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For the trigonometric identity
11. If cos 27° = x, then the value of tan 63° interims of “x” is x/√1 – x²
12. If Θ be an acute angle and 7sin²Θ + 3 cos²Θ= 4, then tan Θ is 1/√3
13. The value of tan 80° × tan 10° + sin² 70° + sin² 20° is 2
14. The value of (sin 47°/cos 43°)² + (cos 43°/sin 47°) – 4 cos²45° is 0
15. If 2 (cos²Θ – sin²Θ) = 1, Θ is a positive acute angle them the value of Θ is 30°
16. If 5 tan Θ = 4, then (5 sin Θ – 3 cos Θ)/(5 sin Θ + 2 cos Θ) is equal to 1/6
17. If sin(x + 20)° = cos (x + 10)° then the value of “x” is 30°
18. The value of (sin 65°)/ (cos 25°) is 1
How do we find the various trigonometric identity?
To solve the various trigonometric identity;
11. Given: cos 27° = x
We know that cos (90 – θ) = sin θ
So, cos 63° = sin 27°
And sin 63° = √1 – cos²27°
Substituting cos 27° = x, we get
sin 63° = √1 – x²
Therefore, Therefore, tan 63° = sin 63° / cos 63° = cos 27° / cos 63° = x / cos 63°.
= x/√1 – x²
12. Given: Θ is an acute angle and 7sin²Θ + 3 cos²Θ= 4
Since Θ is an acute angle, sin²Θ + cos²Θ = 1
Substituting sin²Θ + cos²Θ = 1 into the equation 7sin²Θ + 3 cos²Θ= 4, we get
7 (sin²Θ/ cos²Θ) + 3 = 4/cos²Θ – 4 sec²Θ
⇒ 7tan²Θ + 3 = 4(1 + tan²Θ)
⇒ 7tan²Θ + 3 = 4 + 4 tan²Θ
⇒3 tan²Θ = 1
⇒ tan²Θ = 1/3
⇒ tanΘ = 1/√3
13. For tan 80° × tan 10° + sin² 70° + sin² 20°
⇒ tan 80° = cot (90 – 80)° = cot 10°
⇒ sin 70° = cos (90 – 70) = cos 20°
⇒ cot 10° × tan 10° + cos 20° + sin² 20°
= 1 + 1 = 2
14. (sin 47°/cos 43°)² + (cos 43°/sin 47°) – 4 cos²45°
= (sin 47°/cos43°)² + (cos 43°/sin 47°)² – 4(1/√2)²
= (sin (90° – 43°)/cos43°)² + (cos (90° – 47°)/sin)² = 4(1/2)
= (cos 43°/cos 43°)² + (sin 47°/ sin 47°)² – 2
= 1 + 1 – 2 = 0
15. 2 (cos²Θ – sin²Θ) = 1
cos²Θ – sin²Θ = 1/2
Since Θ is an acute angle, sin²Θ + cos²Θ = 1
Substituting sin²Θ + cos²Θ = 1 into the equation cos²Θ – sin²Θ = 1/2, we get
cos²Θ – (1 – cos²Θ) = 1/2
2cos²Θ = 3/2
cos Θ = √3/2(cos 30° = (√3)/2
= 30°
16. Given: 5 tan Θ = 4
We know that tan Θ = sin Θ / cos Θ
So, 5 sin Θ / cos Θ = 4
5 sin Θ = 4 cos Θ
Dividing both sides of the equation by 5, we get
sin Θ / cos Θ = 4/5
∴ sin Θ = 4/5 cos Θ
given that the expression is (5 sin Θ – 3 cos Θ)/(5 sin Θ + 2 cos Θ)
we substitute sin Θ = 4/5 cos Θ into the equation
⇒(5 × 4/5 cos Θ – 3 cos Θ)/(5 × 4/5 cos Θ + 2 cos Θ)
= (4-3)/(4 + 2) = 1/6
17. Given: sin(x + 20)° = cos (x + 10)°
We know that sin(90 – θ) = cos θ
So, sin(x – 20)° = sin(90 – (3x + 10))°
⇒ (x – 20)° = (90 – (3x + 10))°
⇒ x – 20° = 90° – 3x + 10
⇒ 4 x = 120°
⇒ x = 120°/4
⇒ x = 30°
18. To find the value of (sin 65°) / (cos 25°), we can use the trigonometric identity:
To solve this, we can use the following trigonometric identities:
sin(90 – θ) = cos θ
cos(90 – θ) = sin θ
We can also use the fact that sin²θ + cos²θ = 1.
Rewrite sin (65°) / cos (25°)
⇒ sin (65°) = cos (25°)
∴ cos (25°)/ cos (25°) = 1
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