Simplify Trigonometric Expressions - Questions With Answers
Use trigonometric identities and formulas to simplify trigonometric expressions. The trigonometric identities and formulas on this site might be helpful to solve the questions below.
Question 1:
Simplify the following trigonometric expression:
\(\csc(x) \sin\left(\dfrac{\pi}{2} - x\right)\)
Solution to Question 1:
- Use the identity \(\sin\left(\dfrac{\pi}{2} - x\right) = \cos(x)\) and simplify:
\[
\csc(x) \sin\left(\dfrac{\pi}{2} - x\right) = \csc(x) \cos(x) = \cot(x)
\]
Question 2:
Simplify:
\[
\dfrac{\sin^4 x - \cos^4 x}{\sin^2 x - \cos^2 x}
\]
Solution to Question 2:
- Factor the numerator:
\[
\dfrac{(\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x)}{\sin^2 x - \cos^2 x}
\]
- Simplify using \(\sin^2 x + \cos^2 x = 1\):
\[
\dfrac{(\sin^2 x + \cos^2 x)}{1} = 1
\]
Question 3:
Simplify:
\[
\dfrac{\sec(x) \sin^2 x}{1 + \sec(x)}
\]
Solution to Question 3:
- Substitute \(\sec(x) = \dfrac{1}{\cos(x)}\):
\[
\dfrac{\sin^2 x}{\cos x (1 + \sec(x))} = \dfrac{\sin^2 x}{\cos x + 1}
\]
- Use \(\sin^2 x = 1 - \cos^2 x\), factor and simplify:
\[
\dfrac{1 - \cos^2 x}{\cos x + 1} = \dfrac{(1 - \cos x)(1 + \cos x)}{\cos x + 1} = 1 - \cos x
\]
Question 4:
Simplify:
\[ \sin(-x) \cos\left(\dfrac{\pi}{2} - x\right) \]
Solution to Question 4:
- Use \(\sin(-x) = -\sin x\) and \(\cos\left(\dfrac{\pi}{2} - x\right) = \sin x\):
\[
\sin(-x) \cos\left(\dfrac{\pi}{2} - x\right) = -\sin x \cdot \sin x = -\sin^2 x
\]
Question 5:
Simplify:
\(\sin^2 x - \cos^2 x \sin^2 x\)
Solution to Question 5:
- Factor \(\sin^2 x\):
\[
\sin^2 x - \cos^2 x \sin^2 x = \sin^2 x (1 - \cos^2 x) = \sin^4 x
\]
Question 6:
Simplify:
\(\tan^4 x + 2 \tan^2 x + 1\)
Solution to Question 6:
- Write as a square:
\[
\tan^4 x + 2 \tan^2 x + 1 = (\tan^2 x + 1)^2
\]
- Use \(\tan^2 x + 1 = \sec^2 x\):
\[
(\sec^2 x)^2 = \sec^4 x
\]
Question 7:
Add and simplify:
\(\dfrac{1}{1 + \cos x} + \dfrac{1}{1 - \cos x}\)
Solution to Question 7:
- Use a common denominator:
\[
\dfrac{1}{1 + \cos x} + \dfrac{1}{1 - \cos x} = \dfrac{1 - \cos x + 1 + \cos x}{(1 + \cos x)(1 - \cos x)} \]
Simplify the numerator and expand the denominator and simplify it.
\[ = \dfrac{2}{1 - \cos^2 x} = \dfrac{2}{\sin^2 x} = 2 \csc^2 x
\]
Question 8:
Simplify \(\sqrt{4 - 4 \sin^2 x}\) for \(\dfrac{\pi}{2} < x < \pi\)
Solution to Question 8:
- Factor and use \(\cos^2 x = 1 - \sin^2 x\):
\[
\sqrt{4 - 4 \sin^2 x} = \sqrt{4 (1 - \sin^2 x)} = 2 \sqrt{\cos^2 x} = 2 |\cos x|
\]
- Since \(\dfrac{\pi}{2} < x < \pi\), \(\cos x < 0\), so
\[
\sqrt{4 - 4 \sin^2 x} = -2 \cos x
\]
Question 9:
Simplify:
\(\dfrac{1 - \sin^4 x}{1 + \sin^2 x}\)
Solution to Question 9:
- Factor numerator:
\[
\dfrac{1 - \sin^4 x}{1 + \sin^2 x} = \dfrac{(1 - \sin^2 x)(1 + \sin^2 x)}{1 + \sin^2 x} \] and simoplify \[ = 1 - \sin^2 x = \cos^2 x
\]
Question 10:
Add and simplify:
\( \dfrac{1}{1 + \sin x} + \dfrac{1}{1 - \sin x} \)
Solution to Question 10:
- Use a common denominator:
\[
\dfrac{1}{1 + \sin x} + \dfrac{1}{1 - \sin x} = \dfrac{1 - \sin x + 1 + \sin x}{(1 + \sin x)(1 - \sin x)} \] and simplify \[ = \dfrac{2}{1 - \sin^2 x} = \dfrac{2}{\cos^2 x} = 2 \sec^2 x
\]
Question 11:
Simplify:
\(\cos x - \cos x \sin^2 x\)
Solution to Question 11:
- Factor \(\cos x\):
\[
\cos x - \cos x \sin^2 x = \cos x (1 - \sin^2 x) = \cos^3 x
\]
Question 12:
Simplify:
\(\tan^2 x \cos^2 x + \cot^2 x \sin^2 x\)
Solution to Question 12:
- Use \(\tan x = \dfrac{\sin x}{\cos x}\) and \(\cot x = \dfrac{\cos x}{\sin x}\):
\[
\tan^2 x \cos^2 x + \cot^2 x \sin^2 x = \left(\dfrac{\sin x}{\cos x}\right)^2 \cos^2 x + \left(\dfrac{\cos x}{\sin x}\right)^2 \sin^2 x
\]
- Simplify:
\[
\sin^2 x + \cos^2 x = 1
\]
Question 13:
Simplify:
\(\sec\left(\dfrac{\pi}{2} - x\right) - \tan\left(\dfrac{\pi}{2} - x\right) \sin\left(\dfrac{\pi}{2} - x\right)\)
Solution to Question 13:
- Use the identities \(\sec\left(\dfrac{\pi}{2} - x\right) = \csc x\), \(\tan\left(\dfrac{\pi}{2} - x\right) = \cot x\), and \(\sin\left(\dfrac{\pi}{2} - x\right) = \cos x\):
\[
\sec\left(\dfrac{\pi}{2} - x\right) - \tan\left(\dfrac{\pi}{2} - x\right) \sin\left(\dfrac{\pi}{2} - x\right) = \csc x - \cot x \cos x
\]
\[
= \dfrac{1}{\sin x} - \dfrac{\cos^2 x}{\sin x} = \dfrac{1 - \cos^2 x}{\sin x} = \dfrac{\sin^2 x}{\sin x} = \sin x
\]
More trigonometry questions with solutions and answers on this site.