Trigonometric Calculations: Solve Cos, Sin, Arcsin & Arccos

Hey guys! Let's dive into some trigonometric calculations involving inverse trigonometric functions like arcsin, arccos, and arctan. We'll break down each part step-by-step to make sure we understand how to approach these problems. It might seem daunting at first, but trust me, it's all about applying the right identities and understanding the relationships between these functions. So, let's get started and solve these expressions together!

a) Calculating cos(arcsin(4/5))

When dealing with trigonometric functions of inverse trigonometric functions, it's essential to visualize what's happening. Here, we need to find the cosine of an angle whose sine is 4/5. Think of arcsin(4/5) as an angle, let’s call it θ, such that sin(θ) = 4/5. To find cos(θ), we can use the fundamental trigonometric identity: sin²(θ) + cos²(θ) = 1. This identity is like our trusty sidekick in the world of trigonometry, always there to help us relate sines and cosines. Now, let's plug in the value we know: sin(θ) = 4/5. So, we have (4/5)² + cos²(θ) = 1, which simplifies to 16/25 + cos²(θ) = 1. Our next step is to isolate cos²(θ). We do this by subtracting 16/25 from both sides of the equation, giving us cos²(θ) = 1 - 16/25. This simplifies further to cos²(θ) = 9/25.

Now, we're almost there! To find cos(θ), we take the square root of both sides: cos(θ) = ±√(9/25). This gives us two possible values: cos(θ) = ±3/5. But which one is the correct answer? This is where the range of arcsin comes into play. Remember, arcsin gives us angles in the range of -π/2 to π/2. In this range, cosine is non-negative. Therefore, we choose the positive value. So, cos(θ) = 3/5. This means cos(arcsin(4/5)) = 3/5. And there you have it! We've successfully calculated the first expression. It’s all about breaking down the problem, using the right identities, and paying attention to the ranges of the inverse trigonometric functions. Remember, practice makes perfect, so the more you work through these types of problems, the more comfortable you'll become.

b) Evaluating sin(2arcsin(1/4))

Moving on to the second expression, sin(2arcsin(1/4)), we're dealing with a double angle. This should immediately ring a bell, reminding us of the double angle formulas. The double angle formula for sine is sin(2x) = 2sin(x)cos(x). This formula is going to be our key to unlocking this problem. Let's let θ = arcsin(1/4). This means sin(θ) = 1/4. Now, our expression becomes sin(2θ), which we can rewrite using the double angle formula as 2sin(θ)cos(θ). We already know sin(θ) is 1/4, but we need to find cos(θ). Just like in the previous problem, we can use the Pythagorean identity: sin²(θ) + cos²(θ) = 1. Plugging in sin(θ) = 1/4, we get (1/4)² + cos²(θ) = 1, which simplifies to 1/16 + cos²(θ) = 1. To isolate cos²(θ), we subtract 1/16 from both sides: cos²(θ) = 1 - 1/16, which gives us cos²(θ) = 15/16. Taking the square root of both sides, we get cos(θ) = ±√(15/16). Again, we need to consider the range of arcsin. Since arcsin(1/4) gives us an angle in the range -π/2 to π/2, and cosine is positive in this range, we take the positive root: cos(θ) = √(15)/4. Now we have all the pieces we need. We can plug sin(θ) = 1/4 and cos(θ) = √(15)/4 back into our double angle formula: sin(2θ) = 2sin(θ)cos(θ) = 2 * (1/4) * (√(15)/4). Simplifying this expression, we get sin(2θ) = √(15)/8. Therefore, sin(2arcsin(1/4)) = √(15)/8. We've successfully navigated another trigonometric challenge! Remember, double angle formulas are your friends when you see expressions like sin(2x) or cos(2x), so keep them handy in your trigonometric toolkit.

c) Working Out cos(arcsin(3/5) - arccos(5/15))

