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Question Video: Finding the Value of a Trigonometric Function Using Double-Angle Identities Mathematics • Second Year of Secondary School

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Given that sin π‘₯ + cos π‘₯ = βˆ’7/13 and 0 < π‘₯ < 2πœ‹, determine the possible values of cos 2π‘₯.

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Video Transcript

Given that sin π‘₯ plus cos π‘₯ is equal to negative seven thirteenths and π‘₯ is greater than zero and less than two πœ‹, determine the possible values of cos two π‘₯.

There are many ways of approaching this question. However, we will start by squaring both sides of our equation. We will then use our knowledge of the double angle and Pythagorean trigonometric identities. Squaring the left-hand side, we have sin π‘₯ plus cos π‘₯ all squared, and this is equal to negative seven thirteenths squared. We can distribute the parentheses on the left-hand side as shown. This gives us sin squared π‘₯ plus sin π‘₯ cos π‘₯ plus cos π‘₯ sin π‘₯ plus cos squared π‘₯, which can be simplified to sin squared π‘₯ plus cos squared π‘₯ plus two sin π‘₯ cos π‘₯. Squaring the right-hand side gives us 49 over 169.

We are now in a position where we can use some of our trigonometric identities to rewrite the left-hand side. Firstly, since sin squared πœƒ plus cos squared πœƒ is equal to one, then sin squared π‘₯ plus cos squared π‘₯ is equal to one. We also recall that sin two πœƒ is equal to two sin πœƒ cos πœƒ. The left-hand side of our equation is therefore equal to one plus sin two π‘₯.

We are now in a position where we can find an expression for sin two π‘₯. Subtracting one from both sides, we have sin two π‘₯ is equal to 49 over 169 minus one, and this is equal to negative 120 over 169. As we have shown that sin two π‘₯ is negative and we know that the range of π‘₯ is given as π‘₯ is greater than zero and less than two πœ‹, this means that the solutions of sin two π‘₯ must lie in the third and fourth quadrants. This means that one of the corresponding values of cos π‘₯ will be positive and one will be negative.

Let’s now consider how we can find these two values. Using the Pythagorean identity, we see that sin squared two π‘₯ plus cos squared two π‘₯ is equal to one. We can therefore square both sides of our equation such that sin squared two π‘₯ is equal to 14400 over 28561. We can then replace sin squared two π‘₯ with one minus cos squared two π‘₯. And we are now in a position to rearrange this equation to find our values of cos two π‘₯. Firstly, we have one minus 14400 over 28561 is equal to cos squared two π‘₯. This is equal to 14161 over 28561. Next, we can square root both sides such that cos two π‘₯ is equal to positive or negative square root of 14161 over 28561, which is equal to positive or negative 119 over 169.

If sin π‘₯ plus cos π‘₯ is equal to negative seven thirteenths and π‘₯ is greater than zero and less than two πœ‹, then cos two π‘₯ is equal to 119 over 169 or negative 119 over 169.

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