
Write \[\sin 4x\] in terms of \[\sin x\].
Answer
538.8k+ 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
The number of solutions in x in 02pi for which sqrt class 12 maths CBSE

Write any two methods of preparation of phenol Give class 12 chemistry CBSE

Differentiate between action potential and resting class 12 biology CBSE

Two plane mirrors arranged at right angles to each class 12 physics CBSE

Which of the following molecules is are chiral A I class 12 chemistry CBSE

Name different types of neurons and give one function class 12 biology CBSE

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

Explain zero factorial class 11 maths CBSE

What is 1s 2s 2p 3s 3p class 11 chemistry CBSE

Discuss the various forms of bacteria class 11 biology CBSE

State the laws of reflection of light

An example of chemosynthetic bacteria is A E coli B class 11 biology CBSE

