Question Video: Finding the Variation Function at a Point for a Given Trigonometric Function | Nagwa Question Video: Finding the Variation Function at a Point for a Given Trigonometric Function | Nagwa

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Question Video: Finding the Variation Function at a Point for a Given Trigonometric Function Mathematics • Second Year of Secondary School

Determine the variation function of 𝑓(𝑥) = 𝑎 sin 𝑥 at 𝑥 = 𝜋. If 𝑉(𝜋/2) = 1, find 𝑎.

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Video Transcript

Determine the variation function of 𝑓 of 𝑥 is equal to 𝑎 sin 𝑥 at 𝑥 is equal to 𝜋. If 𝑉 of 𝜋 by two is equal to one, find 𝑎.

There are two parts to this question. We’re asked to find the variation function for a trigonometric function 𝑓 of 𝑥 is 𝑎 sin 𝑥 at 𝑥 is equal to 𝜋 and then to find the value of 𝑎 using the fact that 𝑉 of 𝜋 by two is equal to one. Let’s begin by recalling that for a function 𝑓 of 𝑥, the variation function 𝑉 of ℎ at 𝑥 is equal to 𝛼 is 𝑉 of ℎ is 𝑓 of 𝛼 plus ℎ minus 𝑓 of 𝛼. And that’s where ℎ is the change in 𝑥. In our case, we’re given that 𝑥 is equal to 𝜋, and that’s equal to our 𝛼. This means that our variation function 𝑉 of ℎ is 𝑓 of 𝜋 plus ℎ minus 𝑓 of 𝜋.

We’re going to need to find then 𝑓 of 𝜋 plus ℎ and 𝑓 of 𝜋. So first, substituting 𝑥 is equal to 𝜋 plus ℎ into our function 𝑓, we have 𝑓 of 𝜋 plus ℎ is equal to 𝑎 times the sin of 𝜋 plus ℎ. 𝑓 of 𝜋 is simply 𝑎 sin 𝜋 so that our function 𝑉 of ℎ is 𝑎 sin 𝜋 plus ℎ minus 𝑎 sin 𝜋. In fact, we know that sin 𝜋 is equal to zero, so 𝑎 sin 𝜋 is zero. And we have 𝑉 of ℎ is 𝑎 sin 𝜋 plus ℎ. For our expression sin 𝜋 plus ℎ, we can use the identity the sin of 𝜋 plus or minus capital 𝐴 is equal to negative or positive the sin of capital 𝐴. In our case, where the angle capital 𝐴 corresponds to ℎ, we have negative lowercase 𝑎 times the sin of ℎ. Our variation function 𝑉 of ℎ is therefore negative 𝑎 sin ℎ.

Next, we’re asked to find the value of 𝑎 if 𝑉 of 𝜋 by two is equal to one. We can do this by substituting ℎ is equal to 𝜋 by two into 𝑉 of ℎ and putting this equal to one. We then have 𝑉 of 𝜋 by two is equal to negative 𝑎 sin 𝜋 by two is equal to one. In fact, we know that sin 𝜋 by two is equal to one so that we have negative 𝑎 times one is equal to one. That is, negative 𝑎 is equal to one. And multiplying both sides by negative one, we have 𝑎 equal to negative one. The variation function of 𝑓 of 𝑥 is equal to 𝑎 sin 𝑥 at 𝑥 is equal to 𝜋 is 𝑉 of ℎ is negative 𝑎 sin ℎ. And if 𝑉 of 𝜋 by two is equal to one, then 𝑎 is equal to negative one.

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