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1 Without using a calculator, determine the numerical value of: sin 120 2 Prove that: (2 sin A+1)/(cos A+sin 2A) = sec A

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2 parts
3.1

32\frac{\sqrt{3}}{2}

3.2
Shown:

: 2sin⁡A+1cos⁡A+sin⁡2A=sec⁡A\frac{2 \sin A + 1}{\cos A + \sin 2A} = \sec A

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Part 3

Part 3.1

32\frac{\sqrt{3}}{2}

Without using a calculator, determine the numerical value of sin⁡120∘\sin 120^\circ.

  1. Determine the quadrant and reference angle.

    The angle 120∘120^\circ lies in the second quadrant.
    The reference angle θR\theta_R is 180∘−120∘=60∘180^\circ - 120^\circ = 60^\circ.

  2. Determine the sign of sin⁡120∘\sin 120^\circ.

    In the second quadrant, the sine function is positive.
    So, sin⁡120∘=sin⁡60∘\sin 120^\circ = \sin 60^\circ.

  3. Recall the exact value of sin⁡60∘\sin 60^\circ.

    sin⁡60∘=32\sin 60^\circ = \frac{\sqrt{3}}{2}

    The numerical value of sin⁡120∘\sin 120^\circ is 32\boxed{\frac{\sqrt{3}}{2}}.

Part 3.2

: 2sin⁡A+1cos⁡A+sin⁡2A=sec⁡A\frac{2 \sin A + 1}{\cos A + \sin 2A} = \sec A

Prove that: 2sin⁡A+1cos⁡A+sin⁡2A=sec⁡A\frac{2 \sin A + 1}{\cos A + \sin 2A} = \sec A.

  1. Start with the Left Hand Side (LHS) and apply the double angle identity for sin⁡2A\sin 2A.

    Recall that sin⁡2A=2sin⁡Acos⁡A\sin 2A = 2 \sin A \cos A.

    LHS=2sin⁡A+1cos⁡A+2sin⁡Acos⁡A\text{LHS} = \frac{2 \sin A + 1}{\cos A + 2 \sin A \cos A}

  2. Factor out cos⁡A\cos A from the denominator.

    LHS=2sin⁡A+1cos⁡A(1+2sin⁡A)\text{LHS} = \frac{2 \sin A + 1}{\cos A (1 + 2 \sin A)}

  3. Cancel the common factor (2sin⁡A+1)(2 \sin A + 1) from the numerator and denominator.

    LHS=1cos⁡A\text{LHS} = \frac{1}{\cos A}

  4. Express the result in terms of sec⁡A\sec A.

    Recall that sec⁡A=1cos⁡A\sec A = \frac{1}{\cos A}.

    LHS=sec⁡A\text{LHS} = \sec A

    Since LHS = RHS, the identity is proven.

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