Find the exact value of each expression.
tan (5π/12)

Question

Find the exact value of each expression.

tan (5π/12)

Accepted Answer

Hey, everyone here we are asked to determine the exact value of the trigonometric expression tangent of 19 pi divided by 12. Here we have four answer choice options, answer choice. A negative two minus the square root of three B two plus the square root of three. See the square root of three divided by two B, the square root of two divided by three. For this problem, we need to utilize the unit circle. And if we recall the unit circle, we know that we do not have 19 pi divided by 12 anywhere on it. So that tells us we need to take this value and split it up into the sum of two values which are on the unit circle. And so in doing this, our current expression becomes tangent of pi divided by three plus five pi divided by four. And now we see our current expression is the sum of two terms inside a tangent function. And so this tells us we need to recall the sum identity for tangent functions. And this identity tells us that tangent of A plus B is equivalent to tangent of A plus tangent of B all divided by one minus tangent of A multiplied by tangent of B. And now, recalling the expression we are working with, we have tangent of pi divided by three plus five pi divided by four. And now comparing this expression to the left hand side of the formula, we can identify that the value for A is pi divided by three and the value for B is five pi divided by four. So now utilizing these identified values, we can substitute them to the right hand side of the formula so that we can fully expand our current expression. In doing this, we have tangent of pi divided by three plus tangent of five pi divided by four all divided by one minus tangent of pi divided by three, multiplied by tangent of five pi divided by four. So now that we have fully expanded our expression, we want to begin evaluating it. So we can start by evaluating each of the tangent terms individually. So here we can begin by evaluating tangent of pi divided by three. And to evaluate this expression, we need to recall that tangent is equivalent to sine divided by cosine. So this tells us that tangent of pi divided by three is equivalent to sine of pi divided by three, divided by cosine of pi divided by three. And now we just need to utilize the unit circle to find each of these sine and cosine values. And doing this, our current expression becomes the square root of three divided by two, divided by one half. And now to make this complex fraction easier to solve, we can rewrite it by multiplying by the reciprocal of the denominator. So we have the square root of three divided by two, multiplied by two, divided by one. And now we can easily see that the twos will cancel each other out. And so we are just left with the square root of three. Now moving on to our next tangent expression, we have tangent of five pi divided by four. And now solving just as we did previously, we know that this expression is equivalent to sine of five pi divided by four divided by cosine of five pi divided by four. And now, once again, using the unit circle, we know that this expression is equivalent to negative square root of two divided by two divided by negative square root of two divided by two. And now, in this case, we can see that the numerator value is the exact same as the denominator. So both values cancel each other out and we are just left with positive one. And now that we have determined both of the values for both tangent expressions, we can substitute them back into the large expanded expression we had previously. So so far, we know that tangent of pi divided by three plus five pi divided by four is equivalent to the square root of three plus one, all divided by one minus the square root of three multiplied by one. And now just simplifying this expression, we have the square root of three plus one all divided by one minus the square root of three. And now we are not done simplifying yet, we now need to rationalize the denominator and we do this by multiplying by its conjugate. So we multiply both the numerator and the denominator by the conjugate of the denominator, which is one plus the square root of three. And now fully multiplying out each of the terms. Here, we have the square root of three plus three plus one plus the square root of three all divided by one plus the square root of three minus the square root of three minus three. Now, the next step in simplifying is to just combine our like terms in doing this, we have four plus two multiplied by the square root of three all divided by negative two. And now our last step in simplifying is to notice that both of the terms in the numerator are divisible by negative two. So we can simply divide both of the terms by negative two. And we come up with our final simplified answer, which is negative two minus the square root of three. With this final simplified answer, we are left with answer choice A where again we have negative two minus the square root of three. Thanks for watching. I hope you found this video helpful and I'll see you again next time.

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Find the exact value of each expression.tan (5π/12)

Watch next

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Find the exact value of each expression.
tan (5π/12)