
How do you evaluate the six trigonometric functions given $\theta =0$?
Answer
493.2k+ views
Hint: We first express all the six trigonometric functions. We divide them in primary ratios and their inverse ratios. We also find all possible relations between those ratios. Then we take the angle values of ${{0}^{\circ }}$ for all the six trigonometric functions.
Complete step-by-step solution:
We first complete the list of all the six trigonometric functions.
The main three trigonometric ratio functions are $\sin \theta ,\cos \theta ,\tan \theta $. The inverse of these three functions is $\csc \theta ,\sec \theta ,\cot \theta $. Also, we can express $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$.
Therefore, the relations are $\csc \theta =\dfrac{1}{\sin \theta },\sec \theta =\dfrac{1}{\cos \theta },\cot \theta =\dfrac{1}{\tan \theta }$.
We can also express these ratios with respect to a specific angle $\theta $ of a right-angle triangle and use the sides of that triangle to find the value of the ratio.
A right-angle triangle has three sides and they are base, height, hypotenuse. We express the ratios in $\sin \theta =\dfrac{\text{height}}{\text{hypotenuse}},\cos \theta =\dfrac{\text{base}}{\text{hypotenuse}},\tan \theta =\dfrac{\text{height}}{\text{base}}$.
Similarly, $\csc \theta =\dfrac{\text{hypotenuse}}{\text{height}},\sec \theta =\dfrac{\text{hypotenuse}}{\text{base}},\cot \theta =\dfrac{\text{base}}{\text{height}}$.
Now we express the values of these ratios for the conventional angles of ${{0}^{\circ }}$.
Note: We need to remember that in mathematics, the trigonometric functions are real functions which relate an angle of a right-angled triangle to ratios of two side lengths. They are widely used in all sciences that are related to geometry, such as navigation, solid mechanics, celestial mechanics, geodesy, and many others.
Complete step-by-step solution:
We first complete the list of all the six trigonometric functions.
The main three trigonometric ratio functions are $\sin \theta ,\cos \theta ,\tan \theta $. The inverse of these three functions is $\csc \theta ,\sec \theta ,\cot \theta $. Also, we can express $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$.
Therefore, the relations are $\csc \theta =\dfrac{1}{\sin \theta },\sec \theta =\dfrac{1}{\cos \theta },\cot \theta =\dfrac{1}{\tan \theta }$.
We can also express these ratios with respect to a specific angle $\theta $ of a right-angle triangle and use the sides of that triangle to find the value of the ratio.
A right-angle triangle has three sides and they are base, height, hypotenuse. We express the ratios in $\sin \theta =\dfrac{\text{height}}{\text{hypotenuse}},\cos \theta =\dfrac{\text{base}}{\text{hypotenuse}},\tan \theta =\dfrac{\text{height}}{\text{base}}$.
Similarly, $\csc \theta =\dfrac{\text{hypotenuse}}{\text{height}},\sec \theta =\dfrac{\text{hypotenuse}}{\text{base}},\cot \theta =\dfrac{\text{base}}{\text{height}}$.
Now we express the values of these ratios for the conventional angles of ${{0}^{\circ }}$.
Ratios | angles (in degree) | values |
$\sin \theta $ | 0 | 0 |
$\cos \theta $ | 0 | 1 |
$\tan \theta $ | 0 | 0 |
$\csc \theta $ | 0 | undefined |
$\sec \theta $ | 0 | 1 |
$\cot \theta $ | 0 | undefined |
Note: We need to remember that in mathematics, the trigonometric functions are real functions which relate an angle of a right-angled triangle to ratios of two side lengths. They are widely used in all sciences that are related to geometry, such as navigation, solid mechanics, celestial mechanics, geodesy, and many others.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Trending doubts
1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Explain zero factorial class 11 maths CBSE
