Mathematics ยท Quantitative Aptitude
Trigonometric Identities and Values
70 Questions
Trigonometric identities and values questions require evaluating complex angles using standard formulas. These are critical for quantitative aptitude sections in SSC and various state exams. Success depends on memorizing standard values and applying transformation rules.
angle transformationstrigonometric ratiosstandard angle valuessine cosine products
Trigonometric Identities and Values Questions
What is the value of (\sin 30^\circ) according to Aryabhata?
-
\(\frac{1}{2}\)
-
\(\frac{\sqrt{3}}{2}\)
-
\(\frac{1}{\sqrt{2}}\)
-
\(\frac{\sqrt{2}}{2}\)
A
Correct answer
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?
-
\(\frac{1}{2}\)
-
\(\frac{\sqrt{3}}{2}\)
-
\(\frac{1}{\sqrt{2}}\)
-
\(\frac{\sqrt{2}}{2}\)
B
Correct answer
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 (\sin 45^\circ) according to Aryabhata?
-
\(\frac{1}{2}\)
-
\(\frac{\sqrt{3}}{2}\)
-
\(\frac{1}{\sqrt{2}}\)
-
\(\frac{\sqrt{2}}{2}\)
C
Correct answer
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?
-
\(\frac{1}{2}\)
-
\(\frac{\sqrt{3}}{2}\)
-
\(\frac{1}{\sqrt{2}}\)
-
\(\frac{\sqrt{2}}{2}\)
C
Correct answer
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 (\sin 60^\circ) according to Aryabhata?
-
\(\frac{1}{2}\)
-
\(\frac{\sqrt{3}}{2}\)
-
\(\frac{1}{\sqrt{2}}\)
-
\(\frac{\sqrt{2}}{2}\)
B
Correct answer
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?
-
\(\frac{1}{2}\)
-
\(\frac{\sqrt{3}}{2}\)
-
\(\frac{1}{\sqrt{2}}\)
-
\(\frac{\sqrt{2}}{2}\)
A
Correct answer
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?
-
\(\frac{1}{2}\)
-
\(\frac{\sqrt{3}}{2}\)
-
\(\frac{1}{\sqrt{2}}\)
-
\(\frac{\sqrt{2}}{2}\)
Correct answer
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?
-
\(\frac{1}{2}\)
-
\(\frac{\sqrt{3}}{2}\)
-
\(\frac{1}{\sqrt{2}}\)
-
\(\frac{\sqrt{2}}{2}\)
Correct answer
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?
-
\(\frac{1}{2}\)
-
\(\frac{\sqrt{3}}{2}\)
-
\(\frac{1}{\sqrt{2}}\)
-
\(\frac{\sqrt{2}}{2}\)
Correct answer
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 (\cos 90^\circ) according to Aryabhata?
-
\(\frac{1}{2}\)
-
\(\frac{\sqrt{3}}{2}\)
-
\(\frac{1}{\sqrt{2}}\)
-
\(0\)
D
Correct answer
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 $\sin 30^\circ$?
-
$\frac{1}{2}$
-
$\frac{\sqrt{3}}{2}$
-
$\frac{1}{\sqrt{2}}$
-
$\frac{\sqrt{2}}{2}$
A
Correct answer
Explanation
$\sin 30^\circ = \frac{1}{2}$
What is the value of $\cos 45^\circ$?
-
$\frac{1}{2}$
-
$\frac{\sqrt{3}}{2}$
-
$\frac{1}{\sqrt{2}}$
-
$\frac{\sqrt{2}}{2}$
D
Correct answer
Explanation
$\cos 45^\circ = \frac{\sqrt{2}}{2}$
What is the value of $\tan 60^\circ$?
-
$\frac{1}{2}$
-
$\frac{\sqrt{3}}{2}$
-
$\frac{1}{\sqrt{2}}$
-
$\frac{\sqrt{2}}{2}$
Correct answer
Explanation
$\tan 60^\circ = \sqrt{3}$
What is the value of $\cot 30^\circ$?
-
$\frac{1}{2}$
-
$\frac{\sqrt{3}}{2}$
-
$\frac{1}{\sqrt{2}}$
-
$\frac{\sqrt{2}}{2}$
Correct answer
Explanation
$\cot 30^\circ = \sqrt{3}$
What is the value of $\sin^{-1} \frac{1}{2}$?
-
$30^\circ$
-
$45^\circ$
-
$60^\circ$
-
$75^\circ$
A
Correct answer
Explanation
$\sin^{-1} \frac{1}{2} = 30^\circ$