
What is the value of given trigonometric expression \[\sin {{15}^{\circ }}\]?
(a) \[\dfrac{\sqrt{3}-1}{2\sqrt{2}}\]
(b) \[\dfrac{\sqrt{3}+1}{2\sqrt{2}}\]
(c) \[\dfrac{\sqrt{3}-1}{\sqrt{3}+1}\]
(d) \[\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\]
Answer
603.6k+ views
Hint: Use the trigonometric identity \[\sin \left( x-y \right)=\sin x\cos y-\cos x\sin y\]. Substitute \[x={{45}^{\circ }},y={{30}^{\circ }}\] in the above formula and use the values \[\sin {{45}^{\circ }}=\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}},\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2},\sin {{30}^{\circ }}=\dfrac{1}{2}\] to calculate the value of \[\sin {{15}^{\circ }}\].
Complete step-by-step answer:
We have to calculate the value of \[\sin {{15}^{\circ }}\].
We will use the trigonometric identity \[\sin \left( x-y \right)=\sin x\cos y-\cos x\sin y\].
Substituting \[x={{45}^{\circ }},y={{30}^{\circ }}\] in the above equation, we have \[\sin \left( {{45}^{\circ }}-{{30}^{\circ }} \right)=\sin {{15}^{\circ }}=\sin {{45}^{\circ }}\cos {{30}^{\circ }}-\cos {{45}^{\circ }}\sin {{30}^{\circ }}\]. We can also substitute \[x={{60}^{\circ }},y={{45}^{\circ }}\] in the trigonometric equation. In fact, we can substitute all those x and y whose values are known to us and \[x-y={{15}^{\circ }}\].
We know the values \[\sin {{45}^{\circ }}=\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}},\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2},\sin {{30}^{\circ }}=\dfrac{1}{2}\].
Thus, we have \[\sin {{15}^{\circ }}=\dfrac{1}{\sqrt{2}}\times \dfrac{\sqrt{3}}{2}-\dfrac{1}{\sqrt{2}}\times \dfrac{1}{2}\].
Rearranging the terms, we have \[\sin \left( {{15}^{\circ }} \right)=\dfrac{\sqrt{3}-1}{2\sqrt{2}}\].
Hence, the value of \[\sin \left( {{15}^{\circ }} \right)\] is \[\dfrac{\sqrt{3}-1}{2\sqrt{2}}\], which is option A.
Trigonometric functions are real functions which relate any angle of a right angled triangle to the ratios of any two of its sides. The widely used trigonometric functions are sine, cosine and tangent. However, we can also use their reciprocals, i.e., cosecant, secant and cotangent. We can use geometric definitions to express the value of these functions on various angles using unit circle (circle with radius 1). We also write these trigonometric functions as infinite series or as solutions to differential equations. Thus, allowing us to expand the domain of these functions from the real line to the complex plane. One should be careful while using the trigonometric identities and rearranging the terms to convert from one trigonometric function to the other one.
Note: We can also solve this question by using the trigonometric identity \[\cos 2x=1-2{{\sin }^{2}}x\]. Substitute \[x={{15}^{\circ }}\] in the above equation and use \[\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}\] to simplify the above equation. To find the value of \[\sin \left( {{15}^{\circ }} \right)\], we must also use the identity \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\] and \[{{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)\].
Complete step-by-step answer:
We have to calculate the value of \[\sin {{15}^{\circ }}\].
We will use the trigonometric identity \[\sin \left( x-y \right)=\sin x\cos y-\cos x\sin y\].
Substituting \[x={{45}^{\circ }},y={{30}^{\circ }}\] in the above equation, we have \[\sin \left( {{45}^{\circ }}-{{30}^{\circ }} \right)=\sin {{15}^{\circ }}=\sin {{45}^{\circ }}\cos {{30}^{\circ }}-\cos {{45}^{\circ }}\sin {{30}^{\circ }}\]. We can also substitute \[x={{60}^{\circ }},y={{45}^{\circ }}\] in the trigonometric equation. In fact, we can substitute all those x and y whose values are known to us and \[x-y={{15}^{\circ }}\].
We know the values \[\sin {{45}^{\circ }}=\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}},\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2},\sin {{30}^{\circ }}=\dfrac{1}{2}\].
Thus, we have \[\sin {{15}^{\circ }}=\dfrac{1}{\sqrt{2}}\times \dfrac{\sqrt{3}}{2}-\dfrac{1}{\sqrt{2}}\times \dfrac{1}{2}\].
Rearranging the terms, we have \[\sin \left( {{15}^{\circ }} \right)=\dfrac{\sqrt{3}-1}{2\sqrt{2}}\].
Hence, the value of \[\sin \left( {{15}^{\circ }} \right)\] is \[\dfrac{\sqrt{3}-1}{2\sqrt{2}}\], which is option A.
Trigonometric functions are real functions which relate any angle of a right angled triangle to the ratios of any two of its sides. The widely used trigonometric functions are sine, cosine and tangent. However, we can also use their reciprocals, i.e., cosecant, secant and cotangent. We can use geometric definitions to express the value of these functions on various angles using unit circle (circle with radius 1). We also write these trigonometric functions as infinite series or as solutions to differential equations. Thus, allowing us to expand the domain of these functions from the real line to the complex plane. One should be careful while using the trigonometric identities and rearranging the terms to convert from one trigonometric function to the other one.
Note: We can also solve this question by using the trigonometric identity \[\cos 2x=1-2{{\sin }^{2}}x\]. Substitute \[x={{15}^{\circ }}\] in the above equation and use \[\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}\] to simplify the above equation. To find the value of \[\sin \left( {{15}^{\circ }} \right)\], we must also use the identity \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\] and \[{{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)\].
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Class 12 Question and Answer - Your Ultimate Solutions Guide

Master Class 12 Economics: Engaging Questions & Answers for Success

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

How much time does it take to bleed after eating p class 12 biology CBSE

Explain sex determination in humans with line diag class 12 biology CBSE

When was the first election held in India a 194748 class 12 sst CBSE

