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Solving basic trigonometry equations

November 13, 2010 Leave a comment

Let’s consider the sine function

Sine curve

As you can see, it gives positive values for 0^o<x<180^o and negative for values for 180^o<x<360^o

Now let’s consider the cosine function

cosine curve

Here you can see that is positive for values 0^o<x<90^o\mbox{ and } 270^o<x<360^o and negative for values 90^o<x<270^o

Finally, let’s consider the tangent function.

Tangent curve

From the graph or by dividing the sine values by the cosine values you can see that it is positive for 0^o<x<90^o\mbox{ and } 180^o<x<270^o and negative for 90^o<x<180^o\mbox{ and } 270^o<x<360^o

The Trigonometry Circle

The information above can be summarised with the diagram below

Wheel of fortune

  1. In the first quadrant ‘A’, all the ratios give a positive value
  2. In the second quadrant ‘S’, only sine gives a positive value
  3. In the third quadrant ‘T’, only tangent gives a positive value
  4. In the forth quadrant ‘C’, only cosine gives a positive value

Solving a simple problem

  1. rearrange to ratio(angle) = value
  2. solve for positive value to give the angle in each quadrant
  3. Pick quadrants dependent on whether value was positive or negative
  4. measure round to each angle in an anti-clockwise direction from the positive x axis

Example

Solve sin x+1=0.3\mbox{ for } 0^o\leq x\leq 360^o

Step 1 \implies sin x=-0.7

Step 2 x=sin^{-1}(0.7)=44.4^o (Remember we use positive 0.7, because this will give us the angle in the first quadrant)

Step 3 Since -0.7 is negative, we want the third and forth quadrants (Sine is positive in the ‘ALL’ and ‘SINE’ quadrants)

Step 4 Solutions are x=180^o+44.4^o=224.4^o\mbox{ and } x=360^o-44.4^o=315.6^o


Solve cos^3 x + 3=3.4\mbox{ for } 0^o\leq x\leq 360^o

Step 1 cos^3 x= 0.4

\implies cos x =(0.4)^{\frac{1}{3}}=0.737

Step 2 x=cos^{-1}(0.737)=42.5^o

Step 3 Since 0.737 is positive, we need the first and forth quadrants (Cosine is positive in the ‘All’ and ‘COSINE’ quadrants)

Step 4 Solutions are x=42.5^o\mbox{ and } x=360^o-42.5^o=317.5^o


Solve Solve tan x + 1=5.1\mbox{ for } 0\leq x\leq 2\piStep 1 tan x=4.1Step 2 x=tan^{-1}(4.1)=1.33

Step 3 Since 4.1 is positive we require the first and third quadrants (Tangent is positive in the ‘All’ and ‘TANGENT’ quadrants)

Step 4 Solutions are x=1.33\mbox{ and } x=\pi + 1.33=4.47

Multiple Angles

When dealing with problems involving multiple angles we have to take into account the scaling that we will have to do at the end.  This is done by changing the range of values that we will consider.  For example if we go from 0 to 360 degrees for x we will go from 0 to 720 degrees  (twice round) for 2x.

Examples

Solve sin(2x)=0.9\mbox{ for } 0^o\leq x\leq 360^o\implies 0^o\leq 2x\leq 720^o

Step 2 2x=sin^{-1}(0.9)=64.2^o

Step 3 Since 0.9 is positive, we want the first and second quadrants (Sine is positive in the ‘ALL’ and ‘SINE’ quadrants). But we must go round twice.

Step 4 So 2x=64.2^o, 2x=180^o-64.2^o=115.8^o, 2x=360+64.2^o=424.2^o\mbox{ and }2x=540^o-64.2^o=475.8^o

So dividing by 2 gives x=32.1^o, x=57.9^o, x=212.1^o\mbox{ and }x=237.9^o


Solve tan\left(\frac{x}{2}\right)=-0.5\mbox{ for } 0^o\leq x\leq 360^o\implies 0^o\leq \frac{x}{2}\leq 180^o 

Step 2 \frac{x}{2}=tan^{-1}(0.5)=26.6^o

Step 3 Since -0.5 is negative we want quadrants ‘2’ and ‘4’.  However, we are only going up to 180 degrees so we only want the second quadrant.

Step 4 So \frac{x}{2}=180^o-26.6^o=153.4^o\implies x=306.9^o

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Categories: Trigonometry