The Trigonometric Identities
Solving Trigonometric Identities, is one of the most complex
to deal in Trigonometry. As shown in our presentation;
Ex: Solve for x, if tan3x = 5 tanx
Solving for x, we must consider the problem as equation tan3x = 5 tanx – eq. 1
And tan3x = tan(2x + x)
How to get equation 2? Since 2x+x = 3x Therefore, 3x is separated to get tan3x.
tan2x + tanx
Then, tan3x = tan(2x +x) = ______________ eq. 2
1 – tan2x tanx
From Sum of Angles formula tan (A+B) = tanA + tanB
1- tanA tanB
now,
substitute equation 1 and 2
we have, tan2x + tanx = 5 tanx
1 – tan2x tanx
transpose 1 – tan2x tanx right side,
we have, tan2x + tanx = 5 tanx (1 – tan2xtanx)
distribute 5 tanx = 5 tanx – 5
therefore, tan2x + tanx = 5 tanx – 5 tan2xtan2x
subtract tanx to 5 tanx , tanx – 5 tanx = 4 tanx
hence, tan2x = 4 tanx – 5
note, tan2x have in common therefore, 4 tanx = tan2x (1 + 5 tan2x) - eq.3
from the double angles formula, tan2A = 2 tan A
1 – tan2A
therefore, tan2x = 2 tanx
1 – tan2x - eq. 4
substitute equation 3 and 4
2 tanx (1 + 5 tan2x) = 4 tanx
1- tan2x
transpose 1 – tan2x to the right side of 4 tanx
we have, 2 tanx (1 + 5 tan2x) = 4 tanx (1- tan2x) by using distributive law,
2 tanx + 10 tan3x = 4 tanx – 4 tan3x subtract both of the common terms
2 tanx - 4 tanx = 2 tanx
+10 tan3x – (-4 tan3x) = 14 tan3x
we have, 14 tan3x = 2 tanx
divide 14 tanx to 2 tanx, get 0.142857
hence, tan2x = 0.142857
tanx = √0.142857
tanx = 0.3779642
x = 20.705˚