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Question

Direction: Read the following information and answer the  three  items that follow:

Let α = β = 15°.

What is the value of sin α + cos β ?

This question was previously asked in
NDA 2020 GAT Previous Year Paper (06-Sep-2020)
The correct answer is

√3 / √2

Calculating sin 15° + cos 15°

The problem asks for the value of \(\sin \alpha + \cos \beta\), given that \(\alpha = \beta = 15^\circ\). This means we need to find the value of \(\sin 15^\circ + \cos 15^\circ\).

To solve this, we first need to determine the values of \(\sin 15^\circ\) and \(\cos 15^\circ\). These values can be found using trigonometric identities for differences of angles. We know that \(15^\circ = 45^\circ - 30^\circ\).

Finding the Value of sin 15°

We use the sine subtraction formula: \(\sin(A - B) = \sin A \cos B - \cos A \sin B\).

Let \(A = 45^\circ\) and \(B = 30^\circ\).

So, \(\sin 15^\circ = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ\).

We know the standard values for \(45^\circ\) and \(30^\circ\):

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

Substitute these values into the formula:

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

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

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

Finding the Value of cos 15°

We use the cosine subtraction formula: \(\cos(A - B) = \cos A \cos B + \sin A \sin B\).

Let \(A = 45^\circ\) and \(B = 30^\circ\).

So, \(\cos 15^\circ = \cos(45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ\).

Substitute the standard values:

\(\cos 15^\circ = \left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right)\)

\(\cos 15^\circ = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}}\)

\(\cos 15^\circ = \frac{\sqrt{3}+1}{2\sqrt{2}}\)

Calculating sin 15° + cos 15°

Now we add the values we found for \(\sin 15^\circ\) and \(\cos 15^\circ\):

\(\sin 15^\circ + \cos 15^\circ = \frac{\sqrt{3}-1}{2\sqrt{2}} + \frac{\sqrt{3}+1}{2\sqrt{2}}\)

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

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

\(= \frac{2\sqrt{3}}{2\sqrt{2}}\)

\(= \frac{\sqrt{3}}{\sqrt{2}}\)

Alternatively, we can rationalize the denominator by multiplying the numerator and denominator by \(\sqrt{2}\):

\(\frac{\sqrt{3}}{\sqrt{2}} = \frac{\sqrt{3} \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} = \frac{\sqrt{6}}{2}\)

However, the options are not rationalized, so we should compare our result \(\frac{\sqrt{3}}{\sqrt{2}}\) with the given options.

Comparing our result \(\frac{\sqrt{3}}{\sqrt{2}}\) with the options:

  • Option 1: \(\frac{1}{\sqrt{2}}\) (Incorrect)
  • Option 2: \(\frac{1}{2\sqrt{2}}\) (Incorrect)
  • Option 3: \(\frac{\sqrt{3}}{2\sqrt{2}}\) (Incorrect)
  • Option 4: \(\frac{\sqrt{3}}{\sqrt{2}}\) (Correct)

The value of \(\sin 15^\circ + \cos 15^\circ\) is \(\frac{\sqrt{3}}{\sqrt{2}}\).

Revision Table: Key Trigonometric Values

Angle sin cos tan
\(0^\circ\) 0 1 0
\(30^\circ\) (\(\frac{\pi}{6}\)) \(\frac{1}{2}\) \(\frac{\sqrt{3}}{2}\) \(\frac{1}{\sqrt{3}}\)
\(45^\circ\) (\(\frac{\pi}{4}\)) \(\frac{1}{\sqrt{2}}\) \(\frac{1}{\sqrt{2}}\) 1
\(60^\circ\) (\(\frac{\pi}{3}\)) \(\frac{\sqrt{3}}{2}\) \(\frac{1}{2}\) \(\sqrt{3}\)
\(90^\circ\) (\(\frac{\pi}{2}\)) 1 0 Undefined

Additional Information on Trigonometric Identities

Besides difference formulas, there are many other trigonometric identities that are useful for simplifying expressions and solving equations. Some key identities include:

  • Pythagorean Identity: \(\sin^2 \theta + \cos^2 \theta = 1\)
  • Sum Formulas:
    • \(\sin(A + B) = \sin A \cos B + \cos A \sin B\)
    • \(\cos(A + B) = \cos A \cos B - \sin A \sin B\)
    • \(\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\)
  • Double Angle Formulas:
    • \(\sin(2\theta) = 2 \sin \theta \cos \theta\)
    • \(\cos(2\theta) = \cos^2 \theta - \sin^2 \theta = 2 \cos^2 \theta - 1 = 1 - 2 \sin^2 \theta\)
    • \(\tan(2\theta) = \frac{2 \tan \theta}{1 - \tan^2 \theta}\)
  • Half Angle Formulas:
    • \(\sin\left(\frac{\theta}{2}\right) = \pm\sqrt{\frac{1 - \cos \theta}{2}}\)
    • \(\cos\left(\frac{\theta}{2}\right) = \pm\sqrt{\frac{1 + \cos \theta}{2}}\)
    • \(\tan\left(\frac{\theta}{2}\right) = \frac{1 - \cos \theta}{\sin \theta} = \frac{\sin \theta}{1 + \cos \theta}\)

These identities help in calculating trigonometric values for various angles, including \(15^\circ\), \(75^\circ\), etc., by expressing them as sums or differences of standard angles.

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