Mastering Inverse Trig: sin, cos, tan Explained

Hey guys! Ever feel like you’re wrestling with inverse trigonometric functions? You know, those pesky sin⁻¹ , cos⁻¹ , and tan⁻¹ that seem to have a mind of their own? Well, today we’re going to tackle some common head-scratchers: sin(sin 5π/6) , cos(cos 7π/6) , and tan(tan 3π/4) . We’ll break them down step-by-step, so by the end of this, you’ll be a total pro, or at least feel a lot more confident tackling these bad boys. Let’s dive in!

Understanding Inverse Trigonometric Functions

Before we get our hands dirty with the specific problems, let’s quickly refresh what inverse trigonometric functions actually are. Think of them as the “undo” buttons for the regular trig functions. If sin(x) = y , then sin⁻¹(y) = x . Simple enough, right? But here’s the kicker, guys: these functions have specific ranges . This is super important because it dictates what answer you’ll get. For sine and tangent, the range is typically [-π/2, π/2] (or [-90°, 90°] ), and for cosine, it’s [0, π] (or [0°, 180°] ). Why do they have these ranges? It’s to make them functions – meaning for every input, there’s only one output. Without these restricted ranges, you’d have multiple angles giving you the same sine, cosine, or tangent value, which would break the definition of a function. So, when we’re evaluating sin⁻¹(sin(x)) , we’re looking for an angle within the inverse function’s restricted range that has the same sine value as x .

Now, let’s talk about the specific examples. The first one is sin(sin 5π/6) . You might be tempted to just say 5π/6 , but hold up! We need to consider the range of sin⁻¹ . Remember, it’s [-π/2, π/2] . Is 5π/6 within this range? Nope, it’s not. 5π/6 radians is equal to 150 degrees, which is way outside our [-90°, 90°] window. So, what do we do? We need to find an angle inside the [-π/2, π/2] range that has the same sine value as 5π/6 . Let’s think about the unit circle. The sine value represents the y-coordinate. The angle 5π/6 is in the second quadrant. The reference angle is π - 5π/6 = π/6 . Angles in the first and second quadrants have positive sine values. So, the angle in the first quadrant with the same sine value as 5π/6 is simply its reference angle, which is π/6 . And hey, guess what? π/6 is within our [-π/2, π/2] range! So, sin⁻¹(sin 5π/6) = π/6 . See? It’s all about checking that range, guys!

Tackling Cosine Inverse: cos(cos 7π/6)

Alright, moving on to our next challenge: cos(cos 7π/6) . Similar to the sine function, the inverse cosine function, cos⁻¹ , has a restricted range, which is [0, π] (or [0°, 180°] ). Now, let’s look at our angle, 7π/6 . Is 7π/6 within the [0, π] range? Uh-uh, it’s not. 7π/6 radians is 210 degrees, which falls outside our allowed window. So, just like before, we need to find an angle that is within the [0, π] range and has the same cosine value as 7π/6 . The cosine value relates to the x-coordinate on the unit circle. The angle 7π/6 is located in the third quadrant. In the third quadrant, both sine and cosine are negative. The reference angle for 7π/6 is 7π/6 - π = π/6 . Now, we need to find an angle in the [0, π] range (which covers the first and second quadrants) that has the same cosine value as 7π/6 . Since 7π/6 is in the third quadrant, its cosine is negative. We need to find an angle in the second quadrant (which is within our [0, π] range) that also has a negative cosine value and the same magnitude. The angle in the second quadrant with the same reference angle π/6 is π - π/6 = 5π/6 . Let’s check: cos(5π/6) is indeed negative, and it’s equal to -√3/2 , which is the same cosine value as cos(7π/6) . Crucially, 5π/6 is within the [0, π] range. Therefore, cos⁻¹(cos 7π/6) = 5π/6 . It’s that simple when you keep the range rules in mind!

Wrapping Up with Tangent Inverse: tan(tan 3π/4)

Last but not least, let’s conquer tan(tan 3π/4) . The inverse tangent function, tan⁻¹ , has a range of (-π/2, π/2) (or (-90°, 90°) ). Our angle here is 3π/4 . Is 3π/4 within the (-π/2, π/2) range? Nope, it’s not. 3π/4 radians is 135 degrees, which is outside our acceptable window. So, same drill, different function! We need to find an angle within (-π/2, π/2) that has the same tangent value as 3π/4 . The tangent of an angle is the ratio of its sine to its cosine (y/x). The angle 3π/4 is in the second quadrant. In the second quadrant, sine is positive and cosine is negative, making the tangent negative. The reference angle for 3π/4 is π - 3π/4 = π/4 . We need an angle in the (-π/2, π/2) range that has the same tangent value. This range covers the first and fourth quadrants. Since tan(3π/4) is negative, we need to look for an angle in the fourth quadrant (which is within (-π/2, π/2) ) that also yields a negative tangent. The angle in the fourth quadrant with the reference angle π/4 is -π/4 (or 7π/4 , but -π/4 is within our principal range). Let’s verify: tan(3π/4) is -1 , and tan(-π/4) is also -1 . And critically, -π/4 is within the (-π/2, π/2) range. Bingo! So, tan⁻¹(tan 3π/4) = -π/4 . It’s all about finding that equivalent angle within the principal range, guys. It might seem tricky at first, but once you get the hang of the ranges and reference angles, these problems become super manageable. Keep practicing, and you’ll be a trig master in no time!