TRIGONOMETRY QUESTIONS WITH SOLUTIONS

1. (a) Prove sin2A + sin2B + sin2C = 4sinAsinBsinC for triangle angles

Proof:

For any triangle ABC, A + B + C = π radians (180°).

Using sum-to-product formula:

sin2A + sin2B = 2sin(A+B)cos(A-B)

Thus:

sin2A + sin2B + sin2C = 2sin(A+B)cos(A-B) + sin2C

Since A+B = π - C ⇒ sin(A+B) = sinC

And 2C = 2π - 2(A+B) ⇒ sin2C = -sin(2A+2B)

Now:

= 2sinCcos(A-B) + 2sinCcos(A+B)

= 2sinC[cos(A-B) + cos(A+B)]

Using cosine addition:

= 2sinC[2cosAcosB] = 4sinAsinBsinC

Because cos(A-B) + cos(A+B) = 2cosAcosB

Thus proved: sin2A + sin2B + sin2C = 4sinAsinBsinC

2. (a) Prove secθ + cscθ / (1 + tanθ) = cscθ

We need to prove: (secθ + cscθ)/(1 + tanθ) = cscθ

Proof:

Convert all terms to sine and cosine:

LHS = (1/cosθ + 1/sinθ) / (1 + sinθ/cosθ)

Combine numerator:

= [(sinθ + cosθ)/(sinθcosθ)] / [(cosθ + sinθ)/cosθ]

Simplify complex fraction:

= (sinθ + cosθ)/(sinθcosθ) × cosθ/(sinθ + cosθ)

= 1/sinθ = cscθ = RHS

Thus proved.

3. (b) Write 7sinx - 4cosx in form Rsin(x - α)

We want to express 7sinx - 4cosx as Rsin(x - α)

Solution:

First expand Rsin(x - α):

Rsin(x - α) = Rsinxcosα - Rcosxsinα

Compare coefficients with 7sinx - 4cosx:

Rcosα = 7

Rsinα = 4

Find R:

R² = 7² + 4² = 49 + 16 = 65 ⇒ R = √65

Find α:

tanα = Rsinα/Rcosα = 4/7 ⇒ α ≈ 29.74°

Thus:

7sinx - 4cosx = √65 sin(x - 29.74°)

4. (a) Prove cos(A/2) = √[s(s - a)/bc] for triangle ABC

Where s = (a + b + c)/2 is the semi-perimeter.

Proof:

Start with the cosine half-angle formula:

cos(A/2) = √[(1 + cosA)/2]

From cosine rule: cosA = (b² + c² - a²)/2bc

Thus:

1 + cosA = 1 + (b² + c² - a²)/2bc = [2bc + b² + c² - a²]/2bc

= [(b + c)² - a²]/2bc = [(b + c + a)(b + c - a)]/2bc

Now, s = (a + b + c)/2 ⇒ 2s = a + b + c

Thus:

b + c - a = 2s - 2a = 2(s - a)

Substitute back:

1 + cosA = [2s × 2(s - a)]/2bc = 4s(s - a)/2bc = 2s(s - a)/bc

Therefore:

cos(A/2) = √[(1 + cosA)/2] = √[s(s - a)/bc]

QED.

5. (a) If A + B = C and tanA = p tanB, express sin(A - B)/sinC in terms of p

Solution:

Given A + B = C and tanA = p tanB.

We need to find expression for sin(A - B)/sinC.

First, write sin(A - B) using sine subtraction formula:

sin(A - B) = sinAcosB - cosAsinB

And sinC = sin(A + B) = sinAcosB + cosAsinB

Thus:

sin(A - B)/sinC = [sinAcosB - cosAsinB]/[sinAcosB + cosAsinB]

Divide numerator and denominator by cosAcosB:

= [tanA - tanB]/[tanA + tanB]

Given tanA = p tanB, substitute:

= [p tanB - tanB]/[p tanB + tanB] = tanB(p - 1)/tanB(p + 1)

Simplify:

sin(A - B)/sinC = (p - 1)/(p + 1)

6. (a) Express Sin3θ in terms of Sinθ

Solution:

Using triple angle formula:

sin3θ = sin(2θ + θ) = sin2θcosθ + cos2θsinθ

Using double angle formulas:

= (2sinθcosθ)cosθ + (1 - 2sin²θ)sinθ

= 2sinθcos²θ + sinθ - 2sin³θ

Substitute cos²θ = 1 - sin²θ:

= 2sinθ(1 - sin²θ) + sinθ - 2sin³θ

= 2sinθ - 2sin³θ + sinθ - 2sin³θ = 3sinθ - 4sin³θ

Thus: sin3θ = 3sinθ - 4sin³θ

7. (a) If sin⁻¹x + sin⁻¹y = π/2 and cos⁻¹x - cos⁻¹y = 0, prove x = y = ±1/√2

Proof:

Given equations:

(1) sin⁻¹x + sin⁻¹y = π/2

(2) cos⁻¹x - cos⁻¹y = 0 ⇒ cos⁻¹x = cos⁻¹y ⇒ x = y

Substitute x = y into equation (1):

2sin⁻¹x = π/2 ⇒ sin⁻¹x = π/4 ⇒ x = sin(π/4) = 1/√2

Thus y = 1/√2 also.

But we must consider the range restrictions:

sin⁻¹x is defined for x ∈ [-1,1] and outputs ∈ [-π/2, π/2]

cos⁻¹x is defined for x ∈ [-1,1] and outputs ∈ [0, π]

The negative solution also satisfies:

If x = y = -1/√2:

sin⁻¹(-1/√2) = -π/4

-π/4 + -π/4 = -π/2 ≠ π/2 (doesn't satisfy first equation)

Thus the only solution is:

x = y = 1/√2

Key Trigonometry Concepts Covered:

• Trigonometric identities and proofs
• Sum-to-product and product-to-sum formulas
• Expression of trigonometric functions in alternative forms
• Inverse trigonometric functions
• Applications to triangles and angle relationships