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Middle Years

6.02 Trigonometric identities

Lesson

Pythagorean identity

A very useful expression can be found by applying Pythagoras' theorem to the right-angled triangle formed in a unit circle. Consider the picture below.

 

 

 

 

 

 

 

 

 

This means that the opposite side of the triangle from $\theta$θ has a length of $\sin\theta$sinθ and the adjacent side of the triangle has a length of $\cos\theta$cosθ

Applying Pythagoras' theorem:

 

$a^2+b^2$a2+b2 $=$= $c^2$c2
$\left(\cos\theta\right)^2+\left(\sin\theta\right)^2$(cosθ)2+(sinθ)2 $=$= $1$1

Which can also be written as:

$\cos^2\theta+\sin^2\theta$cos2θ+sin2θ $=$= $1$1

This relationship holds for any angle $\theta$θ.

 

Pythagorean identity
$\cos^2\theta+\sin^2\theta$cos2θ+sin2θ $=$= $1$1

 

Also since $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}=\frac{\sin\theta}{\cos\theta}$tanθ=oppositeadjacent=sinθcosθ 

Tangent identity
$\tan\theta$tanθ $=$= $\frac{\sin\theta}{\cos\theta}$sinθcosθ

Worked examples

Example 1

Evaluate the expression $5\cos^2\left(80^\circ\right)+5\sin^2\left(80^\circ\right)$5cos2(80°)+5sin2(80°).

Although neither $\cos\left(80^\circ\right)$cos(80°) or $\sin\left(80^\circ\right)$sin(80°) can be written in exact form using decimals or surds, we can simplify the expression by recognising the Pythagorean identity.

$5\cos^2\left(80^\circ\right)+5\sin^2\left(80^\circ\right)$5cos2(80°)+5sin2(80°) $=$= $5\left(\cos^2\left(80^\circ\right)+\sin^2\left(80^\circ\right)\right)$5(cos2(80°)+sin2(80°))
  $=$= $5\times1$5×1
  $=$= $5$5
Example 2

Find the exact value of $\cos\theta$cosθ and $\tan\theta$tanθ given that $\sin\theta=\frac{12}{13}$sinθ=1213 and $\theta$θ is in the second quadrant.

Think: We can find $\cos\theta$cosθ using the Pythagorean identity and the additional information about the quadrant will tell us if the ratio is positive or negative. One we know both $\sin\theta$sinθ and $\cos\theta$cosθ we can find $\tan\theta$tanθ using the definition $\tan\theta=\frac{\sin\theta}{\cos\theta}$tanθ=sinθcosθ

Do: Write out the Pythagorean identity and substitute the known value in.

$\cos^2\theta+\sin^2\theta$cos2θ+sin2θ $=$= $1$1
$\cos^2\theta+\left(\frac{12}{13}\right)^2$cos2θ+(1213)2 $=$= $1$1
$\cos^2\theta+\frac{144}{169}$cos2θ+144169 $=$= $1$1
$\cos^2\theta$cos2θ $=$= $1-\frac{144}{169}$1144169
$\cos\theta$cosθ $=$= $\pm\sqrt{\frac{25}{169}}$±25169
  $=$= $\pm\frac{5}{13}$±513

We have been given the additional information that $\theta$θ is in the second quadrant. In the second quadrant sine is positive but cosine and tangent will be negative. Hence, $\cos\theta=\frac{-5}{13}$cosθ=513.

 

Using $\tan\theta=\frac{\sin\theta}{\cos\theta}$tanθ=sinθcosθ, we find:

$\tan\theta$tanθ $=$= $\frac{\sin\theta}{\cos\theta}$sinθcosθ
  $=$= $\frac{12}{13}\div\frac{-5}{13}$1213÷​513
  $=$= $\frac{-12}{5}$125
 

Practice questions

Question 1

Answer the following questions given that $\cos y=-\frac{5}{13}$cosy=513, where $180^\circ180°<y<360°.

  1. In which quadrant does angle $y$y lie?

    Quadrant $I$I

    A

    Quadrant $II$II

    B

    Quadrant $III$III

    C

    Quadrant $IV$IV

    D
  2. Use a Pythagorean identity to find the value of $\tan y$tany.

Question 2

Simplify $\left(\cos\theta-1\right)\left(\cos\theta+1\right)$(cosθ1)(cosθ+1).

 

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