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Trigonometric Identities

Use the two core trigonometric identities to simplify expressions, prove results and find exact ratios while keeping track of signs and restrictions.

The two core identities

\sin^2\theta+\cos^2\theta=1

This holds for every angle and follows from Pythagoras on the unit circle. The notation \sin^2\theta means (\sin\theta)^2, not \sin(\theta^2)

\tan\theta=\frac{\sin\theta}{\cos\theta}

This holds wherever \cos\theta\ne0. An identity holds for every angle for which both sides are defined; an equation may hold only for particular angles.

For a proof, usually start with the more complicated side and transform it into the other. Factor before cancelling, and retain restrictions from every original denominator. Numerical substitutions can check a result but cannot prove an identity.

Worked example 1

Simplify \frac{1-\cos^2\theta}{\sin\theta} and state its restriction.

  1. From the core identity, 1-\cos^2\theta=\sin^2\theta. The original fraction requires \sin\theta\ne0

  2. Your turn. Simplify \sin^2\theta/\sin\theta, for \sin\theta\ne0. You can type theta for \theta

    Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

    The result is \sin\theta for \sin\theta\ne0

  3. Answer: \sin\theta, with \sin\theta\ne0. On 0^\circ\le\theta\le360^\circ, exclude 0^\circ,180^\circ,360^\circ

Worked example 2

Prove that \tan\theta\cos\theta=\sin\theta, stating where it is valid.

  1. Where \cos\theta\ne0, replace tangent by \sin\theta/\cos\theta. The left-hand side becomes \frac{\sin\theta}{\cos\theta}\cos\theta

  2. Your turn. Simplify \frac{\sin\theta}{\cos\theta}\cos\theta. You can type theta for \theta

    Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

    The non-zero cosine factor cancels, leaving \sin\theta

  3. Answer: the identity is proved wherever \cos\theta\ne0. Multiplication by zero cannot make an undefined tangent meaningful.

Worked example 3

Prove that \frac{(1-\sin\theta)(1+\sin\theta)}{\cos^2\theta}=1, wherever the fraction is defined.

  1. The numerator is a difference of squares: (1-\sin\theta)(1+\sin\theta)=1-\sin^2\theta=\cos^2\theta

  2. Your turn. What is \cos^2\theta/\cos^2\theta when \cos\theta\ne0?

    The quotient is 1

  3. Answer: the fraction equals 1 for \cos\theta\ne0. The original denominator excludes angles with zero cosine.

Worked example 4

Deduce 1+\tan^2\theta=\frac1{\cos^2\theta} from the core identities.

  1. Start with \sin^2\theta+\cos^2\theta=1 and divide every term by \cos^2\theta, requiring \cos\theta\ne0
  2. \frac{\sin^2\theta}{\cos^2\theta}+\frac{\cos^2\theta}{\cos^2\theta}=\frac1{\cos^2\theta}. The first quotient is \tan^2\theta

  3. Your turn. What does the second quotient simplify to?

    The second quotient is 1

  4. Answer: 1+\tan^2\theta=1/\cos^2\theta, valid for \cos\theta\ne0. This is a consequence of the two core identities.

Worked example 5

Simplify (\sin\theta+\cos\theta)^2-2\sin\theta\cos\theta.

  1. Expand the square fully: \sin^2\theta+2\sin\theta\cos\theta+\cos^2\theta-2\sin\theta\cos\theta

  2. Your turn. The mixed terms cancel. What is \sin^2\theta+\cos^2\theta?

    The remaining sum is 1

  3. Answer: \sin^2\theta+\cos^2\theta=1. This holds for every angle, with no denominator restrictions.

Worked example 6

\tan\theta=3/4 and 180^\circ<\theta<270^\circ. Find \sin\theta and \cos\theta exactly.

  1. Use 1+\tan^2\theta=1/\cos^2\theta. Then 1+9/16=25/16=1/\cos^2\theta, so \cos^2\theta=16/25

  2. Your turn. Cosine is negative in quadrant III. Find \cos\theta

    \cos\theta=-4/5

  3. \sin\theta=\tan\theta\cos\theta=(3/4)(-4/5)=-3/5
  4. Answer: \sin\theta=-3/5, \cos\theta=-4/5. Their ratio is positive even though both are negative.

Worked example 7

Prove that \frac{1-\cos\theta}{\sin\theta}=\frac{\sin\theta}{1+\cos\theta} for \sin\theta\ne0.

  1. If \sin\theta\ne0, then \cos\theta\ne\pm1, so 1+\cos\theta\ne0. Multiply the left fraction’s numerator and denominator by 1+\cos\theta
  2. This gives \frac{(1-\cos\theta)(1+\cos\theta)}{\sin\theta(1+\cos\theta)}=\frac{\sin^2\theta}{\sin\theta(1+\cos\theta)}

  3. Your turn. Simplify \frac{\sin^2\theta}{\sin\theta(1+\cos\theta)}. You can type theta for \theta

    Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

    Cancelling one sine factor gives \frac{\sin\theta}{1+\cos\theta}

  4. Answer: \frac{\sin\theta}{1+\cos\theta}, the required right-hand side. The stated restriction remains in force.

Common mistakes

  • Squaring the angle instead of the ratio. \sin^2\theta=(\sin\theta)^2, not \sin(\theta^2).
  • Cancelling a term inside a sum. Factorise first; cancel only factors of the whole numerator and denominator.
  • Missing the mixed term in a square. Use (a+b)^2=a^2+2ab+b^2.
  • Choosing a square-root sign without checking. Use the quadrant to select the sign of the trig ratio.
  • Dropping denominator restrictions. Keep the original excluded angles after cancelling.
  • Using a numerical check as proof. Use identities and algebra to show equality for every allowed angle.

Now try it: Trigonometric Identities practice questions

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