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

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

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Determine the variation function of π(π₯) = π sin π₯ at π₯ = π. If π(π/2) = 1, find π.

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### Video Transcript

Determine the variation function of π of π₯ is equal to π sin π₯ at π₯ is equal to π. If π of π by two is equal to one, find π.

There are two parts to this question. Weβre asked to find the variation function for a trigonometric function π of π₯ is π sin π₯ at π₯ is equal to π and then to find the value of π using the fact that π of π by two is equal to one. Letβs begin by recalling that for a function π of π₯, the variation function π of β at π₯ is equal to πΌ is π of β is π of πΌ plus β minus π of πΌ. And thatβs where β is the change in π₯. In our case, weβre given that π₯ is equal to π, and thatβs equal to our πΌ. This means that our variation function π of β is π of π plus β minus π of π.

Weβre going to need to find then π of π plus β and π of π. So first, substituting π₯ is equal to π plus β into our function π, we have π of π plus β is equal to π times the sin of π plus β. π of π is simply π sin π so that our function π of β is π sin π plus β minus π sin π. In fact, we know that sin π is equal to zero, so π sin π is zero. And we have π of β is π sin π plus β. For our expression sin π plus β, we can use the identity the sin of π plus or minus capital π΄ is equal to negative or positive the sin of capital π΄. In our case, where the angle capital π΄ corresponds to β, we have negative lowercase π times the sin of β. Our variation function π of β is therefore negative π sin β.

Next, weβre asked to find the value of π if π of π by two is equal to one. We can do this by substituting β is equal to π by two into π of β and putting this equal to one. We then have π of π by two is equal to negative π 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 π times one is equal to one. That is, negative π is equal to one. And multiplying both sides by negative one, we have π equal to negative one. The variation function of π of π₯ is equal to π sin π₯ at π₯ is equal to π is π of β is negative π sin β. And if π of π by two is equal to one, then π is equal to negative one.

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