Mathematics
Trigonometric Identities and Equations
223 Questions
Solve a variety of questions based on trigonometric identities and mathematical equations. Key areas covered include double angle formulas, the law of sines, and quadrant angles. This material helps students preparing for advanced mathematics exams, UPSC, and state public service commissions.
Double angle formulasLaw of sinesTrigonometric quadrantsLaplace transformSecant functionTaylor series
Trigonometric Identities and Equations Questions
What was Brahmagupta's formula for the sine of an angle?
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$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$
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$\sin \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$
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$\sin \theta = \frac{\text{opposite}}{\text{adjacent}}$
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$\sin \theta = \frac{\text{hypotenuse}}{\text{adjacent}}$
A
Correct answer
Explanation
Brahmagupta's formula for the sine of an angle is $\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$.
What was Brahmagupta's formula for the cosine of an angle?
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$\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$
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$\cos \theta = \frac{\text{opposite}}{\text{hypotenuse}}$
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$\cos \theta = \frac{\text{opposite}}{\text{adjacent}}$
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$\cos \theta = \frac{\text{hypotenuse}}{\text{adjacent}}$
A
Correct answer
Explanation
Brahmagupta's formula for the cosine of an angle is $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$.
What is the general formula for the sine series?
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$sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$
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$sin(x) = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!} + \cdots$
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$sin(x) = x - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$
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$sin(x) = x + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \cdots$
A
Correct answer
Explanation
The general formula for the sine series is $sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$.
What is the general formula for the cosine series?
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$cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$
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$cos(x) = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \cdots$
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$cos(x) = 1 - \frac{x}{2} + \frac{x^2}{4} - \frac{x^3}{6} + \cdots$
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$cos(x) = 1 + \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{6} + \cdots$
A
Correct answer
Explanation
The general formula for the cosine series is $cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$.
What is the value of $sin(\frac{\pi}{2})$ using the sine series?
B
Correct answer
Explanation
Using the sine series, we have $sin(\frac{\pi}{2}) = \frac{\pi}{2} - \frac{(\frac{\pi}{2})^3}{3!} + \frac{(\frac{\pi}{2})^5}{5!} - \frac{(\frac{\pi}{2})^7}{7!} + \cdots = 1$.
What is the value of $cos(\frac{\pi}{2})$ using the cosine series?
A
Correct answer
Explanation
Using the cosine series, we have $cos(\frac{\pi}{2}) = 1 - \frac{(\frac{\pi}{2})^2}{2!} + \frac{(\frac{\pi}{2})^4}{4!} - \frac{(\frac{\pi}{2})^6}{6!} + \cdots = 0$.
What is the value of $sin(\frac{\pi}{3})$ using the sine series?
Correct answer
Explanation
Using the sine series, we have $sin(\frac{\pi}{3}) = \frac{\pi}{3} - \frac{(\frac{\pi}{3})^3}{3!} + \frac{(\frac{\pi}{3})^5}{5!} - \frac{(\frac{\pi}{3})^7}{7!} + \cdots = \frac{\sqrt{3}}{2}$.
What is the value of $cos(\frac{\pi}{3})$ using the cosine series?
D
Correct answer
Explanation
Using the cosine series, we have $cos(\frac{\pi}{3}) = 1 - \frac{(\frac{\pi}{3})^2}{2!} + \frac{(\frac{\pi}{3})^4}{4!} - \frac{(\frac{\pi}{3})^6}{6!} + \cdots = \frac{1}{2}$.
What is the value of $sin(\frac{\pi}{4})$ using the sine series?
Correct answer
Explanation
Using the sine series, we have $sin(\frac{\pi}{4}) = \frac{\pi}{4} - \frac{(\frac{\pi}{4})^3}{3!} + \frac{(\frac{\pi}{4})^5}{5!} - \frac{(\frac{\pi}{4})^7}{7!} + \cdots = \frac{1}{\sqrt{2}}$.
What is the value of $cos(\frac{\pi}{4})$ using the cosine series?
Correct answer
Explanation
Using the cosine series, we have $cos(\frac{\pi}{4}) = 1 - \frac{(\frac{\pi}{4})^2}{2!} + \frac{(\frac{\pi}{4})^4}{4!} - \frac{(\frac{\pi}{4})^6}{6!} + \cdots = \frac{1}{\sqrt{2}}$.
What is the value of $sin(\frac{\pi}{6})$ using the sine series?
D
Correct answer
Explanation
Using the sine series, we have $sin(\frac{\pi}{6}) = \frac{\pi}{6} - \frac{(\frac{\pi}{6})^3}{3!} + \frac{(\frac{\pi}{6})^5}{5!} - \frac{(\frac{\pi}{6})^7}{7!} + \cdots = \frac{1}{2}$.
What is the value of $cos(\frac{\pi}{6})$ using the cosine series?
Correct answer
Explanation
Using the cosine series, we have $cos(\frac{\pi}{6}) = 1 - \frac{(\frac{\pi}{6})^2}{2!} + \frac{(\frac{\pi}{6})^4}{4!} - \frac{(\frac{\pi}{6})^6}{6!} + \cdots = \frac{\sqrt{3}}{2}$.
What is the Laplace transform of the cosine function $\cos(at)$?
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$\frac{s}{s^2+a^2}$
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$\frac{a}{s^2+a^2}$
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$\frac{s}{s^2-a^2}$
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$\frac{a}{s^2-a^2}$
A
Correct answer
Explanation
The Laplace transform of the cosine function $\cos(at)$ is $\frac{s}{s^2+a^2}$.
What does the symbol (\sin) represent?
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Sine
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Cosine
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Tangent
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Cosecant
A
Correct answer
Explanation
(\sin) is a mathematical symbol that represents the sine of an angle. It is used to find the ratio of the length of the opposite side to the length of the hypotenuse of a right triangle.
What does the symbol (\cos) represent?
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Sine
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Cosine
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Tangent
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Cosecant
B
Correct answer
Explanation
(\cos) is a mathematical symbol that represents the cosine of an angle. It is used to find the ratio of the length of the adjacent side to the length of the hypotenuse of a right triangle.