Question
Question: Show that the given trigonometric equation holds true \(\dfrac{{\cos 2A + 2\cos 4A + \cos 6A}}{{\c...
Show that the given trigonometric equation holds true
cosA+2cos3A+cos5Acos2A+2cos4A+cos6A=cosA−sinA.tan3A
Solution
Hint – In this question consider the L.H.S part and apply the trigonometric identitycosC+cosD=2cos(2C+D)cos(2C−D), and do similar simplification in the denominator part. This will help getting to the desired R.H.S.
Complete step-by-step answer:
Proof –
Consider L.H.S
⇒cosA+2cos3A+cos5Acos2A+2cos4A+cos6A
Now rearrange the terms we have,
⇒cos5A+cosA+2cos3Acos6A+cos2A+2cos4A
Now as we know that cosC+cosD=2cos(2C+D)cos(2C−D) so use this property we have,
⇒2cos(25A+A)cos(25A−A)+2cos3A2cos(26A+2A)cos(26A−2A)+2cos4A
Now simplify the above equation we have,
⇒2cos3Acos2A+2cos3A2cos4Acos2A+2cos4A
Now take 2 cos 4A from numerator and 2cos 3A from denominator we have,
⇒2cos3A(cos2A+1)2cos4A(cos2A+1)
⇒cos3Acos4A
Now above equation is also written as
⇒cos3Acos(3A+A)
Now as we know that cos(A+B)=cosAcosB−sinAsinB so use this property we have,
⇒cos3Acos3AcosA−sin3AsinA
⇒cosA−sinA.cos3Asin3A
Now as we know that (sin/cos) = tan so use this property we have,
⇒cosA−sinA.tan3A
= R.H.S
Hence proved.
Note – Tricky point while solving such trigonometric questions is simply to understand which trigonometric identity is to be used, some of them are being mentioned above. Here we have taken out the common part in the numerator separately and the same is being done to the denominator part so as to cancel out terms. It is advised to grasp trigonometric identities, as it helps to save a lot of time.