If cos 47° + sin 47° = k, then what is the value of cos 2 47° - sin 2 47°?

Given :cos 47° + sin 47° = kConcept used :cos 2θ = cos 2 θ - sin 2 θ cos 2 θ + sin 2 θ = 1 2 sin θ cos θ = sin 2θ Calculations :We havecos 47° + sin 47° = k Squaring on both side( cos 47° + sin 47° )2 = k 2We know that,(a + b) 2 = a 2 + 2ab + b 2⇒ cos 247° + sin 2 47° + 2 cos 47° sin 47° = k 21 + 2 cos 47° sin 47° = k 2 (&because; cos 2 θ + sin 2 θ = 1 )2 cos 47° sin 47° = k 2- 1sin 94° = k 2 - 1 (&because; 2 sin θ cos θ = sin 2θ )We know that,sin 2θ + cos 2θ = 1 ⇒ sin 2 94° + cos 2 94° = 1 ⇒ cos 2 94° = 1 - (k 2- 1) 2 [&because; (a - b) 2 = a 2 - 2ab + b 2 ]⇒ cos 2 94° = 1 - (k 4+ 1 - 2k 2)⇒ cos 2 94° = 1 - k 4- 1 + 2k 2⇒ cos 2 94° = 2k 2- k 4⇒ cos 94° = ± k√(2 - k 2),We know that cos 94° lies on the second quadrant, and cos θ is negative in this quadrant. Therefore, cos 94° = - k√(2 - k 2 )⇒ cos 2 × 47 = - k√(2 - k 2 )Since cos 2θ = cos 2 θ - sin 2 θ ⇒ cos 2 47° - sin 2 47° = - k√(2 - k 2 )&there4; Option 2 will be the correct answer.

If cos 47° + sin 47° = k, then what is the value of cos 2 47°

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Question

CDS - 2020

Quantitative Aptitude

If cos 47° + sin 47° = k, then what is the value of cos ² 47° - sin ² 47°?

This question was previously asked in

CDS I 2020 Elementary Mathematics Previous Year Paper (02-Feb-2020)

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\(k\sqrt {2 - {k^2}}\)

\(-k\sqrt {2 - {k^2}}\)

\(k\sqrt {1 - {k^2}}\)

\(-k\sqrt {1 - {k^2}}\)

The correct answer is \(-k\sqrt {2 - {k^2}}\)

Given :

cos 47° + sin 47° = k

Concept used :

cos 2θ = cos ² θ - sin ² θ

cos 2 θ + sin 2 θ = 1

2 sin θ cos θ = sin 2θ

Calculations :

We have

cos 47° + sin 47° = k

Squaring on both side

( cos 47° + sin 47° )² = k ²

We know that,

(a + b) ² = a ² + 2ab + b ²

⇒ cos ²47° + sin ² 47° + 2 cos 47° sin 47° = k ²

1 + 2 cos 47° sin 47° = k ² (∵ cos 2 θ + sin 2 θ = 1 )

2 cos 47° sin 47° = k ²- 1

sin 94° = k 2 - 1 (∵ 2 sin θ cos θ = sin 2θ )

We know that,

sin ²θ + cos ²θ = 1

⇒ sin 2 94° + cos 2 94° = 1

⇒ cos 2 94° = 1 - (k ²- 1) ² [∵ (a - b) 2 = a 2 - 2ab + b 2 ]

⇒ cos 2 94° = 1 - (k ⁴+ 1 - 2k ²)

⇒ cos 2 94° = 1 - k ⁴- 1 + 2k ²

⇒ cos 2 94° = 2k ²- k ⁴

⇒ cos 94° = ± k√(2 - k ²),

We know that cos 94° lies on the second quadrant, and cos θ is

negative in this quadrant.

Therefore,

cos 94° = - k√(2 - k 2 )

⇒ cos 2 × 47 = - k√(2 - k 2 )

Since cos 2θ = cos 2 θ - sin 2 θ

⇒ cos 2 47° - sin 2 47° = - k√(2 - k 2 )

∴ Option 2 will be the correct answer.

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