For the third expression, cos(arcsin(3/5) - arccos(5/15)), we encounter a difference of angles inside the cosine function. This is a clear signal that we need to use the cosine difference formula. The cosine difference formula is cos(A - B) = cos(A)cos(B) + sin(A)sin(B). This formula allows us to break down the cosine of a difference into a sum of products of cosines and sines. Let's set A = arcsin(3/5) and B = arccos(5/15). This means sin(A) = 3/5 and cos(B) = 5/15, which simplifies to 1/3. Now we need to find cos(A) and sin(B). For cos(A), we use the Pythagorean identity: sin²(A) + cos²(A) = 1. Plugging in sin(A) = 3/5, we get (3/5)² + cos²(A) = 1, which simplifies to 9/25 + cos²(A) = 1. Isolating cos²(A), we get cos²(A) = 1 - 9/25 = 16/25. Taking the square root, we have cos(A) = ±4/5. Since arcsin gives us angles in the range -π/2 to π/2, and cosine is positive in this range, we take the positive value: cos(A) = 4/5. Next, we find sin(B). Again, we use the Pythagorean identity: sin²(B) + cos²(B) = 1. Plugging in cos(B) = 1/3, we get sin²(B) + (1/3)² = 1, which simplifies to sin²(B) + 1/9 = 1. Isolating sin²(B), we get sin²(B) = 1 - 1/9 = 8/9. Taking the square root, we have sin(B) = ±√(8/9). Since arccos gives us angles in the range 0 to π, and sine is positive in this range, we take the positive value: sin(B) = √(8)/3 = 2√(2)/3. Now we have all the components we need to apply the cosine difference formula: cos(A - B) = cos(A)cos(B) + sin(A)sin(B). Plugging in our values, we get cos(arcsin(3/5) - arccos(5/15)) = (4/5)(1/3) + (3/5)(2√(2)/3). Simplifying, we get cos(arcsin(3/5) - arccos(5/15)) = 4/15 + 2√(2)/5. To combine these terms, we need a common denominator, so we rewrite 2√(2)/5 as 6√(2)/15. Thus, cos(arcsin(3/5) - arccos(5/15)) = (4 + 6√(2))/15. Great job! We've successfully tackled an expression involving the difference of angles. Remember, the key here is recognizing the structure of the expression and applying the appropriate trigonometric identity.

d) Solving sin(arccos(1/3) + arctan(2/3))

Finally, let's conquer the last expression: sin(arccos(1/3) + arctan(2/3)). This one involves the sine of a sum of angles, so we'll need to use the sine addition formula. The sine addition formula is sin(A + B) = sin(A)cos(B) + cos(A)sin(B). This formula is another essential tool in our trigonometric toolbox. Let's set A = arccos(1/3) and B = arctan(2/3). This means cos(A) = 1/3 and tan(B) = 2/3. Now we need to find sin(A), sin(B), and cos(B). For sin(A), we use the Pythagorean identity: sin²(A) + cos²(A) = 1. Plugging in cos(A) = 1/3, we get sin²(A) + (1/3)² = 1, which simplifies to sin²(A) + 1/9 = 1. Isolating sin²(A), we get sin²(A) = 1 - 1/9 = 8/9. Taking the square root, we have sin(A) = ±√(8/9). Since arccos gives us angles in the range 0 to π, and sine is positive in this range, we take the positive value: sin(A) = √(8)/3 = 2√(2)/3. Next, we need to find sin(B) and cos(B) given tan(B) = 2/3. Remember that tan(B) = sin(B)/cos(B). We can visualize this as a right triangle where the opposite side is 2 and the adjacent side is 3. To find the hypotenuse, we use the Pythagorean theorem: hypotenuse² = opposite² + adjacent² = 2² + 3² = 4 + 9 = 13. So, the hypotenuse is √13. Now we can find sin(B) and cos(B). sin(B) = opposite/hypotenuse = 2/√13 = (2√13)/13, and cos(B) = adjacent/hypotenuse = 3/√13 = (3√13)/13. Now we have all the components we need to apply the sine addition formula: sin(A + B) = sin(A)cos(B) + cos(A)sin(B). Plugging in our values, we get sin(arccos(1/3) + arctan(2/3)) = (2√(2)/3)((3√13)/13) + (1/3)((2√13)/13). Simplifying, we get sin(arccos(1/3) + arctan(2/3)) = (2√(26)/13) + (2√13)/39. To combine these terms, we need a common denominator, which is 39. So, we rewrite 2√(26)/13 as 6√(26)/39. Thus, sin(arccos(1/3) + arctan(2/3)) = (6√(26) + 2√13)/39. And there we have it! We've successfully solved all four trigonometric expressions. Remember, the key to mastering these types of problems is to break them down step by step, identify the relevant trigonometric identities, and carefully apply them. Keep practicing, and you'll become a trigonometry pro in no time!

In conclusion, we've tackled a variety of trigonometric problems involving inverse trigonometric functions and angle sum/difference formulas. These types of calculations might seem complex at first, but with a solid understanding of the fundamental identities and a step-by-step approach, they become much more manageable. Remember to always visualize the angles, use the Pythagorean identity when needed, and keep those angle sum and difference formulas handy. Keep practicing, and you'll be solving trigonometric expressions like a champ!