
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
519.6k+ 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 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Trending doubts
The probability that a leap year will have only 52 class 12 maths CBSE

Describe the poetic devices used in the poem Aunt Jennifers class 12 english CBSE

And such too is the grandeur of the dooms We have imagined class 12 english CBSE

What does the god that failed refer to class 12 english CBSE

Which country did Danny Casey play for class 12 english CBSE

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