notice that our final two function factors, cos of π₯ over six and sin of π₯ over six, share the same argument.

And this can remind us of the double-angle formula for sine, which tells us for any value of π, the sin of two π is equal to two cos of π multiplied by the sin of π. Weβll substitute π is equal to π₯ over six into the double-angle formula. This gives us the sin of two times π₯ over six is equal to two multiplied by the cos of π₯ over six multiplied by the sin of π₯ over six. We can simplify this expression. Two times π₯ over six is equal to π₯ over three. And since weβre looking for an expression of cos π₯ over six times sin of π₯ over six, weβll divide both sides of our equation through by two. This gives us one-half sin of π₯ over three is equal to cos of π₯ over six multiplied by the sin of π₯ over six.

We can then substitute this expression into our function to simplify. This gives us that π¦ is equal to nine cos of π₯ over three multiplied by one-half sin of π₯ over three. And we can simplify this expression. Nine times one-half is nine over two. So π¦ is nine over two times the cos of π₯ over three multiplied by the sin of π₯ over three. And now π¦ is the product of two differentiable functions, so we could differentiate this by using the product rule. However, we can also notice once again we can simplify this expression by using the double-angle formula for sine. This time we need to substitute π is equal to π₯ over three into the double-angle formula. This gives us that the sin of two times π₯ over three is equal to two times the cos of π₯ over three multiplied by the sin of π₯ over three.

And we can simplify this further. Two times π₯ over three is two π₯ over three, and we can divide both sides of the equation through by two. This gives us one-half sin of two π₯ over three is equal to cos of π₯ over three multiplied by the sin of π₯ over three. And now, once again, we can simplify our function by substituting in this expression. This gives us that π¦ is equal to nine over two times one-half sin of two π₯ over three, which we can simplify. Nine over two times one-half is equal to nine over four. Therefore, weβve shown π¦ is equal to nine over four times the sin of two π₯ over three. And weβre now ready to find the first derivative of this function, and we can do this by recalling the following result.

For any real constants π and π, the derivative of π sin of ππ₯ with respect to π₯ is π times π multiplied by the cos of ππ₯. Our value of π is nine over four, and our value of π is two over three. Therefore, substituting these values into our result, we get dπ¦ by dπ₯ is equal to nine over four multiplied by two-thirds times the cos of two-thirds π₯. And we can then simplify this to get our final answer. Nine over four times two over three is equal to three over two. Therefore, we were able to show if π¦ is the function nine cos of π₯ over three times the cos of π₯ over six multiplied by the sin of π₯ over six, then the first derivative of π¦ with respect to π₯ is equal to three over two times the cos of two π₯ over three.

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