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Question: Identify transformations of trigonometric expressions prove the following identities \(\dfrac{{{{\si...

Identify transformations of trigonometric expressions prove the following identities sin⁡22a+4sin⁡4a−4sin⁡2acos⁡2a4−sin⁡22a−4sin⁡2a=tan⁡4a\dfrac{{{{\sin }^2}2a + 4{{\sin }^4}a - 4{{\sin }^2}a{{\cos }^2}a}}{{4 - {{\sin }^2}2a - 4{{\sin }^2}a}} = {\tan ^4}a

Explanation

Solution

In the given question the RHS is given in tan⁡4a{\tan ^4}a so we have to try to convert the LHS part into tan⁡4a{\tan ^4}a or sin⁡4acos⁡4a\dfrac{{{{\sin }^4}a}}{{{{\cos }^4}a}}. For this apply sin⁡2a=2sin⁡acos⁡a\sin 2a = 2\sin a\cos a or sin⁡22a=4sin⁡2acos⁡2a{\sin ^2}2a = 4{\sin ^2}a{\cos ^2}a try to change 2a2a angles in aa . After that apply sin⁡2a+cos⁡2a=1{\sin ^2}a + {\cos ^2}a = 1 to proceed the result .

Complete step-by-step answer:
As in the RHS it is given that tan⁡4a{\tan ^4}a so we have to try to convert the LHS part into tan⁡4a{\tan ^4}a or sin⁡4acos⁡4a\dfrac{{{{\sin }^4}a}}{{{{\cos }^4}a}} .
For this we have to apply trigonometric transformation in the LHS part .
From LHS
sin⁡22a+4sin⁡4a−4sin⁡2acos⁡2a4−sin⁡22a−4sin⁡2a\dfrac{{{{\sin }^2}2a + 4{{\sin }^4}a - 4{{\sin }^2}a{{\cos }^2}a}}{{4 - {{\sin }^2}2a - 4{{\sin }^2}a}}
So we know that sin⁡2a=2sin⁡acos⁡a\sin 2a = 2\sin a\cos a , apply this is in the numerator part , hence we get ,(2sin⁡acos⁡a)2+4sin⁡4a−4sin⁡2acos⁡2a4−sin⁡22a−4sin⁡2a\dfrac{{{{\left( {2\sin a\cos a} \right)}^2} + 4{{\sin }^4}a - 4{{\sin }^2}a{{\cos }^2}a}}{{4 - {{\sin }^2}2a - 4{{\sin }^2}a}}
4sin⁡2acos⁡2a+4sin⁡4a−4sin⁡2acos⁡2a4−sin⁡22a−4sin⁡2a\dfrac{{4{{\sin }^2}a{{\cos }^2}a + 4{{\sin }^4}a - 4{{\sin }^2}a{{\cos }^2}a}}{{4 - {{\sin }^2}2a - 4{{\sin }^2}a}}
Now 4sin⁡2acos⁡2a4{\sin ^2}a{\cos ^2}a will cancel out ,
4sin⁡4a4−sin⁡22a−4sin⁡2a\dfrac{{4{{\sin }^4}a}}{{4 - {{\sin }^2}2a - 4{{\sin }^2}a}}
4sin⁡4a4(1−sin⁡2a)−sin⁡22a\dfrac{{4{{\sin }^4}a}}{{4(1 - {{\sin }^2}a) - {{\sin }^2}2a}}
Now in denominator , we know that sin⁡2a+cos⁡2a=1{\sin ^2}a + {\cos ^2}a = 1 or cos⁡2a=1−sin⁡2a{\cos ^2}a = 1 - {\sin ^2}a apply this in the denominator
4sin⁡4a4cos⁡2a−sin⁡22a\dfrac{{4{{\sin }^4}a}}{{4{{\cos }^2}a - {{\sin }^2}2a}}
Now we know that sin⁡2a=2sin⁡acos⁡a\sin 2a = 2\sin a\cos a and on squaring sin⁡22a=4sin⁡2acos⁡2a{\sin ^2}2a = 4{\sin ^2}a{\cos ^2}a
4sin⁡4a4cos⁡2a−4sin⁡2acos⁡2a\dfrac{{4{{\sin }^4}a}}{{4{{\cos }^2}a - 4{{\sin }^2}a{{\cos }^2}a}}

Take 4cos⁡2a4{\cos ^2}a common in the denominator ,
4sin⁡4a4cos⁡2a(1−sin⁡2a)\dfrac{{4{{\sin }^4}a}}{{4{{\cos }^2}a(1 - {{\sin }^2}a)}}
4sin⁡4a4cos⁡2a.cos⁡2a\dfrac{{4{{\sin }^4}a}}{{4{{\cos }^2}a.{{\cos }^2}a}}

4sin⁡4a4cos⁡4a=tan⁡4a\dfrac{{4{{\sin }^4}a}}{{4{{\cos }^4}a}} = {\tan ^4}a
= RHS
Proved

Note: Students should remember trigonometric formulas, identities and transformation formulas for solving these types of problems.For these types of problems first we have to approach the solution by analysing the R.H.S of the given expression, Use the suitable identities , formula and simplify the L.H.S part to prove the given expression.