Question

Prove the following:
\[\cos 4A=1-8{{\cos }^{2}}A+8{{\cos }^{4}}A\]

Answer 1

Hint: We can start solving by considering the LHS first. First of all, write cos (4A) as cos (2. 2A) and use the formula of \[\cos 2\theta =2{{\cos }^{2}}\theta -1\]. Again write cos (2A) as cos (2. A) and again use the formula \[\cos 2\theta =2{{\cos }^{2}}\theta -1\] and simplify it to get the required answer.

Complete step-by-step solution -
In this question, we have to prove that
\[\cos 4A=1-8{{\cos }^{2}}A+8{{\cos }^{4}}A\]
Let us consider the LHS of the equation given in the question.
LHS = cos 4A
We can write 4A = 2(2A). By using this, we get,
LHS = cos [2(2A)]
We know that \[\cos 2\theta =2{{\cos }^{2}}\theta -1\]. So by taking \[\theta =2A\], we get,
\[LHS=2{{\cos }^{2}}\left( 2A \right)-1\]
Again we know that \[\cos 2\theta =2{{\cos }^{2}}\theta -1\]. So by taking \[\theta =A\], we get,
\[LHS=2{{\left[ 2{{\cos }^{2}}A-1 \right]}^{2}}-1\]
We know that \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\]. By using this in the above equation, we get,
\[LHS=2\left[ {{\left( 2{{\cos }^{2}}A \right)}^{2}}+{{\left( 1 \right)}^{2}}-2\times \left( 2{{\cos }^{2}}A \right)\times 1 \right]-1\]
\[LHS=2\left[ 4{{\cos }^{4}}A+1-4{{\cos }^{2}}A \right]-1\]
\[LHS=8{{\cos }^{4}}A+2-8{{\cos }^{2}}A-1\]
\[LHS=1-8{{\cos }^{2}}A+8{{\cos }^{4}}A\]
So, we get, LHS = RHS
Hence proved.
Therefore, we have proved that
\[\cos 4A=1-8{{\cos }^{2}}A+8{{\cos }^{4}}A\].

Note: In this type of question, where angles are given in multiples of 2 and are reduced to smaller angles, then students must use this approach of converting angles into smaller angles by using half-angle formulas like \[\sin 2\theta =2\sin \theta \cos \theta \text{ and }\cos 2\theta =2{{\cos }^{2}}\theta -1={{\cos }^{2}}\theta -{{\sin }^{2}}\theta =1-2{{\sin }^{2}}\theta \]. Students must remember the different transformations of formulas to prove the respective results accordingly. Also, in this question, students can write cos (4A) as cos (2A + 2A) and use formula cos (A + B) = cos A cos B – sin A sin B twice and use \[{{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta \] to solve this question.