# Solving Trigonometric Equations

## Summary

To solve trigonometric equations, several identities and formulas are used i.e:

• $tan\theta \quad =\quad \frac { sin\theta }{ cos\theta }$
• ${ sin }^{ 2 }\theta \quad +\quad { cos }^{ 2 }\theta \quad =\quad 1$
• ${ cosec }^{ 2 }\theta \quad =\quad 1\quad +\quad { cot }^{ 2 }\theta$
• ${ sec }^{ 2 }\theta \quad =\quad 1\quad +\quad { tan }^{ 2 }\theta$
• $sin2x\quad =\quad 2sin(x)cos(x)$
• $cos2x\quad =\quad { 2cos }^{ 2 }x\quad -\quad 1\quad =\quad { 1\quad -\quad 2sin }^{ 2 }x\quad$

For basic trigonometric equations, we follow the following steps to solve them:
1.    Make sine, cosine or tangent the subject.
2.    Use any method including a calculator to find basic angles.
3.    Using quadrants, find all solutions in the given range.

#### Example #1

Q. Solve the equation $cos\theta \quad =\quad \frac { 1 }{ 2 }$  for the range $0\quad \le \quad \theta \quad \le \quad 2\pi$

Solution:

Using a calculator we can find that the basic angle $\theta \quad =\quad \frac { \pi }{ 3 }$

Now we know that cos is positive in the first quadrant and in the fourth quadrant hence another value of $\theta$ would be: $\theta \quad =\quad 2\pi \quad -\quad \frac { \pi }{ 3 } \quad =\quad \frac { 5\pi }{ 3 }$

Ans: $\theta \quad =\quad \frac { \pi }{ 3 } \quad \quad \quad \quad and\quad \quad \quad \theta \quad =\quad \frac { 5\pi }{ 3 }$

#### Trigonometric Identities

Refer to Fig 1. We know that: $sin\theta \quad =\quad \frac { y }{ r } \quad \quad \Rightarrow \quad 1\quad$ $cos\theta \quad =\quad \frac { x }{ r } \quad \quad \Rightarrow \quad 2$ $tan\theta \quad =\quad \frac { y }{ x } \quad \quad \Rightarrow \quad 3$

Dividing 1 by 2 we get: $tan\theta \quad =\quad \frac { sin\theta }{ cos\theta }$

This is called an ”Identity”.

Using pythagoras theorem: ${ x }^{ 2 }\quad +\quad { y }^{ 2 }\quad =\quad { r }^{ 2 }$

Dividing the whole equation by ${ r }^{ 2 }$: ${ (\frac { x }{ r } ) }^{ 2 }\quad +\quad { (\frac { y }{ r } ) }^{ 2 }\quad =\quad 1$

Substituting the values: ${ (cos\theta ) }^{ 2 }\quad +\quad { (sin\theta ) }^{ 2 }\quad =\quad 1$

We get another identity: ${ sin }^{ 2 }\theta \quad +\quad { cos }^{ 2 }\theta \quad \equiv \quad 1$

where $\equiv$  means ”equivalent to” or ”identical to”.

These identities are very useful when solving many trigonometric equations.

#### Example #2

Q. Prove the identity $tan^{ 2 }\theta \quad -\quad { sin }^{ 2 }\theta \quad =\quad \frac { { sin }^{ 4 }\theta }{ { cos }^{ 2 }\theta }$

Solution:

Taking left hand side only: $=\quad tan^{ 2 }\theta \quad -\quad { sin }^{ 2 }\theta$ $=\quad (\frac { sin\theta }{ cos\theta } )^{ 2 }\quad -\quad { sin }^{ 2 }\theta$ $=\quad \frac { { sin }^{ 2 }\theta }{ { cos }^{ 2 }\theta } \quad -\quad { sin }^{ 2 }\theta$

Take ${ sin }^{ 2 }\theta$  common: $=\quad { sin }^{ 2 }\theta (\frac { 1 }{ { cos }^{ 2 }\theta } \quad -\quad 1)$ $=\quad { sin }^{ 2 }\theta (\frac { 1\quad -\quad { cos }^{ 2 }\theta }{ { cos }^{ 2 }\theta } )$ $=\quad \frac { { sin }^{ 2 }\theta }{ { cos }^{ 2 }\theta } ({ sin }^{ 2 }\theta \quad +\quad { cos }^{ 2 }\theta \quad -\quad cos^{ 2 }\theta )$ $=\quad \frac { { sin }^{ 4 }\theta }{ { cos }^{ 2 }\theta }$     Hence proven!

Similarly, there are a few other formulas that help solve trigonometric equation:

• ${ cosec }^{ 2 }\theta \quad =\quad 1\quad +\quad { cot }^{ 2 }\theta$
• ${ sec }^{ 2 }\theta \quad =\quad 1\quad +\quad { tan }^{ 2 }\theta$
• $sin2x\quad =\quad 2sin(x)cos(x)$
• $cos2x\quad =\quad { 2cos }^{ 2 }x\quad -\quad 1\quad =\quad { 1\quad -\quad 2sin }^{ 2 }x\quad$