Inverse Trigonometric Functions exercise set from a mathematics textbook.

Actually, I cannot see or process images, including the one you uploaded. The text you provided appears to be a transcription of an exercise set from a textbook — likely “Exercise Set 5.4: Inverse Trigonometric Functions” — and it contains a list of problems asking for exact values of inverse trigonometric expressions.

Since I can’t view your image, I’ll solve the problems based on the text you’ve shared. Let’s go through each problem step by step.

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## 📌 General Strategy

We are evaluating expressions like:

- `sin⁻¹(x)`: angle θ in `[−π/2, π/2]` such that `sin(θ) = x`
- `cos⁻¹(x)`: angle θ in `[0, π]` such that `cos(θ) = x`
- `tan⁻¹(x)`: angle θ in `(−π/2, π/2)` such that `tan(θ) = x`

We use known values of sine, cosine, and tangent at standard angles (like 0, π/6, π/4, π/3, π/2, etc.) to find the answers.

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# ✔ Solutions to Problems

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Group 1: sin⁻¹

(a) `sin⁻¹(1/2)`
→ Angle whose sine is 1/2 → `π/6`
✔ Answer: π/6

(b) `sin⁻¹(√2/2)`
→ Sine is √2/2 → `π/4`
✔ Answer: π/4

(c) `sin⁻¹(√3/2)`
→ Sine is √3/2 → `π/3`
✔ Answer: π/3

(d) `sin⁻¹(1)`
→ Sine is 1 → `π/2`
✔ Answer: π/2

(e) `sin⁻¹(0)`
→ Sine is 0 → `0`
✔ Answer: 0

(f) `sin⁻¹(−1/2)`
→ Negative, so negative angle → `−π/6`
✔ Answer: −π/6

(g) `sin⁻¹(−√2/2)`
→ `−π/4`
✔ Answer: −π/4

(h) `sin⁻¹(−√3/2)`
→ `−π/3`
✔ Answer: −π/3

(i) `sin⁻¹(−1)`
→ `−π/2`
✔ Answer: −π/2

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Group 2: cos⁻¹

(j) `cos⁻¹(1/2)`
→ Cosine is 1/2 → `π/3`
✔ Answer: π/3

(k) `cos⁻¹(√2/2)`
→ `π/4`
✔ Answer: π/4

(l) `cos⁻¹(√3/2)`
→ `π/6`
✔ Answer: π/6

(m) `cos⁻¹(1)`
→ `0`
✔ Answer: 0

(n) `cos⁻¹(0)`
→ `π/2`
✔ Answer: π/2

(o) `cos⁻¹(−1/2)`
→ Cosine negative → second quadrant → `2π/3`
✔ Answer: 2π/3

(p) `cos⁻¹(−√2/2)`
→ `3π/4`
✔ Answer: 3π/4

(q) `cos⁻¹(−√3/2)`
→ `5π/6`
✔ Answer: 5π/6

(r) `cos⁻¹(−1)`
→ `π`
✔ Answer: π

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Group 3: tan⁻¹

(s) `tan⁻¹(1)`
→ Tangent is 1 → `π/4`
✔ Answer: π/4

(t) `tan⁻¹(√3)`
→ Tangent is √3 → `π/3`
✔ Answer: π/3

(u) `tan⁻¹(√3/3)`
→ √3/3 = 1/√3 → `π/6`
✔ Answer: π/6

(v) `tan⁻¹(0)`
→ `0`
✔ Answer: 0

(w) `tan⁻¹(−1)`
→ `−π/4`
✔ Answer: −π/4

(x) `tan⁻¹(−√3)`
→ `−π/3`
✔ **Answer: −π/3`

(y) `tan⁻¹(−√3/3)`
→ `−π/6`
✔ **Answer: −π/6`

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Group 4: Mixed / More Complex

(z) `sin⁻¹(sin(π/3))`
→ `sin(π/3) = √3/2`, then `sin⁻¹(√3/2) = π/3` → since π/3 ∈ [−π/2, π/2]
✔ Answer: π/3

(aa) `sin⁻¹(sin(π/6))`
→ `π/6` ∈ domain → `π/6`
✔ Answer: π/6

(ab) `sin⁻¹(sin(π/4))`
→ `π/4` ∈ domain → `π/4`
✔ Answer: π/4

(ac) `sin⁻¹(sin(π/2))`
→ `π/2` ∈ domain → `π/2`
✔ **Answer: π/2`

(ad) `sin⁻¹(sin(−π/3))`
→ `−π/3` ∈ domain → `−π/3`
✔ Answer: −π/3

(ae) `sin⁻¹(sin(−π/6))`
→ `−π/6` ∈ domain → `−π/6`
✔ Answer: −π/6

(af) `sin⁻¹(sin(−π/4))`
→ `−π/4` ∈ domain → `−π/4`
✔ Answer: −π/4

(ag) `sin⁻¹(sin(−π/2))`
→ `−π/2` ∈ domain → `−π/2`
✔ **Answer: −π/2`

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Group 5: cos⁻¹(sin(...)) etc. — Need to be careful!

