Master Inverse Trig: Sin, Cos, Tan Explained

Hey guys! Ever get tripped up by inverse trigonometric functions? You know, the sin⁻¹, cos⁻¹, and tan⁻¹ stuff? Today, we’re diving deep into how to tackle problems like sin⁻¹(sin(5π/6)) , cos⁻¹(cos(7π/6)) , and tan⁻¹(tan(3π/4)) . These might look intimidating, but trust me, once you get the hang of the principles, they become a piece of cake. We’ll break down what inverse trig functions really mean, why they have specific ranges, and how to use those ranges to find the correct answers. So grab your calculators (or just your brains!), and let’s unravel the mysteries of inverse trigonometry together!

Understanding Inverse Trigonometric Functions

So, what exactly are inverse trigonometric functions, guys? Think of them as the opposite of the regular trig functions. If sin(θ) = x , then sin⁻¹(x) = θ . It’s like asking, “What angle gives me this specific sine value?” The same logic applies to cosine and tangent. For example, if cos(θ) = y , then cos⁻¹(y) = θ , and if tan(θ) = z , then tan⁻¹(z) = θ . Pretty straightforward, right? However, there’s a crucial catch: these inverse functions have defined principal value ranges . This is super important because, without them, there would be infinitely many possible angles for a single output value. For sin⁻¹(x) , the range is usually set between -π/2 and π/2 (inclusive). For cos⁻¹(x) , it’s between 0 and π (inclusive). And for tan⁻¹(x) , it’s between -π/2 and π/2 (exclusive of the endpoints). Why these specific ranges? It’s all about ensuring that each input value gives a unique output angle, making the functions well-behaved and invertible. When you’re solving sin⁻¹(sin(θ)) , cos⁻¹(cos(θ)) , or tan⁻¹(tan(θ)) , the goal is to find an angle within the respective inverse function’s range that is coterminal with the original angle θ and has the same trigonometric value. We’ll get into how to do this for our specific examples next. It’s all about using those principal value ranges to your advantage, so keep them in mind!

Solving sin⁻¹(sin(5π/6))

Alright, let’s tackle our first problem: sin⁻¹(sin(5π/6)) . The first thing we do is evaluate the inner part, sin(5π/6) . Now, 5π/6 is in the second quadrant. Remember your unit circle? The sine value is positive in the second quadrant. The reference angle for 5π/6 is π - 5π/6 = π/6. So, sin(5π/6) is the same as sin(π/6) , which is ¹ ⁄ ₂ . Great! Now our problem simplifies to sin⁻¹(1/2) . This is asking, “What angle has a sine value of ¹ ⁄ ₂ ?” And critically, we need to find an angle within the principal value range of arcsine , which is [-π/2, π/2] . We know from our basic trig values that sin(π/6) = 1/2 . And lucky for us, π/6 falls perfectly within the range [-π/2, π/2]. So, the answer to sin⁻¹(sin(5π/6)) is π/6 . Easy peasy, right? The key here was remembering the principal value range for arcsine and using it to select the correct angle. If, for instance, the angle inside hadn’t been in the range, we would have had to find a coterminal angle that was in the range. But for this one, the direct angle worked out, which is always a nice win!

Solving cos⁻¹(cos(7π/6))

Now, let’s move on to cos⁻¹(cos(7π/6)) . Similar to the last one, we first look at cos(7π/6) . The angle 7π/6 is in the third quadrant. On the unit circle, cosine is negative in the third quadrant. The reference angle for 7π/6 is 7π/6 - π = π/6. So, cos(7π/6) is the same as -cos(π/6) . We know that cos(π/6) is √3/2, so cos(7π/6) equals -√3/2 . Now, our problem becomes cos⁻¹(-√3/2) . This means we’re looking for an angle whose cosine is -√3/2, and this angle must be within the principal value range of arccosine , which is [0, π] . We know that cos(π/6) = √3/2 . Since we need a negative cosine value, and our range [0, π] includes the second quadrant (where cosine is negative), we can use our reference angle. The angle in the second quadrant with a reference angle of π/6 is π - π/6 = 5π/6. Let’s check: cos(5π/6) = -cos(π/6) = -√3/2 . And guess what? 5π/6 is within our required range of [0, π]! So, the answer to cos⁻¹(cos(7π/6)) is 5π/6 . Notice how 7π/6 itself is not in the range [0, π]? That’s why we had to find the coterminal angle that is . This step is crucial, guys, always double-check if your final angle is within the specified range!

Solving tan⁻¹(tan(3π/4))

Finally, let’s conquer tan⁻¹(tan(3π/4)) . We start with tan(3π/4) . The angle 3π/4 lies in the second quadrant. Tangent is negative in the second quadrant. The reference angle for 3π/4 is π - 3π/4 = π/4. So, tan(3π/4) is the same as -tan(π/4) . Since tan(π/4) is 1, tan(3π/4) equals -1 . Now, our expression simplifies to tan⁻¹(-1) . We’re searching for an angle whose tangent is -1, and this angle must fall within the principal value range of arctangent , which is (-π/2, π/2) . We know that tan(π/4) = 1 . Because we need a negative tangent value, and our range (-π/2, π/2) includes the fourth quadrant (where tangent is negative), we can use our reference angle. The angle in the fourth quadrant with a reference angle of π/4 is -π/4 (or 7π/4, but -π/4 is simpler and within the range). Let’s check: tan(-π/4) = -tan(π/4) = -1 . And indeed, -π/4 lies within the open interval (-π/2, π/2). So, the answer to tan⁻¹(tan(3π/4)) is -π/4 . Again, the original angle 3π/4 is not in the arctangent’s principal range, so we had to find an equivalent angle that was. It’s all about hitting that target range, folks!

Key Takeaways and Practice

So, what did we learn, guys? The absolute most important thing when dealing with f⁻¹(f(x)) where f is a trig function and f⁻¹ is its inverse is to always consider the principal value range of the inverse function . For sin⁻¹ , it’s [-π/2, π/2]. For cos⁻¹ , it’s [0, π]. For tan⁻¹ , it’s (-π/2, π/2). If the angle inside f(x) is already within this range, then f⁻¹(f(x)) = x . However, if it’s outside the range, you need to find an angle that is within the range but has the same trigonometric value as the original angle. This often involves finding a coterminal angle or using the symmetry properties of the trigonometric functions. Practice is key here! Try these out: sin⁻¹(sin(7π/4)) , cos⁻¹(cos(4π/3)) , and tan⁻¹(tan(-5π/4)) . For sin⁻¹(sin(7π/4)) , 7π/4 is in Q4, sin is negative there, ref angle π/4, so sin(7π/4) = -sin(π/4) = -√2/2. arcsin(-√2/2) in [-π/2, π/2] is -π/4. For cos⁻¹(cos(4π/3)) , 4π/3 is in Q3, cos is negative there, ref angle π/3, so cos(4π/3) = -cos(π/3) = - ¹ ⁄ ₂ . arccos(- ¹ ⁄ ₂ ) in [0, π] is 2π/3. For tan⁻¹(tan(-5π/4)) , -5π/4 is coterminal with 3π/4 (Q2), tan is negative there, ref angle π/4, so tan(-5π/4) = tan(3π/4) = -1. arctan(-1) in (-π/2, π/2) is -π/4. Keep practicing, and these will become second nature. You’ve got this!