Aryabhata's Contributions to Trigonometry

Description: Aryabhata's Contributions to Trigonometry
Number of Questions: 15
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Tags: indian mathematics trigonometry aryabhata
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What is the value of (\sin 30^\circ) according to Aryabhata?

  1. (\frac{1}{2})

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

  3. (\frac{1}{\sqrt{2}})

  4. (\frac{\sqrt{2}}{2})


Correct Option: A
Explanation:

Aryabhata used a table of sines to calculate the values of (\sin \theta) for (\theta = 0^\circ, 1^\circ, 2^\circ, ..., 90^\circ). According to his table, (\sin 30^\circ) is equal to (\frac{1}{2}).

What is the value of (\cos 30^\circ) according to Aryabhata?

  1. (\frac{1}{2})

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

  3. (\frac{1}{\sqrt{2}})

  4. (\frac{\sqrt{2}}{2})


Correct Option: B
Explanation:

Aryabhata used the Pythagorean identity (\sin^2 \theta + \cos^2 \theta = 1) to calculate the values of (\cos \theta). According to his table, (\cos 30^\circ) is equal to (\frac{\sqrt{3}}{2}).

What is the value of (\tan 30^\circ) according to Aryabhata?

  1. (\frac{1}{2})

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

  3. (\frac{1}{\sqrt{2}})

  4. (\frac{\sqrt{2}}{2})


Correct Option:
Explanation:

Aryabhata used the definition of (\tan \theta) as (\frac{\sin \theta}{\cos \theta}) to calculate the values of (\tan \theta). According to his table, (\tan 30^\circ) is equal to (\frac{1}{\sqrt{3}}).

What is the value of (\sin 45^\circ) according to Aryabhata?

  1. (\frac{1}{2})

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

  3. (\frac{1}{\sqrt{2}})

  4. (\frac{\sqrt{2}}{2})


Correct Option: C
Explanation:

Aryabhata used the Pythagorean identity (\sin^2 \theta + \cos^2 \theta = 1) to calculate the values of (\sin \theta). According to his table, (\sin 45^\circ) is equal to (\frac{1}{\sqrt{2}}).

What is the value of (\cos 45^\circ) according to Aryabhata?

  1. (\frac{1}{2})

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

  3. (\frac{1}{\sqrt{2}})

  4. (\frac{\sqrt{2}}{2})


Correct Option: C
Explanation:

Aryabhata used the Pythagorean identity (\sin^2 \theta + \cos^2 \theta = 1) to calculate the values of (\cos \theta). According to his table, (\cos 45^\circ) is equal to (\frac{1}{\sqrt{2}}).

What is the value of (\tan 45^\circ) according to Aryabhata?

  1. (\frac{1}{2})

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

  3. (\frac{1}{\sqrt{2}})

  4. (\frac{\sqrt{2}}{2})


Correct Option:
Explanation:

Aryabhata used the definition of (\tan \theta) as (\frac{\sin \theta}{\cos \theta}) to calculate the values of (\tan \theta). According to his table, (\tan 45^\circ) is equal to (1).

What is the value of (\sin 60^\circ) according to Aryabhata?

  1. (\frac{1}{2})

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

  3. (\frac{1}{\sqrt{2}})

  4. (\frac{\sqrt{2}}{2})


Correct Option: B
Explanation:

Aryabhata used the Pythagorean identity (\sin^2 \theta + \cos^2 \theta = 1) to calculate the values of (\sin \theta). According to his table, (\sin 60^\circ) is equal to (\frac{\sqrt{3}}{2}).

What is the value of (\cos 60^\circ) according to Aryabhata?

  1. (\frac{1}{2})

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

  3. (\frac{1}{\sqrt{2}})

  4. (\frac{\sqrt{2}}{2})


Correct Option: A
Explanation:

Aryabhata used the Pythagorean identity (\sin^2 \theta + \cos^2 \theta = 1) to calculate the values of (\cos \theta). According to his table, (\cos 60^\circ) is equal to (\frac{1}{2}).

What is the value of (\tan 60^\circ) according to Aryabhata?

  1. (\frac{1}{2})

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

  3. (\frac{1}{\sqrt{2}})

  4. (\frac{\sqrt{2}}{2})


Correct Option:
Explanation:

Aryabhata used the definition of (\tan \theta) as (\frac{\sin \theta}{\cos \theta}) to calculate the values of (\tan \theta). According to his table, (\tan 60^\circ) is equal to (\sqrt{3}).

What is the value of (\sin 75^\circ) according to Aryabhata?

  1. (\frac{1}{2})

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

  3. (\frac{1}{\sqrt{2}})

  4. (\frac{\sqrt{2}}{2})


Correct Option:
Explanation:

Aryabhata used the half-angle formula for (\sin \theta) to calculate the values of (\sin \theta) for angles greater than (45^\circ). According to his table, (\sin 75^\circ) is equal to (\frac{\sqrt{6 + \sqrt{3}}}{4}).

What is the value of (\cos 75^\circ) according to Aryabhata?

  1. (\frac{1}{2})

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

  3. (\frac{1}{\sqrt{2}})

  4. (\frac{\sqrt{2}}{2})


Correct Option:
Explanation:

Aryabhata used the half-angle formula for (\cos \theta) to calculate the values of (\cos \theta) for angles greater than (45^\circ). According to his table, (\cos 75^\circ) is equal to (\frac{\sqrt{6 - \sqrt{3}}}{4}).

What is the value of (\tan 75^\circ) according to Aryabhata?

  1. (\frac{1}{2})

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

  3. (\frac{1}{\sqrt{2}})

  4. (\frac{\sqrt{2}}{2})


Correct Option:
Explanation:

Aryabhata used the definition of (\tan \theta) as (\frac{\sin \theta}{\cos \theta}) to calculate the values of (\tan \theta) for angles greater than (45^\circ). According to his table, (\tan 75^\circ) is equal to (\frac{\sqrt{6 + \sqrt{3}}}{\sqrt{6 - \sqrt{3}}}).

What is the value of (\sin 90^\circ) according to Aryabhata?

  1. (\frac{1}{2})

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

  3. (\frac{1}{\sqrt{2}})

  4. (1)


Correct Option: D
Explanation:

Aryabhata defined (\sin 90^\circ) to be equal to (1). This is because (\sin \theta) is the ratio of the opposite side to the hypotenuse, and in a right triangle with an angle of (90^\circ), the opposite side is equal to the hypotenuse.

What is the value of (\cos 90^\circ) according to Aryabhata?

  1. (\frac{1}{2})

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

  3. (\frac{1}{\sqrt{2}})

  4. (0)


Correct Option: D
Explanation:

Aryabhata defined (\cos 90^\circ) to be equal to (0). This is because (\cos \theta) is the ratio of the adjacent side to the hypotenuse, and in a right triangle with an angle of (90^\circ), the adjacent side is equal to (0).

What is the value of (\tan 90^\circ) according to Aryabhata?

  1. (\frac{1}{2})

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

  3. (\frac{1}{\sqrt{2}})

  4. (\text{undefined})


Correct Option: D
Explanation:

Aryabhata did not define the value of (\tan 90^\circ). This is because (\tan \theta) is the ratio of the opposite side to the adjacent side, and in a right triangle with an angle of (90^\circ), the adjacent side is equal to (0). Therefore, (\tan 90^\circ) is undefined.

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