Evaluating the Trigonometric Expression with Given Angles

The problem requires us to calculate the value of the expression 2 cos(θ) sin(γ) using the provided angle measures for θ and γ.

The given values are:

• θ = 135°
• γ = 15°

We need to find the numerical value of 2 cos(135°) sin(15°).

Step-by-Step Calculation for the Trigonometric Value

Step 1: Substitute Angles into the Expression

Substitute the given degrees for θ and γ into the expression:

$$ 2 \cos(135^\circ) \sin(15^\circ) $$

Step 2: Calculate the Value of cos(135°)

To find the value of cos(135°), we note that 135° is in the second quadrant. The reference angle is 180° - 135° = 45°. Since the cosine function is negative in the second quadrant:

$$ \cos(135^\circ) = -\cos(45^\circ) $$

The value of cos(45°) is \(\frac{\sqrt{2}}{2}\). Therefore:

$$ \cos(135^\circ) = -\frac{\sqrt{2}}{2} $$

Step 3: Calculate the Value of sin(15°)

To find the value of sin(15°), we can express 15° as a difference of two standard angles, such as 45° - 30°. We use the sine subtraction identity: sin(A - B) = sin A cos B - cos A sin B.

Let A = 45° and B = 30°.

$$ \sin(15^\circ) = \sin(45^\circ - 30^\circ) = \sin(45^\circ)\cos(30^\circ) - \cos(45^\circ)\sin(30^\circ) $$

Using the known values:

• \(\sin(45^\circ) = \frac{\sqrt{2}}{2}\)
• \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\)
• \(\cos(45^\circ) = \frac{\sqrt{2}}{2}\)
• \(\sin(30^\circ) = \frac{1}{2}\)

Substitute these values into the formula:

$$ \sin(15^\circ) = \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right) $$

$$ \sin(15^\circ) = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4} $$

Step 4: Compute the Final Expression Value

Now, substitute the calculated values of cos(135°) and sin(15°) back into the original expression:

$$ 2 \cos(135^\circ) \sin(15^\circ) = 2 \times \left(-\frac{\sqrt{2}}{2}\right) \times \left(\frac{\sqrt{6} - \sqrt{2}}{4}\right) $$

Simplify the expression:

$$ = -\sqrt{2} \times \left(\frac{\sqrt{6} - \sqrt{2}}{4}\right) $$

$$ = -\frac{\sqrt{2} \cdot \sqrt{6} - \sqrt{2} \cdot \sqrt{2}}{4} $$

$$ = -\frac{\sqrt{12} - 2}{4} $$

Since \(\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}\):

$$ = -\frac{2\sqrt{3} - 2}{4} $$

Factor out 2 from the numerator:

$$ = -\frac{2(\sqrt{3} - 1)}{4} $$

Simplify the fraction:

$$ = -\frac{\sqrt{3} - 1}{2} $$

Distribute the negative sign:

$$ = \frac{1 - \sqrt{3}}{2} $$

Alternative Calculation Using Product-to-Sum Identity

An alternative way to solve this is by using the product-to-sum trigonometric identity: 2 cos A sin B = sin(A + B) - sin(A - B).

In this case, A = θ = 135° and B = γ = 15°.

Applying the identity:

$$ 2 \cos(135^\circ) \sin(15^\circ) = \sin(135^\circ + 15^\circ) - \sin(135^\circ - 15^\circ) $$

$$ = \sin(150^\circ) - \sin(120^\circ) $$

Now, evaluate the sine values:

• sin(150°): This angle is in the second quadrant, with a reference angle of 180° - 150° = 30°. Sine is positive in the second quadrant, so sin(150°) = sin(30°) = \frac{1}{2}.
• sin(120°): This angle is in the second quadrant, with a reference angle of 180° - 120° = 60°. Sine is positive in the second quadrant, so sin(120°) = sin(60°) = \frac{\sqrt{3}}{2}.

Substitute these values back:

$$ \sin(150^\circ) - \sin(120^\circ) = \frac{1}{2} - \frac{\sqrt{3}}{2} = \frac{1 - \sqrt{3}}{2} $$

Final Answer Verification

Both calculation methods yield the result \(\frac{1 - \sqrt{3}}{2}\). This value corresponds to option 2.

The options provided were:

1. \(\frac{\sqrt{3} - 1}{2}\)
2. \(\frac{1 - \sqrt{3}}{2}\)
3. \(2-\sqrt{3}\)
4. \(\sqrt{3}-2\)

Our calculated value matches the second option.