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 is the addition formula for sine?
-
sin(x + y) = sin(x)cos(y) + cos(x)sin(y)
-
sin(x + y) = sin(x)cos(y) - cos(x)sin(y)
-
sin(x - y) = sin(x)cos(y) + cos(x)sin(y)
-
sin(x - y) = sin(x)cos(y) - cos(x)sin(y)
A
Correct answer
Explanation
The addition formula for sine states that the sine of the sum of two angles is equal to the sine of the first angle multiplied by the cosine of the second angle plus the cosine of the first angle multiplied by the sine of the second angle.
What is the addition formula for cosine?
-
cos(x + y) = cos(x)cos(y) - sin(x)sin(y)
-
cos(x + y) = cos(x)cos(y) + sin(x)sin(y)
-
cos(x - y) = cos(x)cos(y) - sin(x)sin(y)
-
cos(x - y) = cos(x)cos(y) + sin(x)sin(y)
A
Correct answer
Explanation
The addition formula for cosine states that the cosine of the sum of two angles is equal to the cosine of the first angle multiplied by the cosine of the second angle minus the sine of the first angle multiplied by the sine of the second angle.
What is the subtraction formula for sine?
-
sin(x - y) = sin(x)cos(y) - cos(x)sin(y)
-
sin(x - y) = sin(x)cos(y) + cos(x)sin(y)
-
cos(x - y) = cos(x)cos(y) - sin(x)sin(y)
-
cos(x - y) = cos(x)cos(y) + sin(x)sin(y)
A
Correct answer
Explanation
The subtraction formula for sine states that the sine of the difference of two angles is equal to the sine of the first angle multiplied by the cosine of the second angle minus the cosine of the first angle multiplied by the sine of the second angle.
What is the subtraction formula for cosine?
-
cos(x - y) = cos(x)cos(y) + sin(x)sin(y)
-
cos(x - y) = cos(x)cos(y) - sin(x)sin(y)
-
sin(x + y) = sin(x)cos(y) + cos(x)sin(y)
-
sin(x + y) = sin(x)cos(y) - cos(x)sin(y)
A
Correct answer
Explanation
The subtraction formula for cosine states that the cosine of the difference of two angles is equal to the cosine of the first angle multiplied by the cosine of the second angle plus the sine of the first angle multiplied by the sine of the second angle.
What is the double-angle formula for sine?
-
sin(2x) = 2sin(x)cos(x)
-
sin(2x) = sin(x) + sin(x)
-
sin(2x) = sin(x) - sin(x)
-
sin(2x) = cos(x) - cos(x)
A
Correct answer
Explanation
The double-angle formula for sine states that the sine of twice an angle is equal to twice the sine of the angle multiplied by the cosine of the angle.
What is the double-angle formula for cosine?
-
cos(2x) = cos^2(x) - sin^2(x)
-
cos(2x) = 2cos^2(x) - 1
-
cos(2x) = 1 - 2sin^2(x)
-
cos(2x) = 2cos(x) - 1
A
Correct answer
Explanation
The double-angle formula for cosine states that the cosine of twice an angle is equal to the square of the cosine of the angle minus the square of the sine of the angle.
What was Brahmagupta's formula for the sine of an angle?
-
$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$
-
$\sin \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$
-
$\sin \theta = \frac{\text{opposite}}{\text{adjacent}}$
-
$\sin \theta = \frac{\text{hypotenuse}}{\text{adjacent}}$
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?
-
$\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$
-
$\cos \theta = \frac{\text{opposite}}{\text{hypotenuse}}$
-
$\cos \theta = \frac{\text{opposite}}{\text{adjacent}}$
-
$\cos \theta = \frac{\text{hypotenuse}}{\text{adjacent}}$
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?
-
$sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$
-
$sin(x) = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!} + \cdots$
-
$sin(x) = x - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$
-
$sin(x) = x + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \cdots$
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?
-
$cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$
-
$cos(x) = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \cdots$
-
$cos(x) = 1 - \frac{x}{2} + \frac{x^2}{4} - \frac{x^3}{6} + \cdots$
-
$cos(x) = 1 + \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{6} + \cdots$
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 $sin(\frac{\pi}{4}) = \frac{\pi}{4} - \frac{(\frac{\pi}{4})^3}{3!} + \frac{(\frac{\pi}{4})^5}{5!} - \frac{(\frac{\pi}{4})^7}{7!} + \cdots = \frac{1}{\sqrt{2}}$.