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Difficulty: HardBasic Trigonometric Ratios, Special Angles, and Identities

If θ\theta is an acute angle such that 1−tan⁡θ1+tan⁡θ=2−3\frac{1 - \tan \theta}{1 + \tan \theta} = 2 - \sqrt{3}, what is the exact value of 2sin⁡θ+3cos⁡θ2\sin \theta + \sqrt{3}\cos \theta?

  1. 52\frac{5}{2}Answer
  2. B
    332\frac{3\sqrt{3}}{2}
  3. C
    52\frac{\sqrt{5}}{2}
  4. D
    32\frac{3}{2}

Answer

The exact value of 2sin⁡θ+3cos⁡θ2\sin \theta + \sqrt{3}\cos \theta is 52\frac{5}{2}.
Cross-multiplying the equation 1−tan⁡θ1+tan⁡θ=2−3\frac{1 - \tan \theta}{1 + \tan \theta} = 2 - \sqrt{3} yields tan⁡θ=13\tan \theta = \frac{1}{\sqrt{3}}, which identifies θ\theta as 30∘30^\circ. Evaluating 2sin⁡30∘+3cos⁡30∘=2(12)+3(32)=1+32=522\sin 30^\circ + \sqrt{3}\cos 30^\circ = 2\left(\frac{1}{2}\right) + \sqrt{3}\left(\frac{\sqrt{3}}{2}\right) = 1 + \frac{3}{2} = \frac{5}{2}.

Step-by-Step Solution

1
Solve the given equation 1−tan⁡θ1+tan⁡θ=2−3\frac{1 - \tan \theta}{1 + \tan \theta} = 2 - \sqrt{3} for tan⁡θ\tan \theta.
1−tan⁡θ=(2−3)(1+tan⁡θ)  ⟹  1−tan⁡θ=2−3+(2−3)tan⁡θ1 - \tan \theta = (2 - \sqrt{3})(1 + \tan \theta) \implies 1 - \tan \theta = 2 - \sqrt{3} + (2 - \sqrt{3})\tan \theta. Rearranging terms gives (3−1)=(3−3)tan⁡θ  ⟹  tan⁡θ=3−13−3=3−13(3−1)=13(\sqrt{3} - 1) = (3 - \sqrt{3})\tan \theta \implies \tan \theta = \frac{\sqrt{3} - 1}{3 - \sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3}(\sqrt{3} - 1)} = \frac{1}{\sqrt{3}}.
Isolating tan⁡θ\tan \theta allows determination of the specific angle θ\theta.
2
Determine the acute angle θ\theta corresponding to tan⁡θ=13\tan \theta = \frac{1}{\sqrt{3}}.
Since θ\theta is acute and tan⁡30∘=13\tan 30^\circ = \frac{1}{\sqrt{3}}, θ=30∘\theta = 30^\circ.
Special angle identities state that tan⁡30∘=13\tan 30^\circ = \frac{1}{\sqrt{3}}.
3
Evaluate the target trigonometric expression 2sin⁡θ+3cos⁡θ2\sin \theta + \sqrt{3}\cos \theta at θ=30∘\theta = 30^\circ.
2sin⁡30∘+3cos⁡30∘=2(12)+3(32)=1+32=522\sin 30^\circ + \sqrt{3}\cos 30^\circ = 2\left(\frac{1}{2}\right) + \sqrt{3}\left(\frac{\sqrt{3}}{2}\right) = 1 + \frac{3}{2} = \frac{5}{2}.
Substituting the exact surd values sin⁡30∘=12\sin 30^\circ = \frac{1}{2} and cos⁡30∘=32\cos 30^\circ = \frac{\sqrt{3}}{2} simplifies directly to the final fraction.

Key Concept

Trigonometric ratio deduction using special angles and surd simplification.
Estimated Time:2m 0s
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Topics

Basic Trigonometric Ratios, Special Angles, and Identities

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