
Write \[\sin 4x\] in terms of \[\sin x\].
Answer
530.7k+ views
Hint: In the given question \[\sin 4x\] needs to be written in terms of \[\sin x\]. So we have to use a multiple angle formula. Use the formula \[\sin 2a\] considering \[a = 2x\], and then convert the “\[\cos \]” terms to “\[\sin \]” terms using the identity \[{\sin ^2}x + {\cos ^2}x = 1\].
Complete step by step solution:
Given expression is \[\sin 4x\], this can be rewritten as \[\sin 2\left( {2x} \right)\].
Let \[2x = a\].
\[\therefore \sin 4x = \sin 2(a)\]
Use the formula \[\sin 2a = 2\sin a \cdot \cos a\]
\[\therefore \sin 2\left( {2x} \right) = 2\sin 2x \cdot \cos 2x\]
Again use the formula \[\sin 2a = 2\sin a \cdot \cos a\]:
\[\therefore \sin 2\left( {2x} \right) = 2 \cdot \left( {2\sin x \cdot \cos x} \right) \cdot \left( {\cos 2x} \right)\]
Use from the formula : \[\cos 2x = {\cos ^2}x - {\sin ^2}x\],
\[ \Rightarrow \cos 2x = 1 - 2{\sin ^2}x\]…..(1) \[[\because {\cos ^2}x = 1 - {\sin ^2}x]\]
Substitute the value of \[\cos 2x\] from (1):
\[\therefore \sin 2\left( {2x} \right) = 2 \cdot \left( {2\sin x \cdot \cos x} \right) \cdot \left( {1 - 2{{\sin }^2}x} \right)\]
\[ \Rightarrow \sin 2(2x) = 4\sin x\cos x - 8{\sin ^3}x\cos x\]
Take \[\cos x\] common:
\[ \Rightarrow \sin 2(2x) = \left( {4\sin x - 8{{\sin }^3}x} \right)\cos x\]
Again, \[{\cos ^2}x = 1 - {\sin ^2}x\]
\[ \Rightarrow \cos x = \sqrt {1 - {{\sin }^2}x} \]
Substitute the value of \[\cos x\]:
\[ \Rightarrow \sin 2(2x) = \left( {4\sin x - 8{{\sin }^3}x} \right)\left( {\sqrt {1 - {{\sin }^2}x} } \right)\]
Hence, \[\sin 4x = \left( {4\sin x - 8{{\sin }^3}x} \right)\left( {\sqrt {1 - {{\sin }^2}x} } \right)\]
Note:
Students must memorise the following identities used in this solution:
\[\sin 2A = 2\sin A\cos A\]
\[\cos 2A = {\cos ^2}A - {\sin ^2}A\]
\[{\sin ^2}A + {\cos ^2}A = 1\]
Some other trigonometric identities and formulas of multiple and sub multiple angles, sums and products, must also be memorized.
While converting angles is the quadrant of the angle is mentioned then the signs of the trigonometric ratios in different quadrants must also be taken care of. For that remember the following:
${1^{st}}$ quadrant: All trigonometric functions are positive.
${2^{nd}}$ quadrant: Sine functions are positive.
${3^{rd}}$ quadrant: Tan functions are positive.
${4^{th}}$ quadrant: Cos functions are positive.
Complete step by step solution:
Given expression is \[\sin 4x\], this can be rewritten as \[\sin 2\left( {2x} \right)\].
Let \[2x = a\].
\[\therefore \sin 4x = \sin 2(a)\]
Use the formula \[\sin 2a = 2\sin a \cdot \cos a\]
\[\therefore \sin 2\left( {2x} \right) = 2\sin 2x \cdot \cos 2x\]
Again use the formula \[\sin 2a = 2\sin a \cdot \cos a\]:
\[\therefore \sin 2\left( {2x} \right) = 2 \cdot \left( {2\sin x \cdot \cos x} \right) \cdot \left( {\cos 2x} \right)\]
Use from the formula : \[\cos 2x = {\cos ^2}x - {\sin ^2}x\],
\[ \Rightarrow \cos 2x = 1 - 2{\sin ^2}x\]…..(1) \[[\because {\cos ^2}x = 1 - {\sin ^2}x]\]
Substitute the value of \[\cos 2x\] from (1):
\[\therefore \sin 2\left( {2x} \right) = 2 \cdot \left( {2\sin x \cdot \cos x} \right) \cdot \left( {1 - 2{{\sin }^2}x} \right)\]
\[ \Rightarrow \sin 2(2x) = 4\sin x\cos x - 8{\sin ^3}x\cos x\]
Take \[\cos x\] common:
\[ \Rightarrow \sin 2(2x) = \left( {4\sin x - 8{{\sin }^3}x} \right)\cos x\]
Again, \[{\cos ^2}x = 1 - {\sin ^2}x\]
\[ \Rightarrow \cos x = \sqrt {1 - {{\sin }^2}x} \]
Substitute the value of \[\cos x\]:
\[ \Rightarrow \sin 2(2x) = \left( {4\sin x - 8{{\sin }^3}x} \right)\left( {\sqrt {1 - {{\sin }^2}x} } \right)\]
Hence, \[\sin 4x = \left( {4\sin x - 8{{\sin }^3}x} \right)\left( {\sqrt {1 - {{\sin }^2}x} } \right)\]
Note:
Students must memorise the following identities used in this solution:
\[\sin 2A = 2\sin A\cos A\]
\[\cos 2A = {\cos ^2}A - {\sin ^2}A\]
\[{\sin ^2}A + {\cos ^2}A = 1\]
Some other trigonometric identities and formulas of multiple and sub multiple angles, sums and products, must also be memorized.
While converting angles is the quadrant of the angle is mentioned then the signs of the trigonometric ratios in different quadrants must also be taken care of. For that remember the following:
${1^{st}}$ quadrant: All trigonometric functions are positive.
${2^{nd}}$ quadrant: Sine functions are positive.
${3^{rd}}$ quadrant: Tan functions are positive.
${4^{th}}$ quadrant: Cos functions are positive.
Recently Updated Pages
Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Master Class 9 English: Engaging Questions & Answers for Success

Master Class 9 Science: Engaging Questions & Answers for Success

Class 9 Question and Answer - Your Ultimate Solutions Guide

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

Master Class 12 Economics: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

Which animal has three hearts class 11 biology CBSE

10 examples of friction in our daily life

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

Difference Between Prokaryotic Cells and Eukaryotic Cells

