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Recap: the unit circle and the basic identities

Trigonometry, grade 11 (5 units) Lesson 1 of 18Learning

Recap: the unit circle and the basic identities

The unit circle

In grade 10 you defined the sine and the cosine with the unit circle – a circle with its centre at the origin and radius 1:

  • The angle α\displaystyle \alpha is measured from the positive x\displaystyle x-axis, anticlockwise.
  • The point the angle gives on the circle is (cos⁡α, sin⁡α)\displaystyle (\cos\alpha,\ \sin\alpha).
  • tan⁡α=sin⁡αcos⁡α\displaystyle {\tan\alpha}={\frac{\sin\alpha}{\cos\alpha}}, when cos⁡α≠0\displaystyle \cos\alpha\neq0.
  • A radian: π\displaystyle \pi radians are 180∘\displaystyle 180^\circ. So π6=30∘\displaystyle {\frac{\pi}{6}}={30^\circ}, π4=45∘\displaystyle {\frac{\pi}{4}}={45^\circ}, π3=60∘\displaystyle {\frac{\pi}{3}}={60^\circ}, π2=90∘\displaystyle {\frac{\pi}{2}}={90^\circ}.
  • sin⁡\displaystyle \sin and cos⁡\displaystyle \cos repeat every 2π\displaystyle 2\pi, and tan⁡\displaystyle \tan every π\displaystyle \pi.

In this course the tools read angles in radians. An angle in degrees is written with the degree sign: 30∘\displaystyle 30^\circ.

The exact values

These values are worth knowing by heart:

α\displaystyle \alpha0\displaystyle 0π6\displaystyle \frac{\pi}{6}π4\displaystyle \frac{\pi}{4}π3\displaystyle \frac{\pi}{3}π2\displaystyle \frac{\pi}{2}
sin⁡α\displaystyle \sin\alpha0\displaystyle 012\displaystyle \frac{1}{2}22\displaystyle \frac{\sqrt{2}}{2}32\displaystyle \frac{\sqrt{3}}{2}1\displaystyle 1
cos⁡α\displaystyle \cos\alpha1\displaystyle 132\displaystyle \frac{\sqrt{3}}{2}22\displaystyle \frac{\sqrt{2}}{2}12\displaystyle \frac{1}{2}0\displaystyle 0
tan⁡α\displaystyle \tan\alpha0\displaystyle 033\displaystyle \frac{\sqrt{3}}{3}1\displaystyle 13\displaystyle \sqrt{3}–

In another quadrant the absolute value is the same, and the sign is the quadrant’s: sin⁡5π6=12\displaystyle {\sin\frac{5\pi}{6}}={\frac{1}{2}}, cos⁡2π3=−12\displaystyle {\cos\frac{2\pi}{3}}={-\frac{1}{2}}.

Identities of symmetry

Two angles whose points on the unit circle mirror each other give the same values, perhaps with another sign:

  • −α\displaystyle -\alpha: a reflection in the x\displaystyle x-axis.
sin⁡(−α)\displaystyle \sin(-\alpha) =\displaystyle = −sin⁡α\displaystyle -\sin\alpha
cos⁡(−α)\displaystyle \cos(-\alpha) =\displaystyle = cos⁡α\displaystyle \cos\alpha
  • π−α\displaystyle \pi-\alpha: a reflection in the y\displaystyle y-axis.
sin⁡(π−α)\displaystyle \sin(\pi-\alpha) =\displaystyle = sin⁡α\displaystyle \sin\alpha
cos⁡(π−α)\displaystyle \cos(\pi-\alpha) =\displaystyle = −cos⁡α\displaystyle -\cos\alpha
  • π2−α\displaystyle \frac{\pi}{2}-\alpha: a reflection in the line y=x\displaystyle {y}={x} – the sine and the cosine swap.
sin⁡(π2−α)\displaystyle \sin\left(\frac{\pi}{2}-\alpha\right) =\displaystyle = cos⁡α\displaystyle \cos\alpha
cos⁡(π2−α)\displaystyle \cos\left(\frac{\pi}{2}-\alpha\right) =\displaystyle = sin⁡α\displaystyle \sin\alpha
sin⁡2α+cos⁡2α=1\displaystyle {\sin^2\alpha+\cos^2\alpha}={1}

The point (cos⁡α, sin⁡α)\displaystyle (\cos\alpha,\ \sin\alpha) is on the unit circle, so by Pythagoras’ theorem:

sin⁡2α+cos⁡2α\displaystyle \sin^2\alpha+\cos^2\alpha =\displaystyle = 1\displaystyle 1

Divide both sides by cos⁡2α\displaystyle \cos^2\alpha and you get another identity:

1+tan⁡2α\displaystyle 1+\tan^2\alpha =\displaystyle = 1cos⁡2α\displaystyle \frac{1}{\cos^2\alpha}

An example: sin⁡α=35\displaystyle {\sin\alpha}={\frac{3}{5}}, and the angle α\displaystyle \alpha is in the second quadrant.

  • cos⁡2α=1−925=1625\displaystyle {\cos^2\alpha}={1-\frac{9}{25}}={\frac{16}{25}}, so cos⁡α=45\displaystyle {\cos\alpha}={\frac{4}{5}} or cos⁡α=−45\displaystyle {\cos\alpha}={-\frac{4}{5}}.
  • In the second quadrant the cosine is negative: cos⁡α=−45\displaystyle {\cos\alpha}={-\frac{4}{5}}.
  • tan⁡α=35÷(−45)=−34\displaystyle {\tan\alpha}={\frac{3}{5}\div\left(-\frac{4}{5}\right)}={-\frac{3}{4}}.
Exercise 1

Work it out without a calculator.

    / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

    / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

    Exercise 2

    Simplify with the identities of symmetry, until you reach 2sin⁡x\displaystyle 2\sin x.

      / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

      / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

      Exercise 3

      Simplify until you reach tan⁡x\displaystyle \tan x.

        / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

        / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

        Exercise 4

        cos⁡α=−513\displaystyle {\cos\alpha}={-\frac{5}{13}}, and the angle α\displaystyle \alpha is in the third quadrant.
        Let s=sin⁡α\displaystyle {s}={\sin\alpha}. Solve the equation that the identity sin⁡2α+cos⁡2α=1\displaystyle {\sin^2\alpha+\cos^2\alpha}={1} gives.

          sin⁡α\displaystyle \sin\alpha

            / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

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            tan⁡α\displaystyle \tan\alpha

              / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

              / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

              Exercise 5

              tan⁡α=2\displaystyle {\tan\alpha}={2}, and the angle α\displaystyle \alpha is in the third quadrant.
              Let c=cos⁡α\displaystyle {c}={\cos\alpha}. Solve the equation that the identity 1+tan⁡2α=1cos⁡2α\displaystyle {1+\tan^2\alpha}={\frac{1}{\cos^2\alpha}} gives, and choose the solution that fits the quadrant.

                cos⁡α\displaystyle \cos\alpha

                  / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

                  / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power