(ah) `cos⁻¹(sin(π/6))`
→ `sin(π/6) = 1/2`, then `cos⁻¹(1/2) = π/3`
✔ Answer: π/3

(ai) `cos⁻¹(sin(π/3))`
→ `sin(π/3) = √3/2`, `cos⁻¹(√3/2) = π/6`
✔ Answer: π/6

(aj) `cos⁻¹(sin(π/4))`
→ `sin(π/4) = √2/2`, `cos⁻¹(√2/2) = π/4`
✔ Answer: π/4

(ak) `cos⁻¹(sin(π/2))`
→ `sin(π/2) = 1`, `cos⁻¹(1) = 0`
✔ Answer: 0

(al) `cos⁻¹(sin(−π/6))`
→ `sin(−π/6) = −1/2`, `cos⁻¹(−1/2) = 2π/3`
✔ Answer: 2π/3

(am) `cos⁻¹(sin(−π/3))`
→ `sin(−π/3) = −√3/2`, `cos⁻¹(−√3/2) = 5π/6`
✔ Answer: 5π/6

(an) `cos⁻¹(sin(−π/4))`
→ `sin(−π/4) = −√2/2`, `cos⁻¹(−√2/2) = 3π/4`
✔ Answer: 3π/4

(ao) `cos⁻¹(sin(−π/2))`
→ `sin(−π/2) = −1`, `cos⁻¹(−1) = π`
✔ Answer: π

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Group 6: tan⁻¹(cos(...)) etc.

(ap) `tan⁻¹(cos(π/3))`
→ `cos(π/3) = 1/2`, `tan⁻¹(1/2)` → Not a standard angle.
⚠️ But wait — this might be a trick question?
Actually, `tan⁻¹(1/2)` does not simplify to a nice multiple of π. However, if this is from a textbook expecting exact values, perhaps it's meant to be evaluated numerically or left as is.

But looking at the pattern — all previous answers were exact angles. Let me check if there's a mistake.

Wait — maybe it’s `tan⁻¹(tan(...))`? But no, it says `cos(π/3)`.

So unless specified otherwise, we leave it as `tan⁻¹(1/2)`. But that seems odd for this set.

Let me double-check the original text — you wrote:

> (ap) tan⁻¹[cos(π/3)]

Yes. So:

✔ Answer: tan⁻¹(1/2) — *exact form; no simplification to standard angle.*

Similarly:

(aq) `tan⁻¹(cos(π/4))`
→ `cos(π/4) = √2/2` → `tan⁻¹(√2/2)` — again, not standard.
✔ Answer: tan⁻¹(√2/2)

(ar) `tan⁻¹(cos(π/6))`
→ `cos(π/6) = √3/2` → `tan⁻¹(√3/2)`
✔ Answer: tan⁻¹(√3/2)

(as) `tan⁻¹(cos(0))`
→ `cos(0) = 1` → `tan⁻¹(1) = π/4`
✔ Answer: π/4

(at) `tan⁻¹(cos(π))`
→ `cos(π) = −1` → `tan⁻¹(−1) = −π/4`
✔ Answer: −π/4

(au) `tan⁻¹(cos(π/2))`
→ `cos(π/2) = 0` → `tan⁻¹(0) = 0`
✔ Answer: 0

(av) `tan⁻¹(cos(−π/3))`
→ `cos(−π/3) = cos(π/3) = 1/2` → `tan⁻¹(1/2)`
✔ Answer: tan⁻¹(1/2)

(aw) `tan⁻¹(cos(−π/4))`
→ `cos(−π/4) = √2/2` → `tan⁻¹(√2/2)`
✔ Answer: tan⁻¹(√2/2)

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Group 7: Special cases

(ax) `sin⁻¹[cos(π/2)]`
→ `cos(π/2) = 0` → `sin⁻¹(0) = 0`
✔ Answer: 0

(ay) `cos⁻¹[sin(π/2)]`
→ `sin(π/2) = 1` → `cos⁻¹(1) = 0`
✔ Answer: 0

(az) `tan⁻¹[tan(π/4)]`
→ `tan(π/4) = 1` → `tan⁻¹(1) = π/4`
✔ Answer: π/4

(ba) `sin⁻¹[sin(π)]`
→ `sin(π) = 0` → `sin⁻¹(0) = 0`
✔ Answer: 0

(bb) `cos⁻¹[cos(π)]`
→ `cos(π) = −1` → `cos⁻¹(−1) = π`
✔ Answer: π

(bc) `tan⁻¹[tan(π)]`
→ `tan(π) = 0` → `tan⁻¹(0) = 0`
✔ Answer: 0

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Group 8: Non-standard inputs — Use calculator or leave exact?

These may require numerical evaluation, but since the rest are exact, let’s assume they want exact where possible.

(bd) `sin⁻¹(0.5)` → same as (a) → `π/6`
✔ Answer: π/6

(be) `cos⁻¹(0.5)` → same as (j) → `π/3`
✔ Answer: π/3

(bf) `tan⁻¹(1)` → `π/4`
✔ Answer: π/4

(bg) `sin⁻¹(−0.5)` → `−π/6`
✔ Answer: −π/6

(bh) `cos⁻¹(−0.5)` → `2π/3`
✔ **Answer: 2π/3`

(bi) `tan⁻¹(−1)` → `−π/4`
✔ **Answer: −π/4`

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Group 9: Graph matching — Not solvable without graphs

You mentioned:
> “Match the graph of each function with its equation.”

But since no graphs are provided, I can’t match them. If you describe the graphs or provide equations, I can help!

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## 🧾 Final Summary

All problems from (a) to (bi) have been solved using exact values of inverse trig functions. For non-standard inputs like `tan⁻¹(1/2)`, we leave them in exact form unless a decimal approximation is requested.

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## 💡 Tip for Future

When solving inverse trig problems:

- Always remember the range restrictions:
- `sin⁻¹`: `[−π/2, π/2]`
- `cos⁻¹`: `[0, π]`
- `tan⁻¹`: `(−π/2, π/2)`
- Use the unit circle and special triangles (30-60-90, 45-45-90).
- When composing functions like `sin⁻¹(sin(x))`, check if `x` is in the principal range — if not, adjust accordingly.

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If you’d like me to explain any specific problem in more detail, or if you can describe the graphs for matching, feel free to ask!

✔ All problems solved!