NZ Level 8 (NZC) Level 3 (NCEA) [In development]
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Pythagorean Identities
Lesson

The theorem of Pythagoras concerning right-angled triangles can be expressed by means of trigonometric ratios. Suppose a right-angled triangle has legs $a$a and $b$b, and hypotenuse $c$c. Let the angle opposite side $a$a be $\alpha$α

We know that $a^2+b^2=c^2$a2+b2=c2 by Pythagoras. If we now divide the equation through by $c^2$c2, we obtain the form $\frac{a^2}{c^2}+\frac{b^2}{c^2}=1$a2c2+b2c2=1

The fractions $\frac{a}{c}$ac and $\frac{b}{c}$bc are recognised as $\sin\alpha$sinα and $\cos\alpha$cosα respectively. Therefore, we can replace the fractions by the trigonometric ratios to obtain:

$\sin^2\alpha+\cos^2\alpha\equiv1$sin2α+cos2α1

The $\equiv$ sign is used because this is an identity, meaning it is true whatever the value of $\alpha$α.

To be convinced that the identity holds for angles of any magnitude, we can inspect the unit circle diagram and apply Pythagoras's theorem to the triangles that are formed. Note that although the values of $\sin\alpha$sinα and $\cos\alpha$cosα may be negative, their squares are not.

Now that we've obtained the fact that $\sin^2\alpha+\cos^2\alpha\equiv1$sin2α+cos2α1 we can divide both sides by $\sin^2\alpha$sin2α to obtain the following identity in terms of $\cot\alpha$cotα and $\csc\alpha$cscα:

$\frac{\sin^2\alpha+\cos^2\alpha}{\sin^2\alpha}\equiv\frac{1}{\sin^2\alpha}$sin2α+cos2αsin2α1sin2α

$\frac{\sin^2\alpha}{\sin^2\alpha}+\frac{\cos^2\alpha}{\sin^2\alpha}\equiv\frac{1}{\sin^2\alpha}$sin2αsin2α+cos2αsin2α1sin2α

$1+\cot^2\alpha\equiv\csc^2\alpha$1+cot2αcsc2α

If we repeat the above process, but now instead divide both sides by $\cos^2\alpha$cos2α, then we obtain the following identity in terms of $\tan\alpha$tanα and $\sec\alpha$secα:

$\frac{\sin^2\alpha+\cos^2\alpha}{\cos^2\alpha}\equiv\frac{1}{\cos^2\alpha}$sin2α+cos2αcos2α1cos2α

$\frac{\sin^2\alpha}{\cos^2\alpha}+\frac{\cos^2\alpha}{\cos^2\alpha}\equiv\frac{1}{\cos^2\alpha}$sin2αcos2α+cos2αcos2α1cos2α

$\tan^2\alpha+1\equiv\sec^2\alpha$tan2α+1sec2α

Example 1

Given that $\cot\theta=\frac{8}{15}$cotθ=815, find the value of $\sin\theta$sinθ

We have $\cot\theta=\frac{\cos\theta}{\sin\theta}=\frac{8}{15}$cotθ=cosθsinθ=815.

So, after multiplying by the denominators, we see that $8\sin\theta=15\cos\theta$8sinθ=15cosθ.

The Pythagorean identity allows us to relate the squares of the $\sin$sin and $\cos$cos ratios. So, we square both sides of this equation to obtain $64\sin^2\theta=225\cos^2\theta$64sin2θ=225cos2θ.  

Now, we make use of the fact that $\cos^2\theta\equiv1-\sin^2\theta$cos2θ1sin2θ, to write $64\sin^2\theta=225\left(1-\sin^2\theta\right)$64sin2θ=225(1sin2θ).

On rearranging, this is $289\sin^2\theta=225$289sin2θ=225 and so, $\sin^2\theta=\frac{225}{289}$sin2θ=225289.

Finally, after taking square roots on both sides, we find

$\sin\theta=\frac{15}{17}$sinθ=1517.

Note that the same result could be obtained more easily by sketching a right-angled triangle with legs of length 8 and 15 units. The hypotenuse would then have length 17 units by Pythagoras's theorem, and so, the sine of the angle whose cotangent is $\frac{8}{15}$815 must be $\frac{15}{17}$1517.

Example 2

The point $\left(x,45\right)$(x,45) lies on a circle with radius $53$53 units. What is the cosine of the angle $\theta$θ made from the horizontal by the radius drawn to the point?

The sine of the angle is $\frac{45}{53}$4553. So, $\left(\frac{45}{53}\right)^2+\cos^2\theta=1$(4553)2+cos2θ=1. Therefore, $\cos^2\theta=1-\left(\frac{45}{53}\right)^2=\frac{784}{2809}$cos2θ=1(4553)2=7842809. On taking the square roots of both sides, we find $\cos\theta=\frac{28}{53}$cosθ=2853.

More Worked Examples

QUESTION 1

The point $\left(x,\frac{4}{5}\right)$(x,45) lies on the unit circle.

Use the identity $\cos^2\left(s\right)+\sin^2\left(s\right)=1$cos2(s)+sin2(s)=1 to find the value of $x$x if $x<0$x<0.

QUESTION 2

Write $\csc\theta$cscθ in terms of $\sin\theta$sinθ.

QUESTION 3

 

Outcomes

M8-6

Manipulate trigonometric expressions

91575

Apply trigonometric methods in solving problems

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