Trigonometry
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The value of 2cos π cos 9π + cos 3π + cos 5π is 13 13 13 13
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2cos π cos 9π + cos 3π + cos 5π 13 13 13 13 = cos 
π + 9π 
+ cos π - 9π + cos 3π + cos 5π 13 13 13 13 13 13
∵ 2 cosA cosB = cos (A + B) + cos (A – B)= cos 10π + cos - 8π + cos 3π + cos 5π 13 13 13 13 = cos 10π + cos 8π + cos 3π + cos 5π 13 13 13 13
∵ cos(-θ) = cosθ= cos π - 3π + cos π - 5π + cos 3π + cos 5π 13 13 13 13 = - cos 3π - cos 5π + cos 3π + cos 5π 13 13 13 13
= 0Correct Option: B
2cos π cos 9π + cos 3π + cos 5π 13 13 13 13 = cos π + 9π + cos π - 9π + cos 3π + cos 5π 13 13 13 13 13 13
∵ 2 cosA cosB = cos (A + B) + cos (A – B)= cos 10π + cos - 8π + cos 3π + cos 5π 13 13 13 13 = cos 10π + cos 8π + cos 3π + cos 5π 13 13 13 13
∵ cos(-θ) = cosθ= cos π - 3π + cos π - 5π + cos 3π + cos 5π 13 13 13 13 = - cos 3π - cos 5π + cos 3π + cos 5π 13 13 13 13
= 0
- If sinθ + cosθ = 1, then the value of sin2θ is equal to
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Here,
sinθ + cosθ = 1
Squaring on both sides, we get
(sinθ + cosθ)² = 1
sin²θ + cos²θ + 2sinθ × cosθ = 1
1 + 2sinθ cosθ = 1
2sinθ× cosθ = 0
And we know that,
sin2θ = 2sinθ cosθ
⇒ sin2θ = 0Correct Option: C
Here,
sinθ + cosθ = 1
Squaring on both sides, we get
(sinθ + cosθ)² = 1
sin²θ + cos²θ + 2sinθ × cosθ = 1
1 + 2sinθ cosθ = 1
2sinθ× cosθ = 0
And we know that,
sin2θ = 2sinθ cosθ
⇒ sin2θ = 0
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The value of cos π - θ cos π - ∅ -sin π - θ sin π - ∅ will be 4 4 4 4
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Here,
cos π - θ cos π - ∅ 4 4 - sin π - θ sin π - ∅ = ? 4 4
Letπ - θ = A 4 π - ∅ = B 4 ⇒ cos π - θ cos π - ∅ 4 4 - sin π - θ sin π - ∅ = ? 4 4 = cos 
π - θ + π - ∅ 
4 4 = cos π - θ - ∅ 2 = cos π - (θ + ∅) 2
= sin (θ + ∅)∵ cos π - θ = sinθ 2 Correct Option: B
Here,
cos π - θ cos π - ∅ 4 4 - sin π - θ sin π - ∅ = ? 4 4
Letπ - θ = A 4 π - ∅ = B 4 ⇒ cos π - θ cos π - ∅ 4 4 - sin π - θ sin π - ∅ = ? 4 4 = cos π - θ + π - ∅ 4 4 = cos π - θ - ∅ 2 = cos π - (θ + ∅) 2
= sin (θ + ∅)∵ cos π - θ = sinθ 2
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If tan θ = x − y , the value of sinθ is equal to [If 0° ≤ θ ≤ 90°] x + y
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Here,
tan θ = x - y x + y
Consider ∆ABC,
Using pythagoras theorem, we get
AC² = AB² + BC²
⇒ AC² = (x + y)² + (x – y)²
= x² + y² + 2xy + x² + y² – 2xy
AC² = 2 (x² + y²)
AC = √2(x² + y²)
As θ lies in first quadrant,
∵ sinq will be +vesinθ = BC AC sinθ = x - y √2(x² + y²) Correct Option: A
Here,
tan θ = x - y x + y
Consider ∆ABC,
Using pythagoras theorem, we get
AC² = AB² + BC²
⇒ AC² = (x + y)² + (x – y)²
= x² + y² + 2xy + x² + y² – 2xy
AC² = 2 (x² + y²)
AC = √2(x² + y²)
As θ lies in first quadrant,
∵ sinq will be +vesinθ = BC AC sinθ = x - y √2(x² + y²)
- If sinC + sinD = x, then the value of x is
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Here, sinC + sinD = x
⇒ x = sinC + sinD⇒ x = 2.sin C + D .cos C - D 2 2
[∵ it is the basic formula of sinC + sinD]Correct Option: D
Here, sinC + sinD = x
⇒ x = sinC + sinD⇒ x = 2.sin C + D .cos C - D 2 2
[∵ it is the basic formula of sinC + sinD]
