If cos 28° + sin 28° = $$k$$​, then cos 17° is equal to:

Correct option is B

Given:

cos 28° + sin 28° = k

Concept Used:

Trigonometric Identities:

sin(A + B) = sin A cos B + cos A sin B

cos(A - B) = cos A cos B + sin A sin B

Solution:

Divide the given equation by $$\sqrt2$$​:

$$\frac{1}{\sqrt2}$$​cos 28° +$$\frac{1}{\sqrt2}$$​sin 28° =$$\frac{k}{\sqrt2}$$​​

cos 45° cos 28° + sin 45° sin 28° = $$\frac{k}{\sqrt2}$$​​

cos (45° - 28°) =$$\frac{k}{\sqrt2}$$​​

cos 17° =$$\frac{k}{\sqrt2}$$​​

Alternate Solution:

Square the given equation:

(cos 28° + sin 28°)² = k²

cos² 28° + sin² 28° + 2 sin 28° cos 28° = k²

1 + sin 56° = k² ( since, sin²​$$\theta$$ + cos²​$$\theta$$=1)

Use sin 56° = cos 34°:

1 + cos 34° = k²

Use the half-angle formula for cosine:

cos 34° = 2 cos² 17° - 1

1 + 2 cos² 17° - 1 = k²

2 cos² 17° = k²

cos² 17° =$$\frac{k²}{2}$$​​

cos 17° = $$\frac{k}{\sqrt2}$$​(since cos 17° is positive)

Therefore, cos 17° =$$\frac{k}{\sqrt2}$$.​