Pythagorean and Reciprocated Trigonometric Equations
Neil Trivedi
Teacher
Pythagorean and Reciprocated Trigonometric Equations
There isn’t much difference between solving simple trigonometric functions and solving with Pythagorean identities and reciprocated trigonometric functions. The only additional thing we need to do is apply Pythagorean identities and re-write the reciprocal functions while rearranging the equations.
There are two common aims:
• Convert all trigonometric functions into a single type of trigonometric function. The function you choose to convert to largely depends on the given equation.
• Obtain an equation, which we can then factorise or apply the quadratic formula to find its factors and then solve for our given variable.
Identities:
Reciprocal Functions:
Example 1:
Solve for ,
Give your answers to significant figures when necessary.
Step 1: Apply the Pythagorean identities to express all the terms in terms of because there is a term that cannot be changed using Pythagorean identities.
Replace with in our equation,
Expand the RHS,
Move all terms on the LHS to the RHS,
Step 2: Let , then solve the quadratic equation by factorising.
Therefore, using the fact that
Step 3: Solve for in each equation.
Trigonometric Functions | Finding the SV |
| PV |
| PV |
| PV |
Find the PV.
(PV)
Find the SV by working out PV.
PV (SV)
Both the PV and SV lie within the range. We can stop here because if we add or subtract from both values, we’d get values that lie outside the range. Therefore,
Find the PV.
(PV)
Find the SV by working out PV.
PV (SV)
Both the PV and SV lie within the range. We can stop here because if we add or subtract from both values, we’d get values that lie outside the range. Therefore,
Therefore, our four solutions are and
No answer provided.
Example 2:
Solve for .
Step 1: Convert the reciprocated trigonometric function and rearrange the equation with Pythagorean identities.
We use the fact that .
Multiply every term by to get rid of the denominators,
Move all terms on the LHS to the RHS,
Factor out ,
Therefore, or .
Step 2: Solve each equation for .
Trigonometric Functions | Finding the SV |
| PV |
| PV |
| PV |
Find the PV.
(PV)
Find the SV by working out PV.
PV (SV)
Neither the PV nor the SV lie within the range so we can move on.
Find the PV.
(PV)
Find the SV by working out PV.
PV (SV)
Both the PV and the SV lie within the range.
Therefore, .
Therefore, the only solutions are .
No answer provided.
Example 3:
Solve .
Give your answers to significant figures.
Step 1: Rearrange the equation and convert the reciprocated trigonometric function.
Dividing both sides by 3.
Using the fact that ,
Reciprocate both sides,
Step 2: The current range is for . We need to modify the range to read . To do this, multiply everything by and then subtract .
Step 3: Solve the equation with the modified range.
Find the PV.
(PV)
Trigonometric Functions | Finding the SV |
| PV |
| PV |
| PV |
Find the SV by working out PV.
PV (SV)
Both the PV and SV are in the modified range. Take the PV and SV and add to both to find the rest of the values in the modified range.
PV
SV (out of range)
We have three values that lie in the modified range. To not lose accuracy, we store these values in our calculator.
Step 4: Un-modify the range to find . Do this by adding to all values, then dividing by .
No answer provided.
Challenging